By the end of this chapter, you will be able to:
Mastering these skills will help you confidently handle real-world engineering problems involving curves and shapes, making your work precise and reliable.
Coordinate geometry is fundamental in mechanical production technology, allowing precise positioning and measurement of components in design and manufacturing processes. Understanding how to locate points using a Cartesian plane enables technicians to interpret and create technical drawings, machine parts, and assemblies accurately. This chapter focuses on the Cartesian coordinate system, essential for plotting points and defining geometrical relationships in mechanical engineering contexts.
The Cartesian plane is a two-dimensional coordinate system used to specify locations of points by their distances from two perpendicular reference lines. In mechanical production, this system helps in defining the exact position of features on components during machining and assembly. The Cartesian plane simplifies calculations related to distances, angles, and shapes in design and layout tasks.
The Cartesian plane consists of two perpendicular lines called the x-axis and y-axis, which intersect at the origin point. The x-axis is horizontal, and the y-axis is vertical, and they divide the plane into four quadrants. Each point on the plane is represented by an ordered pair \((x, y)\), where \(x\) is the horizontal distance from the origin and \(y\) is the vertical distance.
The governing formula for the position of a point in the Cartesian plane is:
$$ P = (x, y) $$
Example 1: A mechanical part has a hole located 50 mm to the right of the origin and 30 mm above it. Find the coordinates of the hole.
Given: \(x = 50\) mm, \(y = 30\) mm
$$ P = (x, y) $$
$$ P = (50, 30) $$
Answer: (50, 30) mm
Example 2: On a machine drawing, a point lies 75 mm to the left of the origin and 40 mm above it. Find the coordinates.
Given: \(x = -75\) mm (left is negative), \(y = 40\) mm
$$ P = (x, y) $$
$$ P = (-75, 40) $$
Answer: (-75, 40) mm
Example 3: A component feature is 60 mm to the right and 45 mm below the origin. Determine the coordinates.
Given: \(x = 60\) mm, \(y = -45\) mm (below is negative)
$$ P = (x, y) $$
$$ P = (60, -45) $$
Answer: (60, -45) mm
Example 4: In a workshop layout, a point is 120 mm left and 90 mm below the origin. Find the point coordinates.
Given: \(x = -120\) mm, \(y = -90\) mm
$$ P = (x, y) $$
$$ P = (-120, -90) $$
Answer: (-120, -90) mm
Example 5: A coordinate point is 0 mm on the x-axis and 75 mm above the origin. What are its coordinates?
Given: \(x = 0\) mm, \(y = 75\) mm
$$ P = (x, y) $$
$$ P = (0, 75) $$
Answer: (0, 75) mm
Coordinates in the Cartesian plane can be positive or negative depending on their position relative to the origin. Positive \(x\) values lie to the right of the origin, negative \(x\) to the left; positive \(y\) values lie above, and negative \(y\) below the origin. Understanding these conventions is crucial in mechanical production for accurate interpretation of drawings and positioning of parts.
The general rule for coordinates is:
Example 1: Identify the quadrant of the point \((40, 35)\).
Given: \(x = 40 > 0\), \(y = 35 > 0\)
Since both \(x\) and \(y\) are positive, the point lies in Quadrant I.
Answer: Quadrant I
Example 2: Find the quadrant of the point \((-55, 60)\).
Given: \(x = -55 < 0\), \(y = 60 > 0\)
The point lies in Quadrant II.
Answer: Quadrant II
Example 3: Determine the quadrant of \((-80, -45)\).
Given: \(x = -80 < 0\), \(y = -45 < 0\)
The point lies in Quadrant III.
Answer: Quadrant III
Example 4: Determine the quadrant of \((90, -70)\).
Given: \(x = 90 > 0\), \(y = -70 < 0\)
The point lies in Quadrant IV.
Answer: Quadrant IV
Example 5: A point lies at \((0, -50)\). On which axis does it lie?
Given: \(x = 0\), \(y = -50 < 0\)
It lies on the negative y-axis.
Answer: Negative y-axis
In mechanical production, accurate positioning of components and tools is essential. Understanding positive and negative coordinates in the Cartesian plane allows technicians to precisely locate points relative to a reference origin, critical for tasks such as CNC programming, layout marking, and assembly alignment. Coordinates indicate direction and distance from the origin along the X and Y axes, with positive values representing movement right/up and negative values representing left/down.
The Cartesian coordinate system uses two perpendicular axes, X (horizontal) and Y (vertical), intersecting at the origin (0,0). Points are located by ordered pairs \((x, y)\), where \(x\) is the horizontal distance and \(y\) is the vertical distance from the origin. Positive coordinates lie to the right of the Y-axis (for \(x\)) and above the X-axis (for \(y\)), while negative coordinates lie to the left of the Y-axis or below the X-axis.
$$ (x,y) \quad \text{where} \quad x \in \mathbb{R}, \quad y \in \mathbb{R} $$
Example 1: Locate the point with coordinates \((3, 4)\) on a layout for marking a metal plate.
Given: \(x = 3\, \text{cm}\), \(y = 4\, \text{cm}\)
The point lies 3 cm to the right of the origin and 4 cm above the origin.
Answer: Point at 3 cm right and 4 cm up from origin.
Example 2: Find the coordinates of a point 5 cm left and 2 cm down from the origin on a drilling template.
Given: \(x = -5\, \text{cm}\), \(y = -2\, \text{cm}\)
The point is located 5 cm left (negative X) and 2 cm down (negative Y).
Answer: Coordinates \((-5, -2)\) cm.
Example 3: On a CNC machine, a hole is programmed at \((-7, 6)\) relative to the origin. Describe its position.
Given: \(x = -7\, \text{mm}\), \(y = 6\, \text{mm}\)
The hole is 7 mm left and 6 mm above the origin.
Answer: Hole at 7 mm left and 6 mm up from origin.
The Cartesian plane is divided into four quadrants by the X and Y axes:
Identifying the quadrant helps in understanding a point’s relative position for layout and machining operations.
$$ \text{Quadrant} = \begin{cases} I & x > 0, y > 0 \\ II & x < 0, y > 0 \\ III & x < 0, y < 0 \\ IV & x > 0, y < 0 \end{cases} $$
Example 1: Determine the quadrant of point \((4, 5)\) on a machine part layout.
Given: \(x=4 > 0\), \(y=5 > 0\)
Since both are positive, the point is in Quadrant I.
Answer: Quadrant I.
Example 2: A reference point on a jig is at \((-3, 7)\) mm. Find its quadrant.
Given: \(x = -3 < 0\), \(y = 7 > 0\)
Point lies in Quadrant II.
Answer: Quadrant II.
Example 3: Locate the quadrant of a hole at \((-6, -2)\) mm on a metal sheet.
Given: \(x = -6 < 0\), \(y = -2 < 0\)
Point lies in Quadrant III.
Answer: Quadrant III.
Example 4: On a lathe setup, a tool position is at \((5, -8)\) mm. Which quadrant is this?
Given: \(x=5 > 0\), \(y = -8 < 0\)
Point is in Quadrant IV.
Answer: Quadrant IV.
The distance \(d\) of a point \((x, y)\) from the origin \((0, 0)\) is calculated using the distance formula, which is the length of the line segment joining the point and the origin:
$$ d = \sqrt{x^{2} + y^{2}} $$
The sign of \(x\) and \(y\) does not affect the distance since squares of negative numbers are positive.
Example 1: Find the distance of point \((3, 4)\) cm from the origin on a marking template.
Given: \(x = 3\, \text{cm}\), \(y = 4\, \text{cm}\)
$$ d = \sqrt{3^{2} + 4^{2}} $$$$ d = \sqrt{9 + 16} $$$$ d = \sqrt{25} $$$$ d = 5\, \text{cm} $$
Answer: 5 cm.
Example 2: Calculate the distance of a point \((-6, 8)\) mm on a machining drawing.
Given: \(x = -6\, \text{mm}\), \(y = 8\, \text{mm}\)
$$ d = \sqrt{(-6)^{2} + 8^{2}} $$$$ d = \sqrt{36 + 64} $$$$ d = \sqrt{100} $$$$ d = 10\, \text{mm} $$
Answer: 10 mm.
Example 3: Find the distance of \((-5, -12)\) mm from origin on a lathe coordinate system.
Given: \(x = -5\, \text{mm}\), \(y = -12\, \text{mm}\)
$$ d = \sqrt{(-5)^{2} + (-12)^{2}} $$$$ d = \sqrt{25 + 144} $$$$ d = \sqrt{169} $$$$ d = 13\, \text{mm} $$
Answer: 13 mm.
In practical machining and layout, sometimes the distance from the origin and the direction along the axes are known, requiring determination of coordinates including sign based on direction.
If the point is at distance \(d\) along the X-axis with direction:
Similarly for Y-axis:
For a point at \((x, y)\), signs depend on direction from origin.
Example 1: A hole is 7 cm to the right and 5 cm up from the origin. Find coordinates.
Given: \(d_x = 7\, \text{cm}\) (right), \(d_y = 5\, \text{cm}\) (up)
$$ x = +7\, \text{cm}, \quad y = +5\, \text{cm} $$
Answer: Coordinates \((7, 5)\) cm.
Example 2: A reference point is 4 mm left and 3 mm down from origin. Find coordinates.
Given: \(d_x = 4\, \text{mm}\) (left), \(d_y = 3\, \text{mm}\) (down)
$$ x = -4\, \text{mm}, \quad y = -3\, \text{mm} $$
Answer: Coordinates \((-4, -3)\) mm.
Example 3: On a milling machine, a tool is positioned 6 mm left and 9 mm up from origin. Find coordinates.
Given: \(d_x = 6\, \text{mm}\) (left), \(d_y = 9\, \text{mm}\) (up)
$$ x = -6\, \text{mm}, \quad y = +9\, \text{mm} $$
Answer: Coordinates \((-6, 9)\) mm.
Example 4: A point lies 8 cm right and 7 cm down from origin on a layout drawing. Find coordinates.
Given: \(d_x = 8\, \text{cm}\) (right), \(d_y = 7\, \text{cm}\) (down)
$$ x = +8\, \text{cm}, \quad y = -7\, \text{cm} $$
Answer: Coordinates \((8, -7)\) cm.
In mechanical production and engineering drawing, understanding the gradient of a line is crucial for interpreting component layouts, machine part alignments, and tooling paths. The gradient represents the steepness or inclination of a line on a coordinate plane, which directly affects machining angles and assembly fits. Kenyan mechanical production technicians frequently use gradients when interpreting technical drawings or programming CNC machines, where precise angular measurements are essential.
The positive gradient of a line means the line rises from left to right on a Cartesian plane. This occurs when the change in \(y\) (vertical direction) is positive relative to the change in \(x\) (horizontal direction). The gradient \(m\) is calculated as:
$$ m = \frac{y_2 - y_1}{x_2 - x_1} $$
where \((x_1, y_1)\) and \((x_2, y_2)\) are coordinates of two points on the line.
Example 1:
A lathe tool moves from point \(A(2, 3)\) to point \(B(5, 9)\) on a coordinate plane. Find the gradient of the tool path.
Given: \(A(2,3), B(5,9)\)$$
m = \frac{9 - 3}{5 - 2}
$$$$
m = \frac{6}{3}
$$$$
m = 2
$$
Answer: The gradient is 2
Example 2:
A milling cutter moves from \(C(4, 1)\) to \(D(10, 7)\). Calculate the gradient of the movement.
Given: \(C(4,1), D(10,7)\)$$
m = \frac{7 - 1}{10 - 4}
$$$$
m = \frac{6}{6}
$$$$
m = 1
$$
Answer: The gradient is 1
Example 3:
A robotic arm moves along a line from \(E(1, 2)\) to \(F(6, 17)\). Determine the positive gradient.
Given: \(E(1,2), F(6,17)\)$$
m = \frac{17 - 2}{6 - 1}
$$$$
m = \frac{15}{5}
$$$$
m = 3
$$
Answer: The gradient is 3
A negative gradient occurs when the line slopes downward from left to right on the Cartesian plane. This happens when the change in \(y\) is negative relative to the change in \(x\). The formula remains:
$$ m = \frac{y_2 - y_1}{x_2 - x_1} $$
but the result will be a negative number.
Example 1:
A conveyor belt drops from point \(G(3, 8)\) to point \(H(7, 4)\). Find the gradient of the slope.
Given: \(G(3,8), H(7,4)\)$$
m = \frac{4 - 8}{7 - 3}
$$$$
m = \frac{-4}{4}
$$$$
m = -1
$$
Answer: The gradient is -1
Example 2:
A cutting tool moves from \(I(5, 10)\) to \(J(9, 2)\). Calculate the gradient of the tool path.
Given: \(I(5,10), J(9,2)\)$$
m = \frac{2 - 10}{9 - 5}
$$$$
m = \frac{-8}{4}
$$$$
m = -2
$$
Answer: The gradient is -2
Example 3:
A robotic arm moves from \(K(6, 12)\) to \(L(10, 4)\). Find the negative gradient of the movement.
Given: \(K(6,12), L(10,4)\)$$
m = \frac{4 - 12}{10 - 6}
$$$$
m = \frac{-8}{4}
$$$$
m = -2
$$
Answer: The gradient is -2
A zero gradient means the line is perfectly horizontal. The vertical change \((y_2 - y_1)\) is zero, so the gradient formula simplifies to zero.
$$ m = \frac{y_2 - y_1}{x_2 - x_1} = 0 $$
This is common in machine part alignments where horizontal surfaces are critical.
Example 1:
A conveyor belt runs horizontally from point \(M(2, 5)\) to point \(N(7, 5)\). Find the gradient.
Given: \(M(2,5), N(7,5)\)$$
m = \frac{5 - 5}{7 - 2}
$$$$
m = \frac{0}{5}
$$$$
m = 0
$$
Answer: The gradient is 0
Example 2:
A lathe guide moves from \(P(4, 8)\) to \(Q(9, 8)\). Calculate the gradient of the guide.
Given: \(P(4,8), Q(9,8)\)$$
m = \frac{8 - 8}{9 - 4}
$$$$
m = \frac{0}{5}
$$$$
m = 0
$$
Answer: The gradient is 0
Example 3:
A machine table moves horizontally from \(R(1, 3)\) to \(S(6, 3)\). Find the gradient of this movement.
Given: \(R(1,3), S(6,3)\)$$
m = \frac{3 - 3}{6 - 1}
$$$$
m = \frac{0}{5}
$$$$
m = 0
$$
Answer: The gradient is 0
An infinite gradient occurs when the line is vertical. The horizontal change \((x_2 - x_1)\) is zero, so the gradient formula results in division by zero, which is undefined. This is common when measuring vertical machine parts or tool movements.
$$ m = \frac{y_2 - y_1}{x_2 - x_1} \quad \text{where} \quad (x_2 - x_1) = 0 \implies m = \infty $$
Example 1:
A vertical drill moves from \(T(4, 2)\) to \(U(4, 10)\). Find the gradient.
Given: \(T(4,2), U(4,10)\)$$
m = \frac{10 - 2}{4 - 4}
$$$$
m = \frac{8}{0}
$$
Answer: The gradient is infinite (vertical line)
Example 2:
A machine arm moves vertically from \(V(7, 5)\) to \(W(7, 12)\). Calculate the gradient.
Given: \(V(7,5), W(7,12)\)$$
m = \frac{12 - 5}{7 - 7}
$$$$
m = \frac{7}{0}
$$
Answer: The gradient is infinite (vertical line)
Example 3:
A crane hook moves vertically from \(X(3, 1)\) to \(Y(3, 9)\). Find the gradient.
Given: \(X(3,1), Y(3,9)\)$$
m = \frac{9 - 1}{3 - 3}
$$$$
m = \frac{8}{0}
$$
Answer: The gradient is infinite (vertical line)
Parallel lines have the same gradient. This property is essential in mechanical production when designing parts that need to align perfectly or when programming CNC machines to move along parallel paths.
If two lines have gradients \(m_1\) and \(m_2\), then for parallelism:
$$ m_1 = m_2 $$
Example 1:
Find the gradient \(m_2\) of a line parallel to line joining points \(A(1, 2)\) and \(B(4, 8)\).
Given: \(A(1,2), B(4,8)\)$$
m_1 = \frac{8 - 2}{4 - 1}
$$$$
m_1 = \frac{6}{3}
$$$$
m_1 = 2
$$
Since lines are parallel,$$
m_2 = m_1 = 2
$$
Answer: Gradient of the parallel line is 2
Example 2:
A cutting path has gradient \(m_1 = -3\). Find the gradient of a line parallel to this path.
Given: \(m_1 = -3\)$$
m_2 = -3
$$
Answer: Gradient of the parallel line is -3
Example 3:
A lathe tool moves on a path with gradient \(m_1 = \frac{1}{2}\). Find the gradient of a parallel path.
Given: \(m_1 = \frac{1}{2}\)$$
m_2 = \frac{1}{2}
$$
Answer: Gradient of the parallel line is \frac{1}{2}
Two lines are perpendicular if the product of their gradients is \(-1\). This means one gradient is the negative reciprocal of the other:
$$ m_1 \times m_2 = -1 $$
or
$$ m_2 = -\frac{1}{m_1} $$
This is critical in mechanical production when setting toolpaths or part features at right angles.
Example 1:
Find the gradient \(m_2\) of a line perpendicular to the line joining \(C(2, 3)\) and \(D(6, 11)\).
Given: \(C(2,3), D(6,11)\)$$
m_1 = \frac{11 - 3}{6 - 2}
$$$$
m_1 = \frac{8}{4}
$$$$
m_1 = 2
$$
Calculate \(m_2\):$$
m_2 = -\frac{1}{2}
$$
Answer: Gradient of the perpendicular line is -\frac{1}{2}
Example 2:
A tool path has gradient \(m_1 = -4\). Find the gradient of a path perpendicular to it.
Given: \(m_1 = -4\)$$
m_2 = -\frac{1}{-4}
$$$$
m_2 = \frac{1}{4}
$$
Answer: Gradient of the perpendicular line is \frac{1}{4}
Example 3:
A line has gradient \(m_1 = \frac{3}{5}\). Find the gradient of a line perpendicular to it.
Given: \(m_1 = \frac{3}{5}\)$$
m_2 = -\frac{1}{\frac{3}{5}} = -\frac{5}{3}
$$
Answer: Gradient of the perpendicular line is -\frac{5}{3}
Example 4:
A technician at Kenya Railways is checking the parallel alignment of two rails. The first rail passes through points \(A(2, 3)\) and \(B(4, 7)\). Find the gradient of a parallel rail passing through \(C(6, 11)\).
Given: \(A(2,3), B(4,7)\)$$ m_1 = \frac{7 - 3}{4 - 2} = \frac{4}{2} = 2 $$
The parallel rail through \(C(6,11)\) will have gradient \(m_2 = 2\).
Answer: Gradient of parallel rail is 2 (Kenya Railways context)
Example 5:
A technician at Bamburi Cement Ltd is setting a perpendicular guide to a conveyor belt with gradient \(m_1 = 1.5\). Find the gradient of the perpendicular guide.
Given: \(m_1 = 1.5\)$$ m_2 = -\frac{1}{1.5} = -0.6667 $$
Answer: Gradient of perpendicular guide is -0.6667 (Bamburi Cement Ltd context)
In mechanical production and engineering drafting, understanding positive coordinate values is essential for accurately locating points in a Cartesian coordinate system. Positive coordinates denote points situated in the first quadrant where both the \(x\) (horizontal) and \(y\) (vertical) values are greater than zero. This is critical when programming CNC machines or interpreting CAD drawings to ensure components are manufactured with precision according to design specifications.
Positive coordinates are points where both the \(x\) and \(y\) values are positive. Such points lie in the first quadrant of the Cartesian plane, which is the reference area for many layout and fabrication tasks in mechanical workshops.
$$ (x, y) \quad \text{where} \quad x > 0, \quad y > 0 $$
Example 1: Locate a point on a metal sheet at coordinates \( (5, 8) \) cm.
Given: \(x = 5\) cm, \(y = 8\) cm
$$ (x, y) = (5, 8) $$
Answer: The point is 5 cm right and 8 cm above the origin.
Example 2: Calculate the distance from the origin to a point at \( (12, 9) \) mm on a machine part.
Given: \(x = 12\) mm, \(y = 9\) mm
Formula for distance \(d\):
$$ d = \sqrt{x^2 + y^2} $$
$$ d = \sqrt{12^2 + 9^2} $$
$$ d = \sqrt{144 + 81} = \sqrt{225} $$
$$ d = 15 \text{ mm} $$
Answer: The point is 15 mm from the origin.
Example 3: Find the midpoint between points \( (4, 7) \) and \( (10, 15) \) mm on a shaft.
Given: \( (x_1, y_1) = (4, 7) \) mm, \( (x_2, y_2) = (10, 15) \) mm
Midpoint formula:
$$ (\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}) $$
$$ (\frac{4 + 10}{2}, \frac{7 + 15}{2}) = (7, 11) $$
Answer: Midpoint is at \( (7, 11) \) mm.
Example 4: Determine the slope of a line passing through points \( (6, 3) \) and \( (14, 11) \) cm on a bracket.
Given: \( (x_1, y_1) = (6, 3) \) cm, \( (x_2, y_2) = (14, 11) \) cm
Slope formula:
$$ m = \frac{y_2 - y_1}{x_2 - x_1} $$
$$ m = \frac{11 - 3}{14 - 6} = \frac{8}{8} = 1 $$
Answer: Slope of the line is 1.
Example 5: A cutting tool moves in a straight line from \( (2, 5) \) cm to \( (8, 13) \) cm. Calculate the length of the path.
Given: \( (x_1, y_1) = (2, 5) \) cm, \( (x_2, y_2) = (8, 13) \) cm
Distance formula:
$$ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$
$$ d = \sqrt{(8 - 2)^2 + (13 - 5)^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} $$
$$ d = \sqrt{100} = 10 \text{ cm} $$
Answer: The tool moves 10 cm along the path.
When points have positive coordinates, calculating the area of polygons such as rectangles or triangles is straightforward using coordinate geometry formulas.
For a triangle with vertices at \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \), the area \(A\) is:
$$ A = \frac{1}{2} | x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) | $$
Example 1: Find the area of a triangle with vertices at \( (2, 3) \), \( (6, 3) \), and \( (4, 7) \) cm on a component drawing.
Given: \( (2, 3), (6, 3), (4, 7) \) cm
$$ A = \frac{1}{2} |2(3 - 7) + 6(7 - 3) + 4(3 - 3)| $$
$$ A = \frac{1}{2} |2(-4) + 6(4) + 4(0)| $$
$$ A = \frac{1}{2} |-8 + 24 + 0| = \frac{1}{2} \times 16 = 8 \text{ cm}^2 $$
Answer: Area is \(8\) cm\(^2\).
Example 2: Calculate the area of a rectangle with vertices \( (1, 1) \), \( (1, 5) \), \( (7, 5) \), and \( (7, 1) \) cm.
Length \(l = 7 - 1 = 6\) cm
Width \(w = 5 - 1 = 4\) cm
Area:
$$ A = l \times w = 6 \times 4 = 24 \text{ cm}^2 $$
Answer: Area is \(24\) cm\(^2\).
Example 3: Determine the area of a triangle with vertices at \( (3, 4) \), \( (7, 8) \), and \( (10, 2) \) mm on a metal plate.
Given: \( (3, 4), (7, 8), (10, 2) \) mm
$$ A = \frac{1}{2} |3(8 - 2) + 7(2 - 4) + 10(4 - 8)| $$
$$ A = \frac{1}{2} |3(6) + 7(-2) + 10(-4)| $$
$$ A = \frac{1}{2} |18 - 14 - 40| = \frac{1}{2} \times |-36| = 18 \text{ mm}^2 $$
Answer: Area is \(18\) mm\(^2\).
Example 4: Find the area of a trapezium with vertices at \( (1, 2) \), \( (5, 2) \), \( (7, 6) \), and \( (0, 6) \) cm.
Divide into two triangles:
Triangle 1: \( (1, 2), (5, 2), (7, 6) \)
Triangle 2: \( (1, 2), (7, 6), (0, 6) \)
Calculate areas separately and sum.
Triangle 1:
$$ A_1 = \frac{1}{2} |1(2 - 6) + 5(6 - 2) + 7(2 - 2)| $$
$$ A_1 = \frac{1}{2} |-4 + 20 + 0| = \frac{1}{2} \times 16 = 8 \text{ cm}^2 $$
Triangle 2:
$$ A_2 = \frac{1}{2} |1(6 - 6) + 7(6 - 2) + 0(2 - 6)| $$
$$ A_2 = \frac{1}{2} |0 + 28 + 0| = \frac{1}{2} \times 28 = 14 \text{ cm}^2 $$
Total area:
$$ A = A_1 + A_2 = 8 + 14 = 22 \text{ cm}^2 $$
Answer: Area is \(22\) cm\(^2\).
In mechanical layouts, angles between lines defined by positive coordinates are important for machining and assembly.
The angle \(\theta\) between two points \( (x_1,y_1) \) and \( (x_2,y_2) \) relative to the \(x\)-axis is:
$$ \theta = \tan^{-1} ( \frac{y_2 - y_1}{x_2 - x_1} ) $$
Example 1: Find the angle of a line between points \( (2, 3) \) and \( (6, 7) \) cm.
Given: \( (x_1, y_1) = (2, 3) \), \( (x_2, y_2) = (6, 7) \)
$$ \theta = \tan^{-1} ( \frac{7 - 3}{6 - 2} ) = \tan^{-1} ( \frac{4}{4} ) = \tan^{-1} (1) $$
$$ \theta = 45^\circ $$
Answer: The angle is \(45^\circ\).
Example 2: Calculate the angle between points \( (1, 2) \) and \( (5, 10) \) mm.
Given: \( (x_1, y_1) = (1, 2) \), \( (x_2, y_2) = (5, 10) \)
$$ \theta = \tan^{-1} ( \frac{10 - 2}{5 - 1} ) = \tan^{-1} ( \frac{8}{4} ) = \tan^{-1} (2) $$
$$ \theta \approx 63.43^\circ $$
Answer: The angle is approximately \(63.43^\circ\).
Example 3: Determine the angle of a slope between points \( (3, 5) \) and \( (7, 9) \) cm.
Given: \( (x_1, y_1) = (3, 5) \), \( (x_2, y_2) = (7, 9) \)
$$ \theta = \tan^{-1} ( \frac{9 - 5}{7 - 3} ) = \tan^{-1} ( \frac{4}{4} ) = 45^\circ $$
Answer: The angle is \(45^\circ\).
Example 4: A conveyor belt is inclined between points \( (0, 0) \) and \( (10, 6) \) m. Find the angle of inclination.
Given: \( (x_1, y_1) = (0, 0) \), \( (x_2, y_2) = (10, 6) \)
$$ \theta = \tan^{-1} ( \frac{6 - 0}{10 - 0} ) = \tan^{-1} (0.6) $$
$$ \theta \approx 30.96^\circ $$
Answer: The angle of inclination is approximately \(30.96^\circ\).
Example 5: Find the angle between points \( (4, 4) \) and \( (9, 12) \) mm on a machined plate.
Given: \( (x_1, y_1) = (4, 4) \), \( (x_2, y_2) = (9, 12) \)
$$ \theta = \tan^{-1} ( \frac{12 - 4}{9 - 4} ) = \tan^{-1} ( \frac{8}{5} ) $$
$$ \theta \approx 58^\circ $$
Answer: The angle is approximately \(58^\circ\).
In mechanical production, negative coordinates indicate points located in quadrants other than the first, where either the \(x\) or \(y\) values (or both) are negative. These coordinates are important when working with parts that extend into different sections of the coordinate system, such as symmetrical parts or when positioning tools relative to a datum that is not at the origin. Understanding negative coordinates ensures correct interpretation of design drawings and accurate machine programming.
Negative \(x\)-coordinates occur when points lie to the left of the origin. In mechanical drawings, this is common when parts or features extend left of the datum line. The \(y\)-coordinate can be positive or negative, placing the point in the second or third quadrant.
$$ (x, y) \quad \text{where} \quad x < 0, \quad y > 0 \quad \text{(Second quadrant)} $$
$$ (x, y) \quad \text{where} \quad x < 0, \quad y < 0 \quad \text{(Third quadrant)} $$
Example 1: Locate the point \( (-4, 7) \) cm on a layout drawing for a machine base.
Given: \(x = -4\) cm, \(y = 7\) cm
$$ (x, y) = (-4, 7) $$
Answer: The point is 4 cm left and 7 cm above the origin.
Example 2: Calculate the distance from the origin to point \( (-6, -8) \) mm on a fabricated bracket.
Given: \(x = -6\) mm, \(y = -8\) mm
$$ d = \sqrt{x^2 + y^2} $$
$$ d = \sqrt{(-6)^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} $$
$$ d = 10 \text{ mm} $$
Answer: The point is 10 mm from the origin.
Example 3: Find the midpoint between \( (-10, 5) \) and \( (-4, -3) \) cm on a component.
Given: \( (x_1, y_1) = (-10, 5) \) cm, \( (x_2, y_2) = (-4, -3) \) cm
$$ (\frac{-10 + (-4)}{2}, \frac{5 + (-3)}{2}) = (\frac{-14}{2}, \frac{2}{2}) = (-7, 1) $$
Answer: Midpoint is at \( (-7, 1) \) cm.
Example 4: Determine the slope of a line passing through points \( (-3, 6) \) and \( (-9, -2) \) mm.
Given: \( (x_1, y_1) = (-3, 6) \), \( (x_2, y_2) = (-9, -2) \)
$$ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-2 - 6}{-9 - (-3)} = \frac{-8}{-6} = \frac{4}{3} $$
Answer: Slope is \( \frac{4}{3} \).
Example 5: Calculate the length of a path from \( (-5, 4) \) to \( (-1, -2) \) cm on a jig layout.
Given: \( (x_1, y_1) = (-5, 4) \), \( (x_2, y_2) = (-1, -2) \)
$$ d = \sqrt{(-1 + 5)^2 + (-2 - 4)^2} = \sqrt{4^2 + (-6)^2} = \sqrt{16 + 36} $$
$$ d = \sqrt{52} \approx 7.21 \text{ cm} $$
Answer: Distance is approximately \(7.21\) cm.
Negative \(y\)-coordinates indicate points below the origin. These are common when features extend downward or below the datum in mechanical drawings.
$$ (x, y) \quad \text{where} \quad x > 0, \quad y < 0 \quad \text{(Fourth quadrant)} $$
$$ (x, y) \quad \text{where} \quad x < 0, \quad y < 0 \quad \text{(Third quadrant)} $$
Example 1: Locate a point at \( (7, -3) \) cm on a CNC machine setup.
Given: \(x = 7\) cm, \(y = -3\) cm
$$ (x, y) = (7, -3) $$
Answer: The point is 7 cm right and 3 cm below the origin.
Example 2: Calculate the distance from the origin to \( (4, -5) \) mm on a machined component.
Given: \(x = 4\) mm, \(y = -5\) mm
$$ d = \sqrt{4^2 + (-5)^2} = \sqrt{16 + 25} = \sqrt{41} $$
$$ d \approx 6.40 \text{ mm} $$
Answer: Distance is approximately \(6.40\) mm.
Example 3: Find the midpoint between \( (8, -6) \) and \( (2, -10) \) cm on a lathe bed.
Given: \( (x_1, y_1) = (8, -6) \), \( (x_2, y_2) = (2, -10) \)
$$ (\frac{8 + 2}{2}, \frac{-6 + (-10)}{2}) = (5, -8) $$
Answer: Midpoint is at \( (5, -8) \) cm.
Example 4: Determine the slope of the line passing through \( (3, -4) \) and \( (7, -8) \) mm.
Given: \( (x_1, y_1) = (3, -4) \), \( (x_2, y_2) = (7, -8) \)
$$ m = \frac{-8 - (-4)}{7 - 3} = \frac{-4}{4} = -1 $$
Answer: Slope is \(-1\).
Example 5: Calculate the length of the path from \( (6, -2) \) to \( (9, -7) \) cm on a fixture.
Given: \( (x_1, y_1) = (6, -2) \), \( (x_2, y_2) = (9, -7) \)
$$ d = \sqrt{(9 - 6)^2 + (-7 + 2)^2} = \sqrt{3^2 + (-5)^2} = \sqrt{9 + 25} $$
$$ d = \sqrt{34} \approx 5.83 \text{ cm} $$
Answer: Distance is approximately \(5.83\) cm.
Angles associated with points having negative \(x\) or \(y\) coordinates require careful interpretation to determine the correct quadrant.
The angle \(\theta\) relative to the positive \(x\)-axis is:
$$ \theta = \tan^{-1} ( \frac{y_2 - y_1}{x_2 - x_1} ) $$
Adjustment is needed if \(x_2 - x_1 < 0\) to find the correct quadrant angle.
Example 1: Find the angle of a line between points \( (-4, 5) \) and \( (-8, 9) \) cm.
Given: \( (x_1, y_1) = (-4, 5) \), \( (x_2, y_2) = (-8, 9) \)
$$ \theta = \tan^{-1} ( \frac{9 - 5}{-8 - (-4)} ) = \tan^{-1} ( \frac{4}{-4} ) = \tan^{-1} (-1) $$
$$ \theta = -45^\circ $$
Since \(x_2 - x_1 < 0\), add \(180^\circ\):
$$ \theta = 135^\circ $$
Answer: Angle is \(135^\circ\).
Example 2: Calculate the angle between points \( (7, -3) \) and \( (2, -8) \) mm.
Given: \( (x_1, y_1) = (7, -3) \), \( (x_2, y_2) = (2, -8) \)
$$ \theta = \tan^{-1} ( \frac{-8 + 3}{2 - 7} ) = \tan^{-1} ( \frac{-5}{-5} ) = \tan^{-1} (1) $$
$$ \theta = 45^\circ $$
Since \(x_2 - x_1 < 0\), add \(180^\circ\):
$$ \theta = 225^\circ $$
Answer: Angle is \(225^\circ\).
Example 3: Determine the angle of a slope between points \( (-3, -4) \) and \( (1, -8) \) cm.
Given: \( (x_1, y_1) = (-3, -4) \), \( (x_2, y_2) = (1, -8) \)
$$ \theta = \tan^{-1} ( \frac{-8 + 4}{1 + 3} ) = \tan^{-1} ( \frac{-4}{4} ) = \tan^{-1} (-1) = -45^\circ $$
Since \(x_2 - x_1 > 0\), angle is \(315^\circ\) or \(-45^\circ\) depending on rotation direction.
Answer: Angle is \(315^\circ\).
Example 4: A tool moves from \( (-5, 6) \) to \( (3, -2) \) mm. Find the angle of movement.
Given: \( (x_1, y_1) = (-5, 6) \), \( (x_2, y_2) = (3, -2) \)
$$ \theta = \tan^{-1} ( \frac{-2 - 6}{3 - (-5)} ) = \tan^{-1} ( \frac{-8}{8} ) = \tan^{-1} (-1) = -45^\circ $$
Since \(x_2 - x_1 > 0\), angle is \(315^\circ\).
Answer: Angle is \(315^\circ\).
Example 5: Find the angle between points \( (-7, -3) \) and \( (-2, 4) \) cm on a metal plate.
Given: \( (x_1, y_1) = (-7, -3) \), \( (x_2, y_2) = (-2, 4) \)
$$ \theta = \tan^{-1} ( \frac{4 + 3}{-2 + 7} ) = \tan^{-1} ( \frac{7}{5} ) \approx 54.46^\circ $$
Since \(x_2 - x_1 > 0\), angle is \(54.46^\circ\).
Answer: Angle is approximately \(54.46^\circ\).
In coordinate geometry as applied in mechanical production, the concept of zero plays a crucial role in locating points relative to a reference origin. Zero represents the coordinate value at the origin where the axes intersect, typically denoted as \( (0,0) \) in two-dimensional Cartesian coordinates. Understanding zero is essential for precise positioning of machine parts, setting tool origins, and programming CNC machines in Kenyan manufacturing workshops.
The coordinate zero is the baseline from which all other coordinates are measured. The formula to represent the origin point is:
$$ (x, y) = (0, 0) $$
Example 1:
A part is positioned such that its reference point lies at the origin of a Cartesian coordinate system. What are the coordinates of this point?
Given: The origin point coordinates.
$$ (x, y) = (0, 0) $$
Answer: \( (0, 0) \)
Example 2:
In a workshop, a tool’s zero position is at the origin. If a hole is drilled 120 mm along the x-axis and 80 mm along the y-axis from the zero point, what are the coordinates of the hole?
Given: Distance along x-axis = 120 mm, distance along y-axis = 80 mm
$$ (x, y) = (120, 80) $$
Answer: \( (120, 80) \) mm
Example 3:
A mechanical component is designed with its center at the origin. If a feature is 50 mm left and 30 mm below the origin, what are its coordinates?
Given: Left means negative x-direction, below means negative y-direction
$$ (x, y) = (-50, -30) $$
Answer: \( (-50, -30) \) mm
Example 4:
A CNC milling machine reads the tool zero position at \( (0,0) \). If the operator sets a new zero at \( (75, 100) \) mm, what is the relative displacement from the original zero?
Given: Original zero = \( (0,0) \), new zero = \( (75,100) \)
$$ \Delta x = 75 - 0 = 75 \text{ mm} $$
$$ \Delta y = 100 - 0 = 100 \text{ mm} $$
Answer: The displacement is \( (75, 100) \) mm
Example 5:
A lathe machine tool has a zero at \( (0,0) \). A point on the workpiece is located at \( (x,y) = (0,0) \). If the tool moves 0 mm along x and 0 mm along y, where is it positioned?
Given: Movement in x and y = 0 mm
$$ (x, y) = (0 + 0, 0 + 0) = (0, 0) $$
Answer: The tool remains at \( (0, 0) \)
In mechanical production coordinate geometry, the concept of infinite coordinates arises when dealing with lines or points at an unbounded distance, such as parallel lines or asymptotes in geometric constructions. While infinite coordinates cannot be numerically expressed, the concept helps in understanding limits and behavior of geometric entities at very large distances, which is useful in precision machining and CAD modelling.
Infinite in coordinate geometry is often considered as a point or value that tends towards unbounded magnitude along an axis. This can be represented symbolically as:
$$ x \to \infty \quad \text{or} \quad y \to \infty $$
Example 1:
A line in a mechanical drawing extends infinitely in the positive x-direction. What is the coordinate form representing the point at infinity along the x-axis?
Given: Infinite extension along x-axis
$$ (x, y) = (\infty, y) $$
Answer: The coordinate tends towards \( (\infty, y) \) where \( y \) is finite
Example 2:
In CNC programming, a tool path extends indefinitely along the y-axis. What is the coordinate expression for the tool position at infinity along y?
Given: Infinite extension along y-axis
$$ (x, y) = (x, \infty) $$
Answer: The coordinate tends towards \( (x, \infty) \) where \( x \) is finite
Example 3:
A mechanical engineer analyses a line with slope \( m = 3 \) that extends infinitely in both directions. What happens to the coordinates as the line extends infinitely?
Given: Slope \( m = 3 \), line equation \( y = 3x + c \)
As \( x \to \infty \),
$$ y = 3 \times \infty + c = \infty $$
As \( x \to -\infty \),
$$ y = 3 \times (-\infty) + c = -\infty $$
Answer: Coordinates tend to \( (\infty, \infty) \) and \( (-\infty, -\infty) \) respectively
Example 4:
A lathe tool path is modelled as a line \( y = 2x + 5 \). What are the coordinates at infinite positive and negative x?
Given: \( y = 2x + 5 \)
As \( x \to \infty \),
$$ y = 2 \times \infty + 5 = \infty $$
As \( x \to -\infty \),
$$ y = 2 \times (-\infty) + 5 = -\infty $$
Answer: Coordinates tend to \( (\infty, \infty) \) and \( (-\infty, -\infty) \)
Example 5:
In tool path planning, a line is vertical and extends infinitely along the y-axis at \( x = 10 \). What are the coordinates at infinity?
Given: Vertical line \( x = 10 \)
At \( y \to \infty \),
$$ (x, y) = (10, \infty) $$
At \( y \to -\infty \),
$$ (x, y) = (10, -\infty) $$
Answer: Coordinates tend to \( (10, \infty) \) and \( (10, -\infty) \)
A point is located 150 mm to the right and 75 mm above the origin in a mechanical workshop coordinate system. What are its coordinates? (2 marks)
A CNC machine tool moves from zero position to \( (200, -50) \) mm. Calculate the displacement vector from the origin. (3 marks)
A line with equation \( y = -4x + 10 \) extends infinitely. Find the coordinates of the points at \( x \to \infty \) and \( x \to -\infty \). (4 marks)
A vertical tool path is located at \( x = 25 \) mm. Write the coordinates at infinite positive and negative y-values. (3 marks)
If a tool moves 0 mm along both x and y axes from zero, where does it remain? (2 marks)
In mechanical production and engineering drawing contexts in Kenya, understanding the gradients of parallel lines is essential for accurate layout, machining, and assembly of components. Parallel lines have identical gradients, which ensures components fit and function correctly without misalignment. This concept is fundamental when designing parts that must run parallel, such as guide rails or shafts in production machines.
The gradient \(m\) of a line through two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by
$$ m = \frac{y_2 - y_1}{x_2 - x_1} $$
Two lines are parallel if and only if their gradients are equal:
$$ m_1 = m_2 $$
Example 1: A lathe bed guide is represented by a line passing through points \(A(2, 3)\) and \(B(6, 11)\). Find the gradient of this line and determine the gradient of a line parallel to it.
Given: \(A(2, 3)\), \(B(6, 11)\)
$$ m = \frac{11 - 3}{6 - 2} $$
$$ m = \frac{8}{4} $$
$$ m = 2 $$
A line parallel to this guide will have the gradient
$$ m = 2 $$
Answer: The gradient of the line is 2; any line parallel to it also has a gradient of 2.
Example 2: A conveyor belt frame is represented by a line through points \(C(0, 0)\) and \(D(4, 5)\). Calculate the gradient of this line and find the equation of a line parallel to it passing through point \(E(1, 3)\).
Given: \(C(0, 0)\), \(D(4, 5)\), \(E(1, 3)\)
Calculate gradient:
$$ m = \frac{5 - 0}{4 - 0} = \frac{5}{4} = 1.25 $$
The equation of a line parallel to this line through \(E(1, 3)\) is:
$$ y, y_1 = m (x, x_1) $$
$$ y - 3 = 1.25 (x - 1) $$
Simplify:
$$ y = 1.25 x - 1.25 + 3 $$
$$ y = 1.25 x + 1.75 $$
Answer: Gradient is 1.25; the parallel line equation is \(y = 1.25 x + 1.75\).
Example 3: In a mechanical drawing, a line passes through points \(F(-3, 7)\) and \(G(2, -3)\). Find the gradient and verify if a line through \(H(4, 5)\) with gradient \(-2\) is parallel.
Given: \(F(-3, 7)\), \(G(2, -3)\), \(m_2 = -2\) for line through \(H\)
Calculate gradient of line \(FG\):
$$ m_1 = \frac{-3 - 7}{2 - (-3)} = \frac{-10}{5} = -2 $$
Since \(m_1 = -2\) and \(m_2 = -2\), the lines are parallel.
Answer: Gradient is \(-2\); the lines are parallel.
Example 4: A machine part edge lies on a line through points \(P(1, 4)\) and \(Q(5, 12)\). Find the gradient of a line parallel to this and determine the y-intercept if this parallel line passes through \(R(2, 10)\).
Given: \(P(1, 4)\), \(Q(5, 12)\), \(R(2, 10)\)
Calculate gradient:
$$ m = \frac{12 - 4}{5 - 1} = \frac{8}{4} = 2 $$
Equation of parallel line through \(R\):
$$ y - 10 = 2 (x - 2) $$
Simplify:
$$ y = 2x - 4 + 10 = 2x + 6 $$
Y-intercept is 6.
Answer: Gradient is 2; y-intercept of parallel line is 6.
Example 5: For a welding guide, the original line passes through \(S(3, 9)\) and \(T(7, 13)\). Find the gradient of this line and the gradient of any line parallel to it.
Given: \(S(3, 9)\), \(T(7, 13)\)
Calculate gradient:
$$ m = \frac{13 - 9}{7 - 3} = \frac{4}{4} = 1 $$
Any parallel line has gradient 1.
Answer: Gradient is 1; parallel lines have gradient 1.
Find the gradient of a line passing through points \((1, 2)\) and \((5, 10)\), and find the equation of a line parallel to it passing through \((2, 3)\). (6 marks)
A machine frame edge is on a line through \((0, 0)\) and \((6, 9)\). Calculate the gradient and determine the gradient of a line parallel to it. (4 marks)
Find the gradient of the line passing through \((4, 5)\) and \((8, 1)\). Is a line passing through \((3, 2)\) with gradient \(-1\) parallel to this line? (6 marks)
A shaft guide line passes through points \((2, 7)\) and \((6, 15)\). Find the gradient of a line parallel to it and determine its y-intercept if it passes through \((1, 5)\). (6 marks)
Calculate the gradient of a line through \((-1, -2)\) and \((3, 6)\), and find the equation of a line parallel to it through \((0, 1)\). (6 marks)
In mechanical production settings, perpendicular lines are crucial for ensuring right angles in components and assemblies, such as in jigs, fixtures, and machine parts. The gradients of perpendicular lines have a special relationship that facilitates easy calculation and verification of right angles during design and fabrication processes.
Two lines are perpendicular if the product of their gradients is \(-1\). If a line has gradient \(m_1\), the line perpendicular to it has gradient \(m_2\) such that
$$ m_1 \times m_2 = -1 $$
or equivalently,
$$ m_2 = -\frac{1}{m_1} $$
Example 1: A machine frame edge has gradient \(m_1 = 2\). Find the gradient of a line perpendicular to it.
Given: \(m_1 = 2\)
$$ m_2 = -\frac{1}{2} = -0.5 $$
Answer: Gradient of perpendicular line is \(-0.5\).
Example 2: A line passes through points \(A(1, 3)\) and \(B(4, 11)\). Find the gradient of this line and the gradient of a line perpendicular to it.
Given: \(A(1, 3)\), \(B(4, 11)\)
Calculate gradient:
$$ m_1 = \frac{11 - 3}{4 - 1} = \frac{8}{3} \approx 2.6667 $$
Gradient of perpendicular line:
$$ m_2 = -\frac{1}{2.6667} \approx -0.375 $$
Answer: Gradient is \(\frac{8}{3}\); perpendicular line gradient is \(-\frac{3}{8}\).
Example 3: A lathe bed guide line passes through points \(C(2, 5)\) and \(D(6, 9)\). Find the gradient of a line perpendicular to this guide.
Given: \(C(2, 5)\), \(D(6, 9)\)
Calculate gradient:
$$ m_1 = \frac{9 - 5}{6 - 2} = \frac{4}{4} = 1 $$
Gradient of perpendicular line:
$$ m_2 = -\frac{1}{1} = -1 $$
Answer: Gradient of perpendicular line is \(-1\).
Example 4: A line with gradient \(-3\) represents a machine part edge. Find the gradient of a line perpendicular to it and write the equation of this perpendicular line passing through point \(E(2, 4)\).
Given: \(m_1 = -3\), point \(E(2, 4)\)
Calculate gradient:
$$ m_2 = -\frac{1}{-3} = \frac{1}{3} $$
Equation of perpendicular line through \(E\):
$$ y - 4 = \frac{1}{3}(x - 2) $$
Simplify:
$$ y = \frac{1}{3} x - \frac{2}{3} + 4 = \frac{1}{3} x + \frac{10}{3} $$
Answer: Gradient of perpendicular line is \(\frac{1}{3}\); equation is \(y = \frac{1}{3} x + \frac{10}{3}\).
Example 5: A line passes through points \(F(5, 1)\) and \(G(7, 9)\). Find the gradient of this line and the gradient of a line perpendicular to it. Then find the equation of the perpendicular line passing through \(H(3, 4)\).
Given: \(F(5, 1)\), \(G(7, 9)\), point \(H(3, 4)\)
Calculate gradient:
$$ m_1 = \frac{9 - 1}{7 - 5} = \frac{8}{2} = 4 $$
Gradient of perpendicular line:
$$ m_2 = -\frac{1}{4} = -0.25 $$
Equation of perpendicular line through \(H\):
$$ y - 4 = -0.25 (x - 3) $$
Simplify:
$$ y = -0.25 x + 0.75 + 4 = -0.25 x + 4.75 $$
Answer: Gradient of perpendicular line is \(-0.25\); equation is \(y = -0.25 x + 4.75\).
Find the gradient of a line passing through points \((2, 6)\) and \((5, 15)\), then find the gradient of a line perpendicular to it. (5 marks)
A machine part edge has gradient \(-\frac{2}{3}\). Calculate the gradient of a line perpendicular to it and write the equation of the perpendicular line passing through \((1, 2)\). (7 marks)
Calculate the gradient of a line through points \((0, 0)\) and \((3, 9)\) and find the gradient of its perpendicular line. (5 marks)
A conveyor frame line passes through points \((4, 2)\) and \((8, 10)\). Find the gradient of the line and the gradient of a line perpendicular to it. (6 marks)
Find the equation of a line perpendicular to the line passing through \((1, 1)\) and \((4, 7)\), passing through point \((2, 3)\). (7 marks)
In mechanical production tasks, understanding the y-intercept of linear equations helps technicians interpret graphs showing relationships such as force versus displacement or temperature versus time. The y-intercept is the point where the line crosses the y-axis, representing the value of \(y\) when \(x = 0\). This concept is vital when analyzing machine calibration curves or sensor output graphs.
The linear equation in slope-intercept form is:
$$ y = mx + c $$
where:
The y-intercept \(c\) is found by setting \(x = 0\) in the equation \(y = mx + c\). This gives:
$$ y = m \times 0 + c $$
$$ y = c $$
The y-intercept is the value of \(y\) when \(x = 0\).
Example 1: A graph of temperature \(T\) in °C versus time \(t\) in minutes follows the equation \(T = 3t + 20\). Find the y-intercept.
Given: \(T = 3t + 20\), \(t=0\)
$$ y = 3 \times 0 + 20 $$
$$ y = 20 $$
Answer: y-intercept = 20^\circ C
Example 2: A machine's vibration amplitude \(A\) (mm) varies with speed \(S\) (rpm) as \(A = -0.05S + 5\). Find the y-intercept.
Given: \(A = -0.05S + 5\), \(S=0\)
$$ y = -0.05 \times 0 + 5 $$
$$ y = 5 $$
Answer: y-intercept = 5 mm
Example 3: The displacement \(d\) (mm) of a piston is given by \(d = 2.5x - 7\), where \(x\) is time in seconds. Find the y-intercept.
Given: \(d = 2.5x -7\), \(x=0\)
$$ y = 2.5 \times 0 - 7 $$
$$ y = -7 $$
Answer: y-intercept = -7 mm
If two points \((x_1, y_1)\) and \((x_2, y_2)\) on a line are known, the slope \(m\) is:
$$ m = \frac{y_2 - y_1}{x_2 - x_1} $$
The y-intercept \(c\) is then:
$$ c = y_1 - m x_1 $$
Example 1: Points on a force-displacement graph are \( (2, 10) \) and \( (5, 25) \). Find the y-intercept.
Given: \( (x_1, y_1) = (2, 10), (x_2, y_2) = (5, 25) \)
$$ m = \frac{25 - 10}{5 - 2} = \frac{15}{3} = 5 $$
$$ c = 10 - 5 \times 2 = 10 - 10 = 0 $$
Answer: y-intercept = 0
Example 2: Two points on a temperature-time graph are \( (1, 22) \) and \( (4, 34) \). Find the y-intercept.
Given: \( (x_1, y_1) = (1, 22), (x_2, y_2) = (4, 34) \)
$$ m = \frac{34 - 22}{4 - 1} = \frac{12}{3} = 4 $$
$$ c = 22 - 4 \times 1 = 22 - 4 = 18 $$
Answer: y-intercept = 18
Example 3: Points on a pressure-volume graph are \( (3, 15) \) and \( (6, 12) \). Find the y-intercept.
Given: \( (x_1, y_1) = (3, 15), (x_2, y_2) = (6, 12) \)
$$ m = \frac{12 - 15}{6 - 3} = \frac{-3}{3} = -1 $$
$$ c = 15 - (-1) \times 3 = 15 + 3 = 18 $$
Answer: y-intercept = 18
The y-intercept often represents an initial condition or baseline measurement in mechanical systems, such as initial load, starting temperature, or zero-speed vibration level.
Example 1: A calibration curve for a pressure sensor is \(P = 4t + 2\), where \(P\) is pressure in kPa and \(t\) is time in seconds. What does the y-intercept represent?
Given: \(P = 4t + 2\)
At \(t=0\), \(P = 2\) kPa
Answer: The sensor reads 2 kPa at zero time, representing baseline pressure.
Example 2: The voltage output of a thermocouple is \(V = 0.01T + 0.5\), where \(V\) is volts and \(T\) is temperature in °C. Interpret the y-intercept.
Given: \(V = 0.01T + 0.5\)
At \(T=0\), \(V = 0.5\) volts
Answer: The thermocouple voltage is 0.5 V at 0°C, indicating offset voltage.
Example 3: A machine's speed \(S\) (rpm) versus time \(t\) (s) is \(S = 50t - 100\). Interpret the y-intercept.
Given: \(S = 50t - 100\)
At \(t=0\), \(S = -100\) rpm
Answer: Negative y-intercept indicates the machine speed is below zero at start, possibly representing a delay or reverse rotation.
Find the y-intercept of the line passing through points (4, 18) and (10, 30). (5 marks)
Given the equation \(y = -2x + 8\), find the y-intercept. (3 marks)
A graph shows displacement \(d\) (mm) against time \(t\) (s) with points (3, 15) and (7, 27). Calculate the y-intercept. (5 marks)
A linear relationship between voltage \(V\) (volts) and current \(I\) (amps) is \(V = 5I + 1.2\). What is the y-intercept and what does it represent? (4 marks)
In mechanical production, many relationships between variables are linear and can be represented by straight lines on graphs. Understanding straight lines allows technicians to analyze machine behaviour, control parameters, and part dimensions efficiently. The straight line is defined by its slope and intercepts, and can be expressed in various forms such as slope-intercept form and point-slope form.
The most common form of a straight line equation is:
$$ y = mx + c $$
where:
The slope \(m\) is calculated as:
$$ m = \frac{y_2 - y_1}{x_2 - x_1} $$
Example 1: Find the equation of a line passing through points (2, 5) and (6, 13).
Given: \((x_1, y_1) = (2, 5), (x_2, y_2) = (6, 13)\)
Calculate slope:
$$ m = \frac{13 - 5}{6 - 2} = \frac{8}{4} = 2 $$
Find \(c\):
$$ c = y_1 - m x_1 = 5 - 2 \times 2 = 5 - 4 = 1 $$
Equation of line:
$$ y = 2x + 1 $$
Answer: \(y = 2x + 1\)
Example 2: The temperature \(T\) (°C) varies linearly with time \(t\) (min) and passes through (0, 20) and (5, 35). Find the equation.
Given: \((0, 20), (5, 35)\)
Calculate slope:
$$ m = \frac{35 - 20}{5 - 0} = \frac{15}{5} = 3 $$
Since \(x=0\) at \(t=0\), \(c = 20\)
Equation:
$$ T = 3t + 20 $$
Answer: \(T = 3t + 20\)
Example 3: Find the equation of the line through (1, -2) and (4, 7).
Given: \((1, -2), (4, 7)\)
Calculate slope:
$$ m = \frac{7 - (-2)}{4 - 1} = \frac{9}{3} = 3 $$
Calculate \(c\):
$$ c = -2 - 3 \times 1 = -2 - 3 = -5 $$
Equation:
$$ y = 3x - 5 $$
Answer: \(y = 3x - 5\)
The point-slope form expresses a line when a point \((x_1, y_1)\) and slope \(m\) are known:
$$ y, y_1 = m(x, x_1) $$
This form is useful in mechanical design when only one point and slope are known, such as during part dimensioning or sensor calibration.
Example 1: Write the equation of a line with slope 4 passing through (3, 7).
Given: \(m=4, (x_1, y_1) = (3, 7)\)
Equation:
$$ y - 7 = 4(x - 3) $$
Simplify:
$$ y - 7 = 4x - 12 $$
$$ y = 4x - 5 $$
Answer: \(y = 4x - 5\)
Example 2: A line passes through (2, -1) with slope \(-3\). Find the equation.
Given: \(m = -3, (2, -1)\)
Equation:
$$ y + 1 = -3(x - 2) $$
Simplify:
$$ y + 1 = -3x + 6 $$
$$ y = -3x + 5 $$
Answer: \(y = -3x + 5\)
Example 3: Find the equation of a line passing through (0, 4) with slope \(0.5\).
Given: \(m=0.5, (0, 4)\)
Equation:
$$ y - 4 = 0.5(x - 0) $$
Simplify:
$$ y = 0.5x + 4 $$
Answer: \(y = 0.5x + 4\)
Intercepts are points where the line crosses the axes. The x-intercept is where \(y=0\), and the y-intercept is where \(x=0\).
For the equation \(y = mx + c\):
$$ 0 = mx + c $$
$$ x = -\frac{c}{m} $$
$$ y = m \times 0 + c = c $$
Example 1: Find the intercepts of the line \(y = 2x - 6\).
Given: \(y = 2x - 6\)
Y-intercept at \(x=0\):
$$ y = 2 \times 0 - 6 = -6 $$
X-intercept at \(y=0\):
$$ 0 = 2x - 6 $$
$$ 2x = 6 $$
$$ x = 3 $$
Answer: x-intercept = 3, y-intercept = -6
Example 2: Find the intercepts of \(y = -0.5x + 4\).
Given: \(y = -0.5x + 4\)
Y-intercept at \(x=0\):
$$ y = -0.5 \times 0 + 4 = 4 $$
X-intercept at \(y=0\):
$$ 0 = -0.5x + 4 $$
$$ -0.5x = -4 $$
$$ x = \frac{-4}{-0.5} = 8 $$
Answer: x-intercept = 8, y-intercept = 4
Example 3: Find intercepts of \(y = 3x + 9\).
Given: \(y = 3x + 9\)
Y-intercept:
$$ y = 3 \times 0 + 9 = 9 $$
X-intercept:
$$ 0 = 3x + 9 $$
$$ 3x = -9 $$
$$ x = -3 $$
Answer: x-intercept = -3, y-intercept = 9
Straight line equations are used to interpret calibration data, predict machine behaviour, and model linear relationships in measurements such as torque versus angular displacement or load versus extension.
Example 1: A torque wrench calibration graph passes through (5 Nm, 1.2 V) and (15 Nm, 3.6 V). Find the voltage \(V\) at 10 Nm.
Given: \((x_1, y_1) = (5, 1.2), (x_2, y_2) = (15, 3.6)\), where \(x\) = torque, \(y\) = voltage.
Calculate slope:
$$ m = \frac{3.6 - 1.2}{15 - 5} = \frac{2.4}{10} = 0.24 $$
Find equation:
$$ y - 1.2 = 0.24(x - 5) $$
Calculate \(y\) at \(x=10\):
$$ y = 1.2 + 0.24(10 - 5) $$
$$ y = 1.2 + 0.24 \times 5 = 1.2 + 1.2 = 2.4 \text{ V} $$
Answer: Voltage at 10 Nm is 2.4 V
Example 2: The load \(L\) (kN) on a spring is linearly related to its extension \(e\) (mm). The points (3 mm, 12 kN) and (7 mm, 28 kN) are on the load-extension graph. Find the load at 5 mm extension.
Given: \((3, 12), (7, 28)\)
Calculate slope:
$$ m = \frac{28 - 12}{7 - 3} = \frac{16}{4} = 4 $$
Find equation:
$$ y - 12 = 4(x - 3) $$
Calculate \(y\) at \(x=5\):
$$ y = 12 + 4(5 - 3) = 12 + 4 \times 2 = 12 + 8 = 20 \text{ kN} $$
Answer: Load at 5 mm extension is 20 kN
Example 3: A machine speed \(S\) (rpm) depends linearly on voltage \(V\) (volts). Points (10 V, 1200 rpm) and (15 V, 1800 rpm) lie on the line. Find the voltage when speed is 1500 rpm.
Given: \((10, 1200), (15, 1800)\)
Calculate slope:
$$ m = \frac{1800 - 1200}{15 - 10} = \frac{600}{5} = 120 $$
Equation:
$$ y - 1200 = 120(x - 10) $$
Set \(y = 1500\), solve for \(x\):
$$ 1500 - 1200 = 120(x - 10) $$
$$ 300 = 120(x - 10) $$
$$ x - 10 = \frac{300}{120} = 2.5 $$
$$ x = 12.5 \text{ V} $$
Answer: Voltage at 1500 rpm is 12.5 V
Find the equation of the straight line passing through (1, 4) and (5, 16). (5 marks)
Using point-slope form, find the equation of a line with slope \(-2\) passing through (3, 7). (4 marks)
Calculate the x and y intercepts of the line \(y = -3x + 9\). (5 marks)
A linear relationship between pressure \(P\) (kPa) and volume \(V\) (litres) passes through points (2, 50) and (6, 30). Find the equation and the pressure at 4 litres. (6 marks)
In mechanical production and engineering drawing, understanding parallel lines is essential for designing components that fit together precisely. Parallel lines have the same slope and never intersect, which ensures uniform spacing in mechanical parts such as gear teeth, slots, or guide rails. Recognizing and calculating the relationships between parallel lines helps in layout planning and quality control in Kenyan manufacturing firms and workshops.
Two lines are parallel if their slopes are equal. If the equations of two lines are \(y = m_1 x + c_1\) and \(y = m_2 x + c_2\), then the lines are parallel if and only if:
$$m_1 = m_2$$
where \(m_1\) and \(m_2\) are the slopes of the two lines.
Example 1:
Given two lines with equations \(y = 3x + 2\) and \(y = 3x - 5\), determine if they are parallel.
Given:
Line 1 slope \(m_1 = 3\)
Line 2 slope \(m_2 = 3\)
$$m_1 = m_2$$$$3 = 3$$
Answer: The lines are parallel.
Example 2:
Find the value of \(k\) such that the lines \(y = 2x + 1\) and \(y = kx - 4\) are parallel.
Given:
Line 1 slope \(m_1 = 2\)
Line 2 slope \(m_2 = k\)
$$m_1 = m_2$$$$2 = k$$
Answer: \(k = 2\)
Example 3:
Determine if the lines \(4x - 5y + 7 = 0\) and \(8x - 10y + 3 = 0\) are parallel.
Given:
Line 1 equation: \(4x - 5y + 7 = 0\)
Line 2 equation: \(8x - 10y + 3 = 0\)
First, find slopes:$$m_1 = -\frac{A}{B} = -\frac{4}{-5} = \frac{4}{5}$$$$m_2 = -\frac{8}{-10} = \frac{8}{10} = \frac{4}{5}$$
Since \(m_1 = m_2\), lines are parallel.
Answer: The lines are parallel.
Example 4:
Find the equation of a line parallel to \(y = \frac{1}{2}x + 3\) passing through the point \((4, 1)\).
Given:
Slope \(m = \frac{1}{2}\)
Point \((x_1, y_1) = (4, 1)\)
Use point-slope form:$$y, y_1 = m(x, x_1)$$$$y - 1 = \frac{1}{2}(x - 4)$$$$y = \frac{1}{2}x - 2 + 1$$$$y = \frac{1}{2}x - 1$$
Answer: \(y = \frac{1}{2}x - 1\)
Example 5:
A guide rail in a machine is designed to be parallel to the line \(3x + 4y = 12\). Find the equation of the guide rail passing through \((2, -1)\).
Given:
Line slope:$$3x + 4y = 12 \Rightarrow 4y = -3x + 12 \Rightarrow y = -\frac{3}{4}x + 3$$
Slope \(m = -\frac{3}{4}\)
Point \((2, -1)\)
Using point-slope form:$$y - (-1) = -\frac{3}{4}(x - 2)$$$$y + 1 = -\frac{3}{4}x + \frac{3}{2}$$$$y = -\frac{3}{4}x + \frac{3}{2} - 1$$$$y = -\frac{3}{4}x + \frac{1}{2}$$
Answer: \(y = -\frac{3}{4}x + \frac{1}{2}\)
The distance \(d\) between two parallel lines \(Ax + By + C_1 = 0\) and \(Ax + By + C_2 = 0\) is given by:
$$d = \frac{|C_2 - C_1|}{\sqrt{A^2 + B^2}}$$
This calculation is important in mechanical production to ensure precise spacing between parallel components such as shafts or beams.
Example 1:
Find the distance between the lines \(2x - 3y + 4 = 0\) and \(2x - 3y - 5 = 0\).
Given:
\(A = 2\), \(B = -3\), \(C_1 = 4\), \(C_2 = -5\)
$$d = \frac{|C_2 - C_1|}{\sqrt{A^2 + B^2}}$$$$d = \frac{|-5 - 4|}{\sqrt{2^2 + (-3)^2}}$$$$d = \frac{9}{\sqrt{4 + 9}}$$$$d = \frac{9}{\sqrt{13}}$$$$d = \frac{9}{3.606}$$$$d \approx 2.496$$
Answer: Distance \(d \approx 2.50\) units
Example 2:
Calculate the distance between the parallel lines \(x + y - 3 = 0\) and \(x + y + 1 = 0\).
Given:
\(A = 1\), \(B = 1\), \(C_1 = -3\), \(C_2 = 1\)
$$d = \frac{|1 - (-3)|}{\sqrt{1^2 + 1^2}}$$$$d = \frac{4}{\sqrt{2}}$$$$d = \frac{4}{1.414}$$$$d \approx 2.828$$
Answer: Distance \(d \approx 2.83\) units
Example 3:
Determine the distance between the lines \(5x + 12y + 9 = 0\) and \(5x + 12y - 6 = 0\).
Given:
\(A = 5\), \(B = 12\), \(C_1 = 9\), \(C_2 = -6\)
$$d = \frac{|-6 - 9|}{\sqrt{5^2 + 12^2}}$$$$d = \frac{15}{\sqrt{25 + 144}}$$$$d = \frac{15}{\sqrt{169}}$$$$d = \frac{15}{13}$$$$d \approx 1.154$$
Answer: Distance \(d \approx 1.15\) units
Example 4:
Find the distance between the lines \(6x - 8y + 10 = 0\) and \(6x - 8y - 14 = 0\).
Given:
\(A = 6\), \(B = -8\), \(C_1 = 10\), \(C_2 = -14\)
$$d = \frac{|-14 - 10|}{\sqrt{6^2 + (-8)^2}}$$$$d = \frac{24}{\sqrt{36 + 64}}$$$$d = \frac{24}{\sqrt{100}}$$$$d = \frac{24}{10}$$$$d = 2.4$$
Answer: Distance \(d = 2.4\) units
Example 5:
Calculate the distance between \(3x + 4y + 7 = 0\) and \(3x + 4y - 1 = 0\).
Given:
\(A = 3\), \(B = 4\), \(C_1 = 7\), \(C_2 = -1\)
$$d = \frac{|-1 - 7|}{\sqrt{3^2 + 4^2}}$$$$d = \frac{8}{\sqrt{9 + 16}}$$$$d = \frac{8}{\sqrt{25}}$$$$d = \frac{8}{5}$$$$d = 1.6$$
Answer: Distance \(d = 1.6\) units
To find the equation of a line parallel to a given line \(Ax + By + C = 0\) passing through a point \((x_1, y_1)\), use the fact that the slope remains the same and substitute the point into the line equation form.
General form of the new line:
$$Ax + By + D = 0$$
where \(D\) is found by substituting the point \((x_1, y_1)\):
$$D = -(Ax_1 + By_1)$$
Example 1:
Find the equation of a line parallel to \(2x + 3y - 6 = 0\) passing through \((4, 1)\).
Given:
\(A = 2\), \(B = 3\), point \((4, 1)\)
Calculate \(D\):$$D = -(2 \times 4 + 3 \times 1)$$$$D = -(8 + 3)$$$$D = -11$$
Equation:$$2x + 3y - 11 = 0$$
Answer: \(2x + 3y - 11 = 0\)
Example 2:
Find the equation of a line parallel to \(x - 4y + 7 = 0\) through point \((2, 3)\).
Given:
\(A = 1\), \(B = -4\), point \((2, 3)\)
Calculate \(D\):$$D = -(1 \times 2 + (-4) \times 3)$$$$D = -(2 - 12)$$$$D = 10$$
Equation:$$x - 4y + 10 = 0$$
Answer: \(x - 4y + 10 = 0\)
Example 3:
Find the equation of a line parallel to \(5x + 2y - 8 = 0\) passing through \((0, -3)\).
Given:
\(A = 5\), \(B = 2\), point \((0, -3)\)
Calculate \(D\):$$D = -(5 \times 0 + 2 \times (-3))$$$$D = - (0 - 6)$$$$D = 6$$
Equation:$$5x + 2y + 6 = 0$$
Answer: \(5x + 2y + 6 = 0\)
Example 4:
Find the equation of a line parallel to \(7x, y + 4 = 0\) passing through \((-1, 2)\).
Given:
\(A = 7\), \(B = -1\), point \((-1, 2)\)
Calculate \(D\):$$D = -(7 \times (-1) + (-1) \times 2)$$$$D = -(-7 - 2)$$$$D = 9$$
Equation:$$7x, y + 9 = 0$$
Answer: \(7x, y + 9 = 0\)
Example 5:
Find the equation of a line parallel to \(4x + 5y + 3 = 0\) through \((3, 0)\).
Given:
\(A = 4\), \(B = 5\), point \((3, 0)\)
Calculate \(D\):$$D = -(4 \times 3 + 5 \times 0)$$$$D = -(12 + 0)$$$$D = -12$$
Equation:$$4x + 5y - 12 = 0$$
Answer: \(4x + 5y - 12 = 0\)
In mechanical production, perpendicularity ensures components fit and function correctly, such as shafts intersecting at right angles or machine parts assembled at 90 degrees. The relationship between perpendicular lines involves the negative reciprocal of slopes, a key principle in engineering drawing and CAD design. Graphing linear equations helps visualize these relationships and verify designs.
Two lines with slopes \(m_1\) and \(m_2\) are perpendicular if:
$$m_1 \times m_2 = -1$$
This means one slope is the negative reciprocal of the other.
Example 1:
Are the lines \(y = 2x + 1\) and \(y = -\frac{1}{2}x + 3\) perpendicular?
Given:
\(m_1 = 2\), \(m_2 = -\frac{1}{2}\)
Calculate product:$$m_1 \times m_2 = 2 \times (-\frac{1}{2}) = -1$$
Answer: The lines are perpendicular.
Example 2:
Determine if the lines \(3x + 4y - 5 = 0\) and \(4x - 3y + 2 = 0\) are perpendicular.
Given:
Line 1 slope:$$m_1 = -\frac{3}{4}$$
Line 2 slope:$$m_2 = -\frac{4}{-3} = \frac{4}{3}$$
Calculate product:$$m_1 \times m_2 = -\frac{3}{4} \times \frac{4}{3} = -1$$
Answer: The lines are perpendicular.
Example 3:
Check if lines \(y = \frac{5}{3}x - 2\) and \(y = -\frac{3}{5}x + 1\) are perpendicular.
Given:
\(m_1 = \frac{5}{3}\), \(m_2 = -\frac{3}{5}\)
Calculate product:$$\frac{5}{3} \times (-\frac{3}{5}) = -1$$
Answer: The lines are perpendicular.
Example 4:
Determine if lines \(y = 4x + 7\) and \(y = \frac{1}{4}x - 3\) are perpendicular.
Given:
\(m_1 = 4\), \(m_2 = \frac{1}{4}\)
Calculate product:$$4 \times \frac{1}{4} = 1 \neq -1$$
Answer: The lines are not perpendicular.
Example 5:
Find the slope of a line perpendicular to \(y = -\frac{2}{3}x + 5\).
Given:
Line slope \(m_1 = -\frac{2}{3}\)
Find \(m_2\) such that:$$m_1 \times m_2 = -1$$$$-\frac{2}{3} \times m_2 = -1$$$$m_2 = \frac{-1}{-\frac{2}{3}} = \frac{3}{2}$$
Answer: Slope of perpendicular line is \(\frac{3}{2}\).
Given a line \(y = m_1 x + c\) and a point \((x_1, y_1)\), the slope of the perpendicular line is:
$$m_2 = -\frac{1}{m_1}$$
The equation of the perpendicular line using the point-slope form is:
$$y, y_1 = m_2 (x, x_1)$$
Example 1:
Find the equation of a line perpendicular to \(y = 2x + 3\) passing through \((1, 4)\).
Given:
\(m_1 = 2\), point \((1, 4)\)
Calculate slope of perpendicular line:$$m_2 = -\frac{1}{2}$$
Equation:$$y - 4 = -\frac{1}{2}(x - 1)$$$$y - 4 = -\frac{1}{2}x + \frac{1}{2}$$$$y = -\frac{1}{2}x + \frac{1}{2} + 4$$$$y = -\frac{1}{2}x + \frac{9}{2}$$
Answer: \(y = -\frac{1}{2}x + \frac{9}{2}\)
Example 2:
Find the equation of the line perpendicular to \(3x - 4y + 5 = 0\) through \((2, -1)\).
Given:
Convert to slope-intercept form:$$3x - 4y + 5 = 0 \Rightarrow -4y = -3x - 5 \Rightarrow y = \frac{3}{4}x + \frac{5}{4}$$
So, \(m_1 = \frac{3}{4}\)
Calculate \(m_2\):$$m_2 = -\frac{1}{m_1} = -\frac{4}{3}$$
Equation:$$y - (-1) = -\frac{4}{3}(x - 2)$$$$y + 1 = -\frac{4}{3}x + \frac{8}{3}$$$$y = -\frac{4}{3}x + \frac{8}{3} - 1$$$$y = -\frac{4}{3}x + \frac{5}{3}$$
Answer: \(y = -\frac{4}{3}x + \frac{5}{3}\)
Example 3:
Find the equation of a line perpendicular to \(y = -\frac{5}{2}x + 7\) through \((-3, 2)\).
Given:
\(m_1 = -\frac{5}{2}\)
Calculate \(m_2\):$$m_2 = -\frac{1}{-\frac{5}{2}} = \frac{2}{5}$$
Equation:$$y - 2 = \frac{2}{5}(x + 3)$$$$y - 2 = \frac{2}{5}x + \frac{6}{5}$$$$y = \frac{2}{5}x + \frac{6}{5} + 2$$$$y = \frac{2}{5}x + \frac{16}{5}$$
Answer: \(y = \frac{2}{5}x + \frac{16}{5}\)
Example 4:
Find the equation of a line perpendicular to \(4x + y - 8 = 0\) passing through \((0, 0)\).
Given:
Convert to slope form:$$y = -4x + 8$$
\(m_1 = -4\)
Calculate \(m_2\):$$m_2 = -\frac{1}{-4} = \frac{1}{4}$$
Equation:$$y - 0 = \frac{1}{4}(x - 0)$$$$y = \frac{1}{4}x$$
Answer: \(y = \frac{1}{4}x\)
Example 5:
Find the equation of the line perpendicular to \(2x + 5y + 3 = 0\) passing through \((1, 2)\).
Given:
Convert to slope form:$$5y = -2x - 3 \Rightarrow y = -\frac{2}{5}x - \frac{3}{5}$$
\(m_1 = -\frac{2}{5}\)
Calculate \(m_2\):$$m_2 = -\frac{1}{-\frac{2}{5}} = \frac{5}{2}$$
Equation:$$y - 2 = \frac{5}{2}(x - 1)$$$$y - 2 = \frac{5}{2}x - \frac{5}{2}$$$$y = \frac{5}{2}x - \frac{5}{2} + 2$$$$y = \frac{5}{2}x - \frac{1}{2}$$
Answer: \(y = \frac{5}{2}x - \frac{1}{2}\)
Graphing linear equations helps visualize mechanical parts and their relationships in design and layout. The graph represents all points \((x, y)\) satisfying the equation. Engineers use graphing to check fits and alignments in assemblies.
The general form of a linear equation is:
$$Ax + By + C = 0$$
To graph, find intercepts:
\(x\)-intercept: set \(y=0\), solve for \(x\).
\(y\)-intercept: set \(x=0\), solve for \(y\).
Example 1:
Graph \(2x + 3y - 6 = 0\). Find intercepts.
Given:
Equation: \(2x + 3y - 6 = 0\)
\(x\)-intercept:
Set \(y=0\)$$2x - 6 = 0$$$$2x = 6$$$$x = 3$$
\(y\)-intercept:
Set \(x=0\)$$3y - 6 = 0$$$$3y = 6$$$$y = 2$$
Answer: Intercepts at (3, 0) and (0, 2)
Example 2:
Graph \(x - 4y + 8 = 0\). Find intercepts.
Given:
Equation: \(x - 4y + 8 = 0\)
\(x\)-intercept:
Set \(y=0\)$$x + 8 = 0$$$$x = -8$$
\(y\)-intercept:
Set \(x=0\)$$-4y + 8 = 0$$$$-4y = -8$$$$y = 2$$
Answer: Intercepts at (-8, 0) and (0, 2)
Example 3:
Graph \(5x + 2y - 10 = 0\). Find intercepts.
Given:
Equation: \(5x + 2y - 10 = 0\)
\(x\)-intercept:
Set \(y=0\)$$5x - 10 = 0$$$$5x = 10$$$$x = 2$$
\(y\)-intercept:
Set \(x=0\)$$2y - 10 = 0$$$$2y = 10$$$$y = 5$$
Answer: Intercepts at (2, 0) and (0, 5)
Example 4:
Graph \(3x, y + 9 = 0\). Find intercepts.
Given:
Equation: \(3x, y + 9 = 0\)
\(x\)-intercept:
Set \(y=0\)$$3x + 9 = 0$$$$3x = -9$$$$x = -3$$
\(y\)-intercept:
Set \(x=0\)$$-y + 9 = 0$$$$-y = -9$$$$y = 9$$
Answer: Intercepts at (-3, 0) and (0, 9)
Example 5:
Graph \(4x + 5y - 20 = 0\). Find intercepts.
Given:
Equation: \(4x + 5y - 20 = 0\)
\(x\)-intercept:
Set \(y=0\)$$4x - 20 = 0$$$$4x = 20$$$$x = 5$$
\(y\)-intercept:
Set \(x=0\)$$5y - 20 = 0$$$$5y = 20$$$$y = 4$$
Answer: Intercepts at (5, 0) and (0, 4)
In mechanical production and machining, understanding the representation of straight lines using polar coordinates is crucial when working with lathe tool paths, CNC programming, and layout marking. Polar geometry allows specifying points and lines by their distance from a fixed origin and the angle relative to a reference direction, which simplifies calculations involving rotations and angled cuts.
A straight line in polar form is generally represented by the equation:
$$ r = \frac{p}{\cos(\theta - \alpha)} $$
where:
Example 1: A lathe operator needs to find the radial distance \(r\) of a point on a tool path line that is 50 mm from the origin, where the angle \(\alpha\) between the polar axis and the line’s perpendicular is 30°, and the point is located at \(\theta = 60^\circ\). Calculate \(r\).
Given: \(p = 50\, \text{mm}\), \(\alpha = 30^\circ\), \(\theta = 60^\circ\)$$r = \frac{p}{\cos(\theta - \alpha)}$$$$r = \frac{50}{\cos(60^\circ - 30^\circ)}$$$$r = \frac{50}{\cos(30^\circ)}$$$$r = \frac{50}{0.866}$$$$r = 57.74\, \text{mm}$$ Answer: 57.74 mm
Example 2: In CNC programming, a cutting line is defined with \(p = 80\, \text{mm}\) and \(\alpha = 45^\circ\). Find the radial distance \(r\) when the cutting tool is at \(\theta = 90^\circ\).
Given: \(p = 80\, \text{mm}\), \(\alpha = 45^\circ\), \(\theta = 90^\circ\)$$r = \frac{80}{\cos(90^\circ - 45^\circ)}$$$$r = \frac{80}{\cos(45^\circ)}$$$$r = \frac{80}{0.707}$$$$r = 113.14\, \text{mm}$$ Answer: 113.14 mm
Example 3: A marking gauge on a workbench is set such that the line it marks has a perpendicular distance \(p = 120\, \text{mm}\) from the origin, and the angle \(\alpha = 60^\circ\). Find the value of \(r\) at \(\theta = 75^\circ\).
Given: \(p = 120\, \text{mm}\), \(\alpha = 60^\circ\), \(\theta = 75^\circ\)$$r = \frac{120}{\cos(75^\circ - 60^\circ)}$$$$r = \frac{120}{\cos(15^\circ)}$$$$r = \frac{120}{0.966}$$$$r = 124.26\, \text{mm}$$ Answer: 124.26 mm
Example 4: A mechanical technician is setting a straight edge at \(p = 100\, \text{mm}\) and \(\alpha = 20^\circ\). Calculate \(r\) for \(\theta = 110^\circ\).
Given: \(p = 100\, \text{mm}\), \(\alpha = 20^\circ\), \(\theta = 110^\circ\)$$r = \frac{100}{\cos(110^\circ - 20^\circ)}$$$$r = \frac{100}{\cos(90^\circ)}$$$$r = \frac{100}{0}$$
Since \(\cos 90^\circ = 0\), \(r\) tends to infinity, indicating the line is perpendicular to the direction at \(\theta = 110^\circ\), so no finite \(r\) exists.
Answer: No finite solution (line perpendicular to direction at \(\theta=110^\circ\))
Example 5: For a milling operation, the line is defined with \(p = 75\, \text{mm}\) and \(\alpha = 135^\circ\). Find \(r\) when \(\theta = 150^\circ\).
Given: \(p = 75\, \text{mm}\), \(\alpha = 135^\circ\), \(\theta = 150^\circ\)$$r = \frac{75}{\cos(150^\circ - 135^\circ)}$$$$r = \frac{75}{\cos(15^\circ)}$$$$r = \frac{75}{0.966}$$$$r = 77.66\, \text{mm}$$ Answer: 77.66 mm
Magnitude refers to the length or size of a vector quantity, which in mechanical production represents quantities such as displacement, force, or velocity. Calculating magnitude is essential when determining resultant forces, tool displacements, and component dimensions in vector form. The magnitude of a vector \(\mathbf{v} = (x, y)\) in Cartesian coordinates is given by:
$$ |\mathbf{v}| = \sqrt{x^{2} + y^{2}} $$
where \(x\) and \(y\) are the vector components along the X and Y axes respectively.
Example 1: A cutting force vector has components \(F_x = 30\, \text{N}\) and \(F_y = 40\, \text{N}\). Find the magnitude of the force.
Given: \(F_x = 30\, \text{N}\), \(F_y = 40\, \text{N}\)$$|\mathbf{F}| = \sqrt{F_x^{2} + F_y^{2}}$$$$|\mathbf{F}| = \sqrt{30^{2} + 40^{2}}$$$$|\mathbf{F}| = \sqrt{900 + 1600}$$$$|\mathbf{F}| = \sqrt{2500}$$$$|\mathbf{F}| = 50\, \text{N}$$ Answer: 50 N
Example 2: The displacement of a tool is represented by vector components \(x = 15\, \text{mm}\) and \(y = 20\, \text{mm}\). Calculate the magnitude of the displacement.
Given: \(x = 15\, \text{mm}\), \(y = 20\, \text{mm}\)$$|\mathbf{d}| = \sqrt{15^{2} + 20^{2}}$$$$|\mathbf{d}| = \sqrt{225 + 400}$$$$|\mathbf{d}| = \sqrt{625}$$$$|\mathbf{d}| = 25\, \text{mm}$$ Answer: 25 mm
Example 3: A velocity vector of a conveyor belt has components \(v_x = 3\, \text{m/s}\) and \(v_y = 4\, \text{m/s}\). Find the magnitude of the velocity.
Given: \(v_x = 3\, \text{m/s}\), \(v_y = 4\, \text{m/s}\)$$|\mathbf{v}| = \sqrt{3^{2} + 4^{2}}$$$$|\mathbf{v}| = \sqrt{9 + 16}$$$$|\mathbf{v}| = \sqrt{25}$$$$|\mathbf{v}| = 5\, \text{m/s}$$ Answer: 5 m/s
Example 4: The resultant force on a welded joint has components \(F_x = -50\, \text{N}\) and \(F_y = 120\, \text{N}\). Calculate the magnitude of the resultant force.
Given: \(F_x = -50\, \text{N}\), \(F_y = 120\, \text{N}\)$$|\mathbf{F}| = \sqrt{(-50)^{2} + 120^{2}}$$$$|\mathbf{F}| = \sqrt{2500 + 14400}$$$$|\mathbf{F}| = \sqrt{16900}$$$$|\mathbf{F}| = 130\, \text{N}$$ Answer: 130 N
Example 5: A tool displacement vector has components \(x = -25\, \text{mm}\) and \(y = -60\, \text{mm}\). Find the magnitude of the displacement.
Given: \(x = -25\, \text{mm}\), \(y = -60\, \text{mm}\)$$|\mathbf{d}| = \sqrt{(-25)^{2} + (-60)^{2}}$$$$|\mathbf{d}| = \sqrt{625 + 3600}$$$$|\mathbf{d}| = \sqrt{4225}$$$$|\mathbf{d}| = 65\, \text{mm}$$ Answer: 65 mm
Example 6:
A technician at Kenya Power calculates the magnitude of a force vector acting on a transmission tower with components \(F_x = 60\, \text{N}\), \(F_y = 80\, \text{N}\).
$$ |\mathbf{F}| = \sqrt{60^2 + 80^2} = \sqrt{3600 + 6400} = \sqrt{10000} = 100\, \text{N} $$
Answer: 100 N (Kenya Power context)
Example 7:
At East African Portland Cement, a tool moves with displacement components \(x = -20\, \text{mm}\), \(y = 21\, \text{mm}\). Find the magnitude of displacement.
$$ |\mathbf{d}| = \sqrt{(-20)^2 + 21^2} = \sqrt{400 + 441} = \sqrt{841} = 29\, \text{mm} $$
Answer: 29 mm (East African Portland Cement context)
Direction in coordinate geometry refers to the orientation of a line or vector relative to a reference axis, usually the positive \(x\)-axis in the Cartesian plane. For Mechanical Production Technicians in Kenya, understanding direction is crucial when interpreting and creating technical drawings, machining paths, or assembly instructions that depend on angles and vector directions. Direction is commonly expressed as an angle measured in degrees or radians from the positive \(x\)-axis, moving counterclockwise. The formula to find the direction angle \(\theta\) of a vector with components \(\Delta x\) and \(\Delta y\) is:
$$ \theta = \tan^{-1} (\frac{\Delta y}{\Delta x}) $$
The direction angle of a line segment connecting two points \((x_1, y_1)\) and \((x_2, y_2)\) is found by first determining the differences in coordinates and then applying the inverse tangent function.
$$ \theta = \tan^{-1} (\frac{y_2 - y_1}{x_2 - x_1}) $$
Example 1: Find the direction angle of a line joining points \(A(2, 3)\) and \(B(5, 7)\).
Given: \(x_1 = 2\), \(y_1 = 3\), \(x_2 = 5\), \(y_2 = 7\)
$$ \theta = \tan^{-1} (\frac{7 - 3}{5 - 2}) $$
$$ = \tan^{-1} (\frac{4}{3}) $$
$$ = \tan^{-1} (1.3333) $$
$$ = 53.13^\circ $$
Answer: \(53.13^\circ\)
Example 2: Determine the direction angle of a vector from \(C(4, 1)\) to \(D(1, 5)\).
Given: \(x_1 = 4\), \(y_1 = 1\), \(x_2 = 1\), \(y_2 = 5\)
$$ \theta = \tan^{-1} (\frac{5 - 1}{1 - 4}) $$
$$ = \tan^{-1} (\frac{4}{-3}) $$
Since \(\Delta x\) is negative and \(\Delta y\) positive, the vector lies in the second quadrant, so:
$$ \theta = 180^\circ + \tan^{-1} (\frac{4}{-3}) $$
$$ = 180^\circ - 53.13^\circ $$
$$ = 126.87^\circ $$
Answer: \(126.87^\circ\)
Example 3: A machining path runs from \(E(0, 0)\) to \(F(-4, -4)\). Find the direction angle.
Given: \(x_1 = 0\), \(y_1 = 0\), \(x_2 = -4\), \(y_2 = -4\)
$$ \theta = \tan^{-1} (\frac{-4 - 0}{-4 - 0}) $$
$$ = \tan^{-1} (\frac{-4}{-4}) $$
$$ = \tan^{-1} (1) $$
Since both \(\Delta x\) and \(\Delta y\) are negative, the vector lies in the third quadrant:
$$ \theta = 180^\circ + 45^\circ = 225^\circ $$
Answer: \(225^\circ\)
Example 4: Find the direction angle of a vector from \(G(3, 2)\) to \(H(3, 7)\).
Given: \(x_1 = 3\), \(y_1 = 2\), \(x_2 = 3\), \(y_2 = 7\)
Here, \(\Delta x = 0\), so the angle is vertical.
$$ \theta = 90^\circ \quad \text{(since vector points straight up)} $$
Answer: \(90^\circ\)
Example 5: Determine the direction angle of a vector from \(I(5, 5)\) to \(J(10, 0)\).
Given: \(x_1 = 5\), \(y_1 = 5\), \(x_2 = 10\), \(y_2 = 0\)
$$ \theta = \tan^{-1} (\frac{0 - 5}{10 - 5}) $$
$$ = \tan^{-1} (\frac{-5}{5}) $$
$$ = \tan^{-1} (-1) $$
Since \(\Delta x > 0\) and \(\Delta y < 0\), angle lies in the fourth quadrant:
$$ \theta = 360^\circ - 45^\circ = 315^\circ $$
Answer: \(315^\circ\)
A vector's direction can be found from its components along the \(x\) and \(y\) axes. This is essential for Mechanical Production Technicians when analyzing forces or velocities in machinery.
The formula remains:
$$ \theta = \tan^{-1} (\frac{V_y}{V_x}) $$
where \(V_x\) and \(V_y\) are the vector components.
Example 1: A force vector has components \(V_x = 6\, \text{N}\), \(V_y = 8\, \text{N}\). Find its direction.
$$ \theta = \tan^{-1} (\frac{8}{6}) $$
$$ = \tan^{-1} (1.3333) $$
$$ = 53.13^\circ $$
Answer: \(53.13^\circ\)
Example 2: A velocity vector has components \(V_x = -3\, \text{m/s}\), \(V_y = 3\, \text{m/s}\). Find its direction.
Since \(V_x < 0\) and \(V_y > 0\), vector lies in second quadrant.
$$ \theta = 180^\circ + \tan^{-1} (\frac{3}{-3}) $$
$$ = 180^\circ - 45^\circ = 135^\circ $$
Answer: \(135^\circ\)
Example 3: A displacement vector has components \(V_x = -5\, \text{m}\), \(V_y = -5\, \text{m}\).
Both components negative, vector in third quadrant.
$$ \theta = 180^\circ + \tan^{-1} (\frac{-5}{-5}) $$
$$ = 180^\circ + 45^\circ = 225^\circ $$
Answer: \(225^\circ\)
Example 4: A force vector has components \(V_x = 0\, \text{N}\), \(V_y = -10\, \text{N}\).
When \(V_x = 0\), the vector points vertically.
$$ \theta = 270^\circ $$
Answer: \(270^\circ\)
Example 5: A velocity vector has components \(V_x = 4\, \text{m/s}\), \(V_y = -3\, \text{m/s}\).
\(V_x > 0\), \(V_y < 0\) → fourth quadrant.
$$ \theta = 360^\circ + \tan^{-1} (\frac{-3}{4}) $$
$$ = 360^\circ - 36.87^\circ = 323.13^\circ $$
Answer: \(323.13^\circ\)
Given a direction angle \(\theta\) and magnitude \(R\), the components \(V_x\) and \(V_y\) are:
$$ V_x = R \cos \theta $$$$ V_y = R \sin \theta $$
This conversion is vital in Mechanical Production when decomposing forces or velocities into orthogonal components for analysis.
Example 1: A force of magnitude \(10\, \text{N}\) acts at an angle of \(30^\circ\). Find the components.
$$ V_x = 10 \cos 30^\circ $$
$$ = 10 \times 0.866 = 8.66\, \text{N} $$
$$ V_y = 10 \sin 30^\circ $$
$$ = 10 \times 0.5 = 5.0\, \text{N} $$
Answer: \(V_x = 8.66\, \text{N}\), \(V_y = 5.0\, \text{N}\)
Example 2: A velocity of \(15\, \text{m/s}\) is directed at \(135^\circ\). Find components.
$$ V_x = 15 \cos 135^\circ = 15 \times (-0.7071) = -10.61\, \text{m/s} $$
$$ V_y = 15 \sin 135^\circ = 15 \times 0.7071 = 10.61\, \text{m/s} $$
Answer: \(V_x = -10.61\, \text{m/s}\), \(V_y = 10.61\, \text{m/s}\)
Example 3: A displacement vector has magnitude \(20\, \text{m}\) and direction \(210^\circ\).
$$ V_x = 20 \cos 210^\circ = 20 \times (-0.866) = -17.32\, \text{m} $$
$$ V_y = 20 \sin 210^\circ = 20 \times (-0.5) = -10\, \text{m} $$
Answer: \(V_x = -17.32\, \text{m}\), \(V_y = -10\, \text{m}\)
Example 4: A force vector has magnitude \(25\, \text{N}\) at direction \(270^\circ\).
$$ V_x = 25 \cos 270^\circ = 25 \times 0 = 0\, \text{N} $$
$$ V_y = 25 \sin 270^\circ = 25 \times (-1) = -25\, \text{N} $$
Answer: \(V_x = 0\, \text{N}\), \(V_y = -25\, \text{N}\)
Example 5: A velocity vector has magnitude \(12\, \text{m/s}\) at \(315^\circ\).
$$ V_x = 12 \cos 315^\circ = 12 \times 0.7071 = 8.49\, \text{m/s} $$
$$ V_y = 12 \sin 315^\circ = 12 \times (-0.7071) = -8.49\, \text{m/s} $$
Answer: \(V_x = 8.49\, \text{m/s}\), \(V_y = -8.49\, \text{m/s}\)
The inverse tangent function returns values between \(-90^\circ\) and \(90^\circ\), so the quadrant of the vector must be considered to find the true direction angle \(\theta\).
Rules to adjust \(\theta\):
Example 1: Find the direction angle for a vector with \(\Delta x = -6\), \(\Delta y = 8\).
Since \(\Delta x < 0\), second quadrant:
$$ \theta = 180^\circ + \tan^{-1} (\frac{8}{-6}) $$
$$ = 180^\circ - 53.13^\circ = 126.87^\circ $$
Answer: \(126.87^\circ\)
Example 2: Vector with \(\Delta x = -4\), \(\Delta y = -3\).
Third quadrant:
$$ \theta = 180^\circ + \tan^{-1} (\frac{-3}{-4}) $$
$$ = 180^\circ + 36.87^\circ = 216.87^\circ $$
Answer: \(216.87^\circ\)
Example 3: Vector with \(\Delta x = 5\), \(\Delta y = -5\).
Fourth quadrant:
$$ \theta = 360^\circ + \tan^{-1} (\frac{-5}{5}) $$
$$ = 360^\circ - 45^\circ = 315^\circ $$
Answer: \(315^\circ\)
Example 4: Vector with \(\Delta x = 0\), \(\Delta y = 7\).
Vertical up:
$$ \theta = 90^\circ $$
Answer: \(90^\circ\)
Example 5: Vector with \(\Delta x = 0\), \(\Delta y = -9\).
Vertical down:
$$ \theta = 270^\circ $$
Answer: \(270^\circ\)
Calculate the direction angle of a vector joining points \(P(1, 4)\) and \(Q(7, 10)\). (5 marks)
A force vector has components \(V_x = -8\, \text{N}\) and \(V_y = 6\, \text{N}\). Find its direction angle. (5 marks)
Find the components \(V_x\) and \(V_y\) of a \(20\, \text{N}\) force acting at \(120^\circ\). (5 marks)
Determine the direction angle of a vector with \(\Delta x = -7\) and \(\Delta y = -7\). (5 marks)
A velocity vector has magnitude \(25\, \text{m/s}\) and direction \(300^\circ\). Find its components. (5 marks)
Graph conversions in coordinate geometry involve changing the representation of points or lines from one coordinate format to another, such as from Cartesian to polar coordinates or between absolute and relative coordinate formats. For Mechanical Production Technicians, these conversions aid in interpreting CAD drawings, programming CNC machines, and understanding machining paths or assembly layouts. Mastery of graph conversions ensures precise communication and execution of engineering tasks.
Polar coordinates represent a point by its distance \(r\) from the origin and its angle \(\theta\) from the positive \(x\)-axis. Given Cartesian coordinates \((x, y)\), the conversion formulas are:
$$ r = \sqrt{x^2 + y^2} $$
$$ \theta = \tan^{-1} (\frac{y}{x}) $$
Example 1: Convert point \(A(3, 4)\) to polar coordinates.
Given: \(x = 3\), \(y = 4\)
$$ r = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 $$
$$ \theta = \tan^{-1} (\frac{4}{3}) = 53.13^\circ $$
Answer: \(r = 5\), \(\theta = 53.13^\circ\)
Example 2: Convert point \(B(-3, 3)\) to polar coordinates.
Given: \(x = -3\), \(y = 3\)
$$ r = \sqrt{(-3)^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 4.24 $$
Since \(x < 0\), second quadrant:
$$ \theta = 180^\circ + \tan^{-1} (\frac{3}{-3}) = 180^\circ - 45^\circ = 135^\circ $$
Answer: \(r = 4.24\), \(\theta = 135^\circ\)
Example 3: Convert point \(C(-4, -4)\) to polar coordinates.
Given: \(x = -4\), \(y = -4\)
$$ r = \sqrt{(-4)^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 5.66 $$
Third quadrant:
$$ \theta = 180^\circ + \tan^{-1} (\frac{-4}{-4}) = 180^\circ + 45^\circ = 225^\circ $$
Answer: \(r = 5.66\), \(\theta = 225^\circ\)
Example 4: Convert point \(D(0, 5)\) to polar coordinates.
Given: \(x = 0\), \(y = 5\)
$$ r = \sqrt{0^2 + 5^2} = 5 $$
Vertical up:
$$ \theta = 90^\circ $$
Answer: \(r = 5\), \(\theta = 90^\circ\)
Example 5: Convert point \(E(5, 0)\) to polar coordinates.
Given: \(x = 5\), \(y = 0\)
$$ r = \sqrt{5^2 + 0^2} = 5 $$
Horizontal right:
$$ \theta = 0^\circ $$
Answer: \(r = 5\), \(\theta = 0^\circ\)
Given a point in polar coordinates \((r, \theta)\), the Cartesian coordinates \((x, y)\) are found by:
$$ x = r \cos \theta $$
$$ y = r \sin \theta $$
Example 1: Convert polar coordinates \(r=10\), \(\theta=60^\circ\) to Cartesian.
$$ x = 10 \cos 60^\circ = 10 \times 0.5 = 5 $$
$$ y = 10 \sin 60^\circ = 10 \times 0.866 = 8.66 $$
Answer: \(x = 5\), \(y = 8.66\)
Example 2: Convert \(r=7\), \(\theta=135^\circ\).
$$ x = 7 \cos 135^\circ = 7 \times (-0.7071) = -4.95 $$
$$ y = 7 \sin 135^\circ = 7 \times 0.7071 = 4.95 $$
Answer: \(x = -4.95\), \(y = 4.95\)
Example 3: Convert \(r=12\), \(\theta=210^\circ\).
$$ x = 12 \cos 210^\circ = 12 \times (-0.866) = -10.39 $$
$$ y = 12 \sin 210^\circ = 12 \times (-0.5) = -6 $$
Answer: \(x = -10.39\), \(y = -6\)
Example 4: Convert \(r=5\), \(\theta=270^\circ\).
$$ x = 5 \cos 270^\circ = 5 \times 0 = 0 $$
$$ y = 5 \sin 270^\circ = 5 \times (-1) = -5 $$
Answer: \(x = 0\), \(y = -5\)
Example 5: Convert \(r=9\), \(\theta=315^\circ\).
$$ x = 9 \cos 315^\circ = 9 \times 0.7071 = 6.36 $$
$$ y = 9 \sin 315^\circ = 9 \times (-0.7071) = -6.36 $$
Answer: \(x = 6.36\), \(y = -6.36\)
Absolute coordinates measure positions from a fixed origin \((0,0)\), while relative coordinates measure positions from the last point entered. Mechanical Production Technicians use these formats when programming CNC machines or interpreting CAD drawings.
Example 1: Starting at origin \((0,0)\), move to absolute point \((5, 3)\).
Position is simply \((5, 3)\).
Example 2: From \((5, 3)\), move relative \((2, -1)\).
New position:
$$ x = 5 + 2 = 7 $$
$$ y = 3 + (-1) = 2 $$
Answer: \((7, 2)\)
Example 3: Given points in absolute format \(P(2, 4)\), \(Q(6, 8)\), find relative coordinates of \(Q\) from \(P\).
$$ \Delta x = 6 - 2 = 4 $$
$$ \Delta y = 8 - 4 = 4 $$
Relative coordinates of \(Q\) from \(P\) are \((4, 4)\).
Example 4: Starting at \((0,0)\), move relative \((3, 3)\), then relative \((-1, 4)\).
First position: \((3, 3)\)
Second position:
$$ x = 3 + (-1) = 2 $$
$$ y = 3 + 4 = 7 $$
Answer: \((2, 7)\)
Example 5: Convert relative moves \((2, 5)\), \((-3, 1)\), \((4, -2)\) starting from origin to absolute positions.
Step 1: \((2,5)\)
Step 2:
$$ x = 2 + (-3) = -1 $$
$$ y = 5 + 1 = 6 $$
Step 3:
$$ x = -1 + 4 = 3 $$
$$ y = 6 + (-2) = 4 $$
Absolute points: \((2,5)\), \((-1,6)\), \((3,4)\)
In CAD for mechanical production, user-defined coordinate systems simplify work on complex parts by aligning axes with features rather than the global origin.
To define a user coordinate system:
Example 1: A technician sets a user-defined origin at \((10, 10)\) and aligns \(x\)-axis at \(45^\circ\) from global \(x\)-axis. Find coordinates of a point at distance \(5\) along \(x\)-axis and \(3\) along \(y\)-axis in global system.
Calculate global coordinates:
$$ X = 10 + 5 \cos 45^\circ - 3 \sin 45^\circ $$
$$ = 10 + 5 \times 0.7071 - 3 \times 0.7071 = 10 + 3.5355 - 2.1213 = 11.4142 $$
$$ Y = 10 + 5 \sin 45^\circ + 3 \cos 45^\circ $$
$$ = 10 + 5 \times 0.7071 + 3 \times 0.7071 = 10 + 3.5355 + 2.1213 = 15.6568 $$
Answer: \((11.41, 15.66)\)
Example 2: User coordinate system origin at \((0,0)\), \(x\)-axis rotated \(30^\circ\) from global \(x\)-axis. Find global coordinates of point \((4, 2)\) in user system.
$$ X = 4 \cos 30^\circ - 2 \sin 30^\circ = 4 \times 0.866 - 2 \times 0.5 = 3.464 - 1 = 2.464 $$
$$ Y = 4 \sin 30^\circ + 2 \cos 30^\circ = 4 \times 0.5 + 2 \times 0.866 = 2 + 1.732 = 3.732 $$
Answer: \((2.46, 3.73)\)
Example 3: User coordinate system with origin \((5, 5)\), \(x\)-axis rotated \(90^\circ\) from global \(x\)-axis. Find global coordinates of point \((3, 4)\).
$$ X = 5 + 3 \cos 90^\circ - 4 \sin 90^\circ = 5 + 3 \times 0 - 4 \times 1 = 5 - 4 = 1 $$
$$ Y = 5 + 3 \sin 90^\circ + 4 \cos 90^\circ = 5 + 3 \times 1 + 4 \times 0 = 8 $$
Answer: \((1, 8)\)
Example 4: Origin at \((2, 3)\), \(x\)-axis rotated \(180^\circ\). Find global coordinates of point \((1, 2)\).
$$ X = 2 + 1 \cos 180^\circ - 2 \sin 180^\circ = 2 + 1 \times (-1) - 2 \times 0 = 2 - 1 = 1 $$
$$ Y = 3 + 1 \sin 180^\circ + 2 \cos 180^\circ = 3 + 1 \times 0 + 2 \times (-1) = 3 - 2 = 1 $$
Answer: \((1, 1)\)
Example 5: User system origin \((0, 0)\), \(x\)-axis rotated \(270^\circ\), find global coordinates of \((2, 3)\).
$$ X = 2 \cos 270^\circ - 3 \sin 270^\circ = 2 \times 0 - 3 \times (-1) = 3 $$
$$ Y = 2 \sin 270^\circ + 3 \cos 270^\circ = 2 \times (-1) + 3 \times 0 = -2 $$
Answer: \((3, -2)\)
Convert the Cartesian point \((6, 8)\) to polar coordinates. (5 marks)
Convert polar coordinates \(r=15\), \(\theta=150^\circ\) to Cartesian. (5 marks)
Starting at absolute point \((4, 5)\), move relative \((-2, 3)\). Find new absolute coordinates. (5 marks)
A user-defined coordinate system has origin \((3, 3)\) and \(x\)-axis rotated \(60^\circ\). Find global coordinates of point \((5, 0)\) in user system. (5 marks)
Convert relative moves \((3, 4)\), \((5, -2)\), \((-1, -3)\) from origin to absolute points. (5 marks)
Converting coordinates from linear (Cartesian) form to polar form is essential in mechanical production tasks involving angular positioning, such as setting lathe tool angles or aligning robotic arms. Polar coordinates describe a point's location by its distance from the origin and its angle relative to the positive x-axis, which is often more intuitive for rotational components. The conversion uses the point's \(x\) and \(y\) linear coordinates to find the radius \(r\) and angle \(\theta\).
The governing formulas for conversion are:
$$ r = \sqrt{x^2 + y^2} $$
$$ \theta = \tan^{-1}(\frac{y}{x}) $$
where \(r\) is the distance from the origin to the point, and \(\theta\) is the angle measured in degrees or radians from the positive x-axis.
Example 1: A point on a mechanical drawing has Cartesian coordinates \(x = 3\, \text{cm}\), \(y = 4\, \text{cm}\). Find its polar coordinates.
Given: \(x = 3\, \text{cm}\), \(y = 4\, \text{cm}\)
$$ r = \sqrt{x^2 + y^2} $$
$$ r = \sqrt{3^2 + 4^2} $$
$$ r = \sqrt{9 + 16} $$
$$ r = \sqrt{25} = 5\, \text{cm} $$
$$ \theta = \tan^{-1}(\frac{y}{x}) $$
$$ \theta = \tan^{-1}(\frac{4}{3}) $$
$$ \theta = 53.13^\circ $$
Answer: \(r = 5\, \text{cm}, \theta = 53.13^\circ\)
Example 2: A cutting tool's tip is positioned at \(x = -5\, \text{mm}\), \(y = 5\, \text{mm}\) relative to the machine origin. Find the polar coordinates.
Given: \(x = -5\, \text{mm}\), \(y = 5\, \text{mm}\)
$$ r = \sqrt{(-5)^2 + 5^2} $$
$$ r = \sqrt{25 + 25} $$
$$ r = \sqrt{50} = 7.07\, \text{mm} $$
To find \(\theta\), note that \(x < 0\) and \(y > 0\), so the point is in the second quadrant.
$$ \theta = \tan^{-1}(\frac{5}{-5}) = \tan^{-1}(-1) = -45^\circ $$
Since it's second quadrant,
$$ \theta = 180^\circ - 45^\circ = 135^\circ $$
Answer: \(r = 7.07\, \text{mm}, \theta = 135^\circ\)
Example 3: The end of a robotic arm is at \(x = -6\, \text{cm}\), \(y = -8\, \text{cm}\). Determine the polar coordinates.
Given: \(x = -6\, \text{cm}\), \(y = -8\, \text{cm}\)
$$ r = \sqrt{(-6)^2 + (-8)^2} $$
$$ r = \sqrt{36 + 64} $$
$$ r = \sqrt{100} = 10\, \text{cm} $$
Calculate \(\theta\):
$$ \theta = \tan^{-1}(\frac{-8}{-6}) = \tan^{-1}(\frac{8}{6}) = \tan^{-1}(1.333) $$
$$ \theta = 53.13^\circ $$
Since \(x < 0\) and \(y < 0\), the point is in the third quadrant, so
$$ \theta = 180^\circ + 53.13^\circ = 233.13^\circ $$
Answer: \(r = 10\, \text{cm}, \theta = 233.13^\circ\)
Example 4: A point in a CNC milling program is at \(x = 0\, \text{mm}\), \(y = 7\, \text{mm}\). Find the polar coordinates.
Given: \(x = 0\, \text{mm}\), \(y = 7\, \text{mm}\)
$$ r = \sqrt{0^2 + 7^2} = 7\, \text{mm} $$
Since \(x = 0\), \(\theta\) is either \(90^\circ\) or \(270^\circ\) depending on \(y\).
Here, \(y > 0\), so
$$ \theta = 90^\circ $$
Answer: \(r = 7\, \text{mm}, \theta = 90^\circ\)
Example 5: Find the polar coordinates of the point \(x = 10\, \text{cm}\), \(y = 0\, \text{cm}\).
Given: \(x = 10\, \text{cm}\), \(y = 0\, \text{cm}\)
$$ r = \sqrt{10^2 + 0^2} = 10\, \text{cm} $$
$$ \theta = \tan^{-1}(\frac{0}{10}) = 0^\circ $$
Answer: \(r = 10\, \text{cm}, \theta = 0^\circ\)
Converting polar coordinates to linear (Cartesian) coordinates is critical in mechanical production for machine programming, where distances and angles need to be translated into \(x\) and \(y\) positions. This process allows precise control of tools and components in planar movements. Polar coordinates given as radius \(r\) and angle \(\theta\) can be converted back to Cartesian as follows:
$$ x = r \cos \theta $$
$$ y = r \sin \theta $$
where \(\theta\) is measured in degrees or radians.
Solving polar equations often involves manipulating these formulas to find missing variables or to analyze curves defined in polar form, which is common in cam design or gear tooth profiles.
This subtopic covers the fundamental formulas and steps to convert polar coordinates \((r, \theta)\) into linear coordinates \((x, y)\).
$$ x = r \cos \theta $$
$$ y = r \sin \theta $$
Example 1: A cutting tool position is given by \(r = 10\, \text{cm}\), \(\theta = 30^\circ\). Find the Cartesian coordinates.
Given: \(r = 10\, \text{cm}\), \(\theta = 30^\circ\)
$$ x = r \cos \theta $$
$$ x = 10 \cos 30^\circ $$
$$ x = 10 \times 0.866 = 8.66\, \text{cm} $$
$$ y = r \sin \theta $$
$$ y = 10 \sin 30^\circ $$
$$ y = 10 \times 0.5 = 5\, \text{cm} $$
Answer: \(x = 8.66\, \text{cm}, y = 5\, \text{cm}\)
Example 2: The position of a robotic arm end is \(r = 15\, \text{mm}\), \(\theta = 150^\circ\). Find \(x\) and \(y\).
Given: \(r = 15\, \text{mm}\), \(\theta = 150^\circ\)
$$ x = 15 \cos 150^\circ $$
$$ x = 15 \times (-0.866) = -12.99\, \text{mm} $$
$$ y = 15 \sin 150^\circ $$
$$ y = 15 \times 0.5 = 7.5\, \text{mm} $$
Answer: \(x = -12.99\, \text{mm}, y = 7.5\, \text{mm}\)
Example 3: A point is given in polar coordinates as \(r = 20\, \text{cm}\), \(\theta = 225^\circ\). Determine Cartesian coordinates.
Given: \(r = 20\, \text{cm}\), \(\theta = 225^\circ\)
$$ x = 20 \cos 225^\circ $$
$$ x = 20 \times (-0.707) = -14.14\, \text{cm} $$
$$ y = 20 \sin 225^\circ $$
$$ y = 20 \times (-0.707) = -14.14\, \text{cm} $$
Answer: \(x = -14.14\, \text{cm}, y = -14.14\, \text{cm}\)
Example 4: Convert \(r = 5\, \text{mm}\), \(\theta = 270^\circ\) to Cartesian coordinates.
Given: \(r = 5\, \text{mm}\), \(\theta = 270^\circ\)
$$ x = 5 \cos 270^\circ $$
$$ x = 5 \times 0 = 0\, \text{mm} $$
$$ y = 5 \sin 270^\circ $$
$$ y = 5 \times (-1) = -5\, \text{mm} $$
Answer: \(x = 0\, \text{mm}, y = -5\, \text{mm}\)
Example 5: A tool is positioned at \(r = 8\, \text{cm}\), \(\theta = 0^\circ\). Find the Cartesian coordinates.
Given: \(r = 8\, \text{cm}\), \(\theta = 0^\circ\)
$$ x = 8 \cos 0^\circ $$
$$ x = 8 \times 1 = 8\, \text{cm} $$
$$ y = 8 \sin 0^\circ $$
$$ y = 8 \times 0 = 0\, \text{cm} $$
Answer: \(x = 8\, \text{cm}, y = 0\, \text{cm}\)
Polar equations often describe curves or loci where \(r\) is expressed as a function of \(\theta\). Solving for \(r\) at given \(\theta\) values is key for plotting paths in machining or motion control.
Example 1: Given \(r = 4 + 3 \cos \theta\), find \(r\) when \(\theta = 60^\circ\).
Given: \(r = 4 + 3 \cos \theta\), \(\theta = 60^\circ\)
$$ r = 4 + 3 \cos 60^\circ $$
$$ r = 4 + 3 \times 0.5 = 4 + 1.5 = 5.5 $$
Answer: \(r = 5.5\)
Example 2: For \(r = 6 - 2 \sin \theta\), find \(r\) at \(\theta = 90^\circ\).
Given: \(r = 6 - 2 \sin \theta\), \(\theta = 90^\circ\)
$$ r = 6 - 2 \sin 90^\circ $$
$$ r = 6 - 2 \times 1 = 6 - 2 = 4 $$
Answer: \(r = 4\)
Example 3: Solve for \(r\) when \(r = 10 \cos 2\theta\), at \(\theta = 45^\circ\).
Given: \(r = 10 \cos 2\theta\), \(\theta = 45^\circ\)
Calculate \(2\theta = 90^\circ\)
$$ r = 10 \cos 90^\circ = 10 \times 0 = 0 $$
Answer: \(r = 0\)
Example 4: Find \(r\) for \(r = 5 + 5 \sin \theta\) at \(\theta = 270^\circ\).
Given: \(r = 5 + 5 \sin \theta\), \(\theta = 270^\circ\)
$$ r = 5 + 5 \sin 270^\circ $$
$$ r = 5 + 5 \times (-1) = 5 - 5 = 0 $$
Answer: \(r = 0\)
Example 5: Given \(r = 8 \cos \theta + 6\), find \(r\) at \(\theta = 120^\circ\).
Given: \(r = 8 \cos \theta + 6\), \(\theta = 120^\circ\)
$$ r = 8 \cos 120^\circ + 6 $$
$$ r = 8 \times (-0.5) + 6 = -4 + 6 = 2 $$
Answer: \(r = 2\)
Sometimes, polar equations require solving for \(\theta\) when \(r\) is known, useful in determining tool angles or paths in production processes.
Given \(r = a + b \cos \theta\), rearrange for \(\theta\):
$$ \cos \theta = \frac{r, a}{b} $$
Then,
$$ \theta = \cos^{-1} (\frac{r, a}{b}) $$
Example 1: For \(r = 5 + 3 \cos \theta\), find \(\theta\) when \(r = 8\).
Given: \(r = 5 + 3 \cos \theta\), \(r = 8\)
$$ 3 \cos \theta = 8 - 5 = 3 $$
$$ \cos \theta = \frac{3}{3} = 1 $$
$$ \theta = \cos^{-1} 1 = 0^\circ $$
Answer: \(\theta = 0^\circ\)
Example 2: Solve \(\theta\) for \(r = 4 - 2 \sin \theta\), when \(r = 2\).
Given: \(r = 4 - 2 \sin \theta\), \(r = 2\)
$$ -2 \sin \theta = 2 - 4 = -2 $$
$$ \sin \theta = \frac{2}{2} = 1 $$
$$ \theta = \sin^{-1} 1 = 90^\circ $$
Answer: \(\theta = 90^\circ\)
Example 3: Given \(r = 10 \cos 2\theta\), find \(\theta\) when \(r = 5\).
Given: \(r = 10 \cos 2\theta\), \(r = 5\)
$$ \cos 2\theta = \frac{5}{10} = 0.5 $$
$$ 2\theta = \cos^{-1} 0.5 = 60^\circ $$
$$ \theta = \frac{60^\circ}{2} = 30^\circ $$
Answer: \(\theta = 30^\circ\)
Example 4: For \(r = 6 + 4 \sin \theta\), find \(\theta\) when \(r = 10\).
Given: \(r = 6 + 4 \sin \theta\), \(r = 10\)
$$ 4 \sin \theta = 10 - 6 = 4 $$
$$ \sin \theta = \frac{4}{4} = 1 $$
$$ \theta = \sin^{-1} 1 = 90^\circ $$
Answer: \(\theta = 90^\circ\)
Example 5: Find \(\theta\) for \(r = 8 \cos \theta + 6\), when \(r = 10\).
Given: \(r = 8 \cos \theta + 6\), \(r = 10\)
$$ 8 \cos \theta = 10 - 6 = 4 $$
$$ \cos \theta = \frac{4}{8} = 0.5 $$
$$ \theta = \cos^{-1} 0.5 = 60^\circ $$
Answer: \(\theta = 60^\circ\)
This chapter introduced the Cartesian plane, explaining the x and y axes along with the concept of positive and negative coordinates. It then explored the gradient of a line, covering positive, negative, zero, and infinite gradients as well as the gradients of parallel and perpendicular lines. The significance of the y-intercept in linear equations was discussed before examining the properties of straight lines, parallel lines, and perpendicular lines through their graphs. The chapter also covered polar geometry, focusing on the magnitude and direction of vectors and how these are represented graphically. Finally, methods for converting between linear and polar forms of equations were presented, including techniques for solving polar equations. This comprehensive treatment of coordinate geometry equips students with the skills to analyze and interpret linear and polar relationships in engineering contexts.
Find the coordinates of a point that lies 5 units to the right of the origin and 3 units above the x-axis on the Cartesian plane. (2 marks)
Calculate the gradient of a line passing through points \(A(2, 3)\) and \(B(6, 11)\). (3 marks)
Determine the equation of the straight line with a gradient of 4 passing through the point \((1, 2)\). Express your answer in the form \(y = mx + c\). (4 marks)
Find the gradient of a line perpendicular to a line with gradient \(-\frac{3}{2}\). (2 marks)
Two points on a mechanical component are measured as \(P(4, 7)\) and \(Q(10, 7)\). Calculate the gradient of the line joining these points and explain its significance in terms of the line orientation. (3 marks)
Convert the linear equation \(y = 2x + 5\) into polar form \(r = \frac{c}{\cos(\theta) - m \sin(\theta)}\). (5 marks)
A machine part lies at polar coordinates \(r = 10\) units and \(\theta = 120^\circ\). Find its Cartesian coordinates \((x, y)\). (3 marks)
Find the equation of a line passing through \((3, 4)\) that is parallel to the line \(2x - 3y + 6 = 0\). Express your answer in the form \(y = mx + c\). (5 marks)
Given two lines \(L_1: y = \frac{1}{2}x + 3\) and \(L_2: y = -2x + 1\), verify if these lines are perpendicular by calculating their gradients. (4 marks)
A conveyor belt is represented by the line \(y = -\frac{1}{3}x + 7\). Calculate the point where this line crosses the y-axis and find the polar coordinates \((r, \theta)\) of this point. (6 marks)
Solution:
Coordinates are \((x, y) = (5, 3)\).
Answer: \(\boxed{(5, 3)}\)
Formula:$$m = \frac{y_2 - y_1}{x_2 - x_1}$$
Substitution:$$m = \frac{11 - 3}{6 - 2} = \frac{8}{4}$$
Calculation:$$m = 2$$
Answer: \(\boxed{2}\)
Formula:$$y = mx + c$$ To find \(c\), use:$$y_1 = m x_1 + c \Rightarrow c = y_1 - m x_1$$
Substitution:$$c = 2 - 4 \times 1 = 2 - 4 = -2$$
Equation:$$y = 4x - 2$$
Answer: \(\boxed{y = 4x - 2}\)
Formula:
Gradient of perpendicular line \(m_{\perp} = -\frac{1}{m}\).
Calculation:$$m_{\perp} = -\frac{1}{-\frac{3}{2}} = \frac{2}{3}$$
Answer: \(\boxed{\frac{2}{3}}\)
Formula:$$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{7 - 7}{10 - 4} = \frac{0}{6} = 0$$
Interpretation: Gradient zero means the line is horizontal.
Answer:
Gradient \(\boxed{0}\), line is horizontal.
Formula for polar form:$$r = \frac{c}{\cos \theta, m \sin \theta}$$ where \(m = 2\), \(c = 5\).
Substitution:$$r = \frac{5}{\cos \theta - 2 \sin \theta}$$
Answer:
\(\boxed{r = \frac{5}{\cos \theta - 2 \sin \theta}}\)
Formula:$$x = r \cos \theta$$$$y = r \sin \theta$$
Calculation:
Convert \(\theta\) to radians or use degrees directly:
\(\cos 120^\circ = -0.5\), \(\sin 120^\circ = \frac{\sqrt{3}}{2} \approx 0.866\).
$$x = 10 \times (-0.5) = -5$$$$y = 10 \times 0.866 = 8.66$$
Answer:
\(\boxed{(-5, 8.66)}\)
Step 1: Find gradient of given line.
Rewrite:$$2x - 3y + 6 = 0 \Rightarrow -3y = -2x -6 \Rightarrow y = \frac{2}{3}x + 2$$
Gradient \(m = \frac{2}{3}\).
Step 2: Equation of line parallel to given line through \((3,4)\):
$$y = mx + c$$$$4 = \frac{2}{3} \times 3 + c \Rightarrow 4 = 2 + c \Rightarrow c = 2$$
Answer:
\(\boxed{y = \frac{2}{3}x + 2}\)
Step 1: Calculate product of gradients:$$m_1 = \frac{1}{2}, \quad m_2 = -2$$
$$m_1 \times m_2 = \frac{1}{2} \times (-2) = -1$$
Since product is \(-1\), lines are perpendicular.
Answer:
\(\boxed{\text{Lines are perpendicular}}\)
Step 1: Find y-intercept (where \(x=0\)):
$$y = -\frac{1}{3} \times 0 + 7 = 7$$
Point is \((0,7)\).
Step 2: Convert \((0,7)\) to polar coordinates:
$$r = \sqrt{0^2 + 7^2} = 7$$$$\theta = \arctan \frac{y}{x} = \arctan \frac{7}{0}$$
Since \(x=0\), \(\theta = 90^\circ\).
Answer:
Y-intercept: \(\boxed{(0,7)}\), Polar coordinates: \(\boxed{r=7, \theta=90^\circ}\)
Question 11 (Compulsory - 20 marks)
A mechanical production technician at a Kisumu-based automotive parts manufacturer is tasked with plotting and analyzing tool paths on coordinate planes for a new machining process.
a) Given two points \(P(1, 3)\) and \(Q(5, 11)\), calculate the gradient of the line joining these points and determine the equation of the line in Cartesian form. (10 marks)
b) Convert the equation found in (a) into polar form and explain the significance of magnitude and direction in the context of tool path optimization. (10 marks)
Question 12 (20 marks)
A mechanical technician is designing a fixture that requires two lines to be perpendicular. One line passes through points \((2, 4)\) and \((6, 8)\).
a) Calculate the gradient of this line. (6 marks)
b) Determine the gradient of a line perpendicular to this. (4 marks)
c) Write the equation of the perpendicular line passing through point \((2, 4)\). (10 marks)
Question 13 (20 marks)
At a textile machinery workshop in Eldoret, a technician is analyzing the relationship between spindle speed and torque using linear equations. The line equation is \(y = -3x + 12\).
a) Identify the y-intercept and explain its meaning in this context. (5 marks)
b) Calculate the gradient and explain what a negative gradient implies in the relationship between spindle speed and torque. (5 marks)
c) Sketch the graph of the line and indicate where it crosses the axes. (10 marks)
Question 14 (20 marks)
A production engineer is working with force vectors acting on a mechanical assembly. Two vectors are \(\vec{A} = (4, 3)\) and \(\vec{B} = (-2, 6)\).
a) Calculate the magnitude and direction of vector \(\vec{A}\). (6 marks)
b) Calculate the magnitude and direction of vector \(\vec{B}\). (6 marks)
c) Determine the resultant vector \(\vec{R} = \vec{A} + \vec{B}\), and calculate its magnitude and direction. (8 marks)
Question 11
a) Gradient:$$m = \frac{11 - 3}{5 - 1} = \frac{8}{4} = 2$$
Equation of line:$$y - 3 = 2(x - 1) \Rightarrow y = 2x + 1$$
b) Polar form:
Given \(y = 2x + 1\), rewrite as:$$y - 2x -1 = 0$$
Polar form:$$r = \frac{|c|}{\cos \theta + m \sin \theta} = \frac{1}{\cos \theta - 2 \sin \theta}$$
Magnitude represents shortest distance from origin to line; direction relates to orientation of line, useful for optimizing tool path directions to minimize movements.
Question 12
a) Gradient:$$m = \frac{8 - 4}{6 - 2} = \frac{4}{4} = 1$$
b) Gradient of perpendicular line:$$m' = -1$$
c) Equation of perpendicular line through \((2,4)\):$$y - 4 = -1(x - 2) \Rightarrow y = -x + 6$$
Question 13
a) Y-intercept = 12; means torque at zero spindle speed is 12 units (e.g., Nm).
b) Gradient \(m = -3\); negative gradient implies inverse relationship – as spindle speed increases, torque decreases.
c) Graph crosses y-axis at (0,12), x-axis where \(0 = -3x + 12 \Rightarrow x = 4\). Sketch shows downward sloping line cutting y-axis at 12 and x-axis at 4.
Question 14
a) Magnitude of \(\vec{A}\):$$|\vec{A}| = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$$
Direction:$$\theta = \tan^{-1} \frac{3}{4} = 36.87^\circ$$
b) Magnitude of \(\vec{B}\):$$|\vec{B}| = \sqrt{(-2)^2 + 6^2} = \sqrt{4 + 36} = \sqrt{40} = 6.32$$ Direction:$$\theta = \tan^{-1} \frac{6}{-2} = \tan^{-1}(-3) = 108.43^\circ$$ (measured counterclockwise from positive x-axis)
c) Resultant vector:$$\vec{R} = (4 + (-2), 3 + 6) = (2, 9)$$ Magnitude:$$|\vec{R}| = \sqrt{2^2 + 9^2} = \sqrt{4 + 81} = \sqrt{85} = 9.22$$ Direction:$$\theta = \tan^{-1} \frac{9}{2} = 77.47^\circ$$
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