Coordinate geometry is central to mechatronics technology, enabling precise design, positioning, and movement control of robotic arms, sensors, and automated systems. Mastery of Cartesian geometry allows technicians to model spatial relationships and trajectories in two dimensions, critical for programming CNC machines and robotic manipulators used in Kenyan manufacturing and automation industries. This chapter develops skills to analyze points, lines, and their relationships on the Cartesian plane, focusing on gradients and intercepts that define linear motion paths and sensor alignments.
Cartesian geometry forms the foundation for describing positions and shapes in a plane using coordinates. This system is essential in mechatronics for defining movement paths, sensor locations, and component placements within automated systems. Understanding the Cartesian plane, gradients of lines, and intercepts allows technicians to predict and control mechanical and electrical system behavior precisely.
The Cartesian plane is a two-dimensional coordinate system defined by two perpendicular axes intersecting at the origin. Each point on the plane is represented by an ordered pair \((x, y)\), where \(x\) is the horizontal coordinate and \(y\) is the vertical coordinate. This system is fundamental for mapping positions and trajectories in mechatronic devices such as robotic arms and automated guided vehicles.
The coordinate axes divide the plane into four quadrants, which help in understanding the sign and position of points relative to the origin. The horizontal axis is called the \(x\)-axis, and the vertical axis is the \(y\)-axis. The origin, where the axes intersect, is at \((0, 0)\).
$$\text{Point } P = (x, y)$$
Example 1:
A sensor is placed at position \((3, 4)\) on a robotic arm's working plane. Find the coordinates of the sensor.
Given:
Coordinates of the sensor are directly given as \((x, y) = (3, 4)\).
Answer: (3, 4)
Example 2:
Locate a point on the Cartesian plane where \(x = -5\) and \(y = 2\).
Given:
\(x = -5\), \(y = 2\)
The point is \(P(-5, 2)\).
Answer: (-5, 2)
Example 3:
Find the coordinates of the midpoint between points \(A(2, 3)\) and \(B(8, 7)\).
Given:
\(A = (2, 3)\), \(B = (8, 7)\)
Formula for midpoint \(M\):
$$M = (\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})$$
Substitute:
$$M = (\frac{2 + 8}{2}, \frac{3 + 7}{2})$$
Calculate:
$$M = (\frac{10}{2}, \frac{10}{2})$$
$$M = (5, 5)$$
Answer: (5, 5)
Example 4:
A robotic platform moves to point \(C(6, -3)\). What quadrant does this point lie in?
Given:
\(x = 6 > 0\), \(y = -3 < 0\)
Quadrants are numbered as:
Since \(x > 0\) and \(y < 0\), point \(C\) is in Quadrant IV.
Answer: Quadrant IV
Example 5:
Calculate the distance from the origin to point \(D(-4, -3)\).
Given:
\(D = (-4, -3)\)
Distance formula:
$$d = \sqrt{(x - 0)^2 + (y - 0)^2}$$
Substitute:
$$d = \sqrt{(-4)^2 + (-3)^2}$$
Calculate:
$$d = \sqrt{16 + 9}$$
$$d = \sqrt{25}$$
$$d = 5$$
Answer: 5 units
The gradient (or slope) of a line in the Cartesian plane measures its steepness and direction. It is defined as the ratio of the vertical change (rise) to the horizontal change (run) between two points on the line. The gradient is vital in mechatronics for determining the angle and speed of linear actuator movement or sensor alignment.
The formula for the gradient \(m\) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is:
$$m = \frac{y_2 - y_1}{x_2 - x_1}$$
Example 1:
Find the gradient of the line passing through points \(A(1, 2)\) and \(B(4, 6)\).
Given:
\(A = (1, 2)\), \(B = (4, 6)\)
Formula:
$$m = \frac{y_2 - y_1}{x_2 - x_1}$$
Substitution:
$$m = \frac{6 - 2}{4 - 1}$$
Calculation:
$$m = \frac{4}{3}$$
Answer: \(\frac{4}{3}\)
Example 2:
Calculate the gradient of a line between points \(C(-2, 5)\) and \(D(3, -1)\).
Given:
\(C = (-2, 5)\), \(D = (3, -1)\)
Formula:
$$m = \frac{-1 - 5}{3 - (-2)}$$
Substitution:
$$m = \frac{-6}{5}$$
Answer: \(-\frac{6}{5}\)
Example 3:
A sensor moves along a line with gradient 0.5. If it passes through point \(E(2, 3)\), find the coordinates of a point \(F\) on the line where \(x = 6\).
Given:
Gradient \(m = 0.5\), point \(E(2, 3)\), \(x_F = 6\)
Formula for line:
$$y, y_1 = m(x, x_1)$$
Substitute:
$$y - 3 = 0.5(x - 2)$$
Calculate \(y\) when \(x=6\):
$$y - 3 = 0.5(6 - 2)$$
$$y - 3 = 0.5 \times 4$$
$$y - 3 = 2$$
$$y = 5$$
Answer: Point \(F(6, 5)\)
Example 4:
Determine the gradient of a vertical line passing through point \(G(4, 1)\).
Given:
Vertical line means \(x = 4\) for all \(y\).
Gradient formula:
$$m = \frac{y_2 - y_1}{x_2 - x_1}$$
Since \(x_2 - x_1 = 0\), gradient is undefined.
Answer: Gradient is undefined (vertical line)
Example 5:
Find the gradient of a horizontal line through point \(H(0, 7)\).
Given:
Horizontal line means \(y = 7\) for all \(x\).
Gradient formula:
$$m = \frac{y_2 - y_1}{x_2 - x_1}$$
Since \(y_2 - y_1 = 0\), gradient is zero.
Answer: Gradient = 0
The \(y\)-intercept of a line is the point where the line crosses the \(y\)-axis. It is an important parameter in linear equations describing motion or sensor response in mechatronic systems. The \(y\)-intercept indicates the initial position or output value when the input (represented by \(x\)) is zero.
The equation of a line in slope-intercept form is:
$$y = mx + c$$
where \(m\) is the gradient and \(c\) is the \(y\)-intercept.
Example 1:
Find the \(y\)-intercept of the line \(y = 3x + 4\).
Given:
Equation \(y = 3x + 4\)
The \(y\)-intercept is the constant term \(c = 4\).
Answer: \(y\)-intercept = 4
Example 2:
Determine the equation of the line with gradient \(m = 2\) and \(y\)-intercept \(c = -3\).
Given:
\(m = 2\), \(c = -3\)
Equation:
$$y = 2x - 3$$
Answer: \(y = 2x - 3\)
Example 3:
Find the \(y\)-intercept of the line passing through points \(A(1, 5)\) and \(B(3, 9)\).
Given:
Points \(A(1, 5)\), \(B(3, 9)\)
Step 1: Calculate gradient \(m\):
$$m = \frac{9 - 5}{3 - 1} = \frac{4}{2} = 2$$
Step 2: Use point-slope form to find \(c\):
$$y = mx + c$$ Substitute \(x=1\), \(y=5\), \(m=2\):
$$5 = 2(1) + c$$
$$5 = 2 + c$$
$$c = 3$$
Answer: \(y\)-intercept = 3
Example 4:
A robotic sensor’s output is modelled by the line \(y = -0.5x + 7\). What is the sensor output when \(x = 0\)?
Given:
Equation \(y = -0.5x + 7\), \(x=0\)
Calculate \(y\):
$$y = -0.5(0) + 7 = 7$$
Answer: Sensor output at \(x=0\) is 7
Example 5:
Find the equation of a line with gradient \(-1\) passing through the point \((4, 2)\).
Given:
\(m = -1\), point \((4, 2)\)
Step 1: Use point-slope form:
$$y, y_1 = m(x, x_1)$$
$$y - 2 = -1(x - 4)$$
Step 2: Expand:
$$y - 2 = -x + 4$$
Step 3: Solve for \(y\):
$$y = -x + 6$$
The \(y\)-intercept \(c = 6\).
Answer: \(y = -x + 6\), \(y\)-intercept = 6
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Create a free accountThis chapter introduced the fundamentals of coordinate geometry, beginning with the Cartesian plane defined by the x and y axes and the concept of positive and negative coordinates. The gradient of a line was explored in detail, including positive, negative, zero, and infinite gradients, along with the relationships between gradients of parallel and perpendicular lines. The importance of the y-intercept in defining linear equations was then examined. Linear equations were studied further with a focus on straight lines and the conditions that determine parallelism and perpendicularity. Graphs of linear equations were illustrated through straight lines on the Cartesian plane. The chapter then shifted to polar geometry, explaining the concepts of magnitude and direction and how to graph points in polar form. Finally, methods for converting between Cartesian and polar coordinates were presented, along with techniques for solving equations expressed in polar form.
Find the gradient of a line passing through the points \(A(2, 3)\) and \(B(6, 11)\). (2 marks)
Determine the equation of the line with gradient \(\frac{1}{2}\) and passing through the point \(P(4, 5)\). (3 marks)