By the end of this chapter, you will be able to:
Mastering these skills will help you confidently handle complex shapes and designs in real engineering tasks, making your work precise and reliable.
Coordinate geometry is fundamental in welding engineering as it enables precise positioning and measurement of components for fabrications, joints, and assemblies. Understanding Cartesian geometry and linear equations helps welding professionals accurately interpret drawings, set out workpieces, and verify alignments. This chapter develops skills to solve coordinate-based problems relevant to welding tasks encountered in Kenyan workshops and industrial sites.
Cartesian geometry provides a framework to locate points in a plane using two perpendicular axes, labelled x and y. This system is essential for welding professionals to determine exact positions of welds, holes and cut lines on metal plates or structures. By mastering coordinate calculations, welders can improve accuracy in layout and reduce material wastage.
The Cartesian plane is a flat surface on which every point is fixed by two numbers measured from a common reference point. It works like the grid a fabricator marks on a steel plate before drilling: each hole is described by how far it is across and how far it is up from one datum corner. Understanding the plane, its axes and the signs of its coordinates is the foundation for gradients, intercepts and all later work on lines.
The Cartesian plane is built from two perpendicular number lines. The horizontal line is the x-axis and the vertical line is the y-axis. They cross at the origin, written (0, 0), which acts as the reference point from which every other position is measured. On a fabrication drawing the origin is usually a datum corner or a marked centre line of the plate, so that all hole and weld positions can be read from one fixed point.
Each point is described by an ordered pair (x, y). The first number, x, is the horizontal distance from the origin along the x-axis. The second number, y, is the vertical distance from the origin along the y-axis. The order matters: (5, 3) and (3, 5) are two different points.
Key features of the axes:
Example 1: Locate a point on a metal plate where the x-coordinate is 5 cm and the y-coordinate is 3 cm.
Given: x = 5 cm, y = 3 cm
Move 5 cm along the x-axis from the origin, then 3 cm parallel to the y-axis. The point is (5, 3).
Answer: The point is 5 cm to the right of and 3 cm above the origin.
Procedure for plotting a point:
The sign of a coordinate shows the direction of the point from the origin. A positive x value lies to the right of the y-axis and a negative x value lies to the left. A positive y value lies above the x-axis and a negative y value lies below it. A coordinate of zero means the point lies on that axis.
The two axes divide the plane into four quadrants, numbered I to IV in an anticlockwise direction starting from the top right.
| Quadrant | Sign of x | Sign of y | Example point |
|---|---|---|---|
| I | Positive | Positive | (4, 2) |
| II | Negative | Positive | (-4, 2) |
| III | Negative | Negative | (-4, -2) |
| IV | Positive | Negative | (4, -2) |
A point with x = 0, such as (0, 6), lies on the y-axis. A point with y = 0, such as (-3, 0), lies on the x-axis. Such points do not belong to any quadrant.
Example 1: Find the quadrant of the point (-4, 7).
Given: x = -4, y = 7
Since x < 0 and y > 0, the point lies in quadrant II.
Answer: Quadrant II
Example 2: Determine the coordinates of a point located 6 cm left and 2 cm below the origin.
Given: left implies negative x, below implies negative y
The coordinates are (-6, -2).
Answer: (-6, -2)
Example 3: On a welding layout, a hole is marked at (8, -5). Find its position relative to the origin.
Given: x = 8, y = -5
The positive x value means right and the negative y value means below.
Answer: 8 cm right of and 5 cm below the origin, in quadrant IV
Example 4: If a weld point is at (-3, -4), which quadrant is it in?
Given: x = -3, y = -4
Both coordinates are negative, so the point lies in quadrant III.
Answer: Quadrant III
Example 5: A drilling mark is at (0, -7). State where it lies.
Given: x = 0, y = -7
The x value is zero, so the point is on the y-axis, 7 cm below the origin.
Answer: On the y-axis, 7 cm below the origin
The distance between two points (x₁, y₁) and (x₂, y₂) in the Cartesian plane is found by applying the distance formula, which comes from the Pythagorean theorem:
d = √((x₂ - x₁)² + (y₂ - y₁)²)
This calculation helps welding engineers determine lengths between holes, edges or weld points on components. Squaring removes the effect of negative signs, so the order in which the points are taken does not change the result.
Example 1: Calculate the distance between points (2, 3) and (5, 7) on a metal sheet.
Given: (x₁, y₁) = (2, 3), (x₂, y₂) = (5, 7)
d = √((x₂ - x₁)² + (y₂ - y₁)²)
d = √((5 - 2)² + (7 - 3)²)
d = √(3² + 4²) = √(9 + 16) = √25
d = 5 cm
Answer: 5 cm
Example 2: Find the distance between (-4, 1) and (3, -2).
Given: (x₁, y₁) = (-4, 1), (x₂, y₂) = (3, -2)
d = √((x₂ - x₁)² + (y₂ - y₁)²)
d = √((3 - (-4))² + (-2 - 1)²)
d = √(7² + (-3)²) = √(49 + 9) = √58
d ≈ 7.62 cm
Answer: 7.62 cm
Example 3: On a welded frame, find the length between points (0, 0) and (6, 8).
Given: (x₁, y₁) = (0, 0), (x₂, y₂) = (6, 8)
d = √((x₂ - x₁)² + (y₂ - y₁)²)
d = √((6 - 0)² + (8 - 0)²)
d = √(36 + 64) = √100
d = 10 cm
Answer: 10 cm
Example 4: Determine the distance between points (-5, -3) and (-1, 4) on a weld layout.
Given: (x₁, y₁) = (-5, -3), (x₂, y₂) = (-1, 4)
d = √((x₂ - x₁)² + (y₂ - y₁)²)
d = √((-1 - (-5))² + (4 - (-3))²)
d = √(4² + 7²) = √(16 + 49) = √65
d ≈ 8.06 cm
Answer: 8.06 cm
Example 5: Calculate the distance between points (2, -1) and (7, 3) on a metal profile.
Given: (x₁, y₁) = (2, -1), (x₂, y₂) = (7, 3)
d = √((x₂ - x₁)² + (y₂ - y₁)²)
d = √((7 - 2)² + (3 - (-1))²)
d = √(5² + 4²) = √(25 + 16) = √41
d ≈ 6.40 cm
Answer: 6.40 cm
The midpoint of a line segment is the point exactly halfway between its two end points. It is useful for locating weld centres or symmetry points in fabrication. The midpoint M of the segment joining (x₁, y₁) and (x₂, y₂) is found by averaging the x values and the y values:
M = ((x₁ + x₂) ÷ 2, (y₁ + y₂) ÷ 2)
Example 1: Find the midpoint between points (4, 6) and (8, 10).
Given: (x₁, y₁) = (4, 6), (x₂, y₂) = (8, 10)
Mx = (4 + 8) ÷ 2 = 12 ÷ 2 = 6
My = (6 + 10) ÷ 2 = 16 ÷ 2 = 8
Answer: The midpoint is (6, 8).
Example 2: Determine the midpoint of the segment between (-2, 3) and (4, -1).
Given: (x₁, y₁) = (-2, 3), (x₂, y₂) = (4, -1)
Mx = (-2 + 4) ÷ 2 = 2 ÷ 2 = 1
My = (3 + (-1)) ÷ 2 = 2 ÷ 2 = 1
Answer: The midpoint is (1, 1).
Example 3: On a steel plate, find the midpoint between points (0, 0) and (5, 12).
Given: (x₁, y₁) = (0, 0), (x₂, y₂) = (5, 12)
Mx = (0 + 5) ÷ 2 = 2.5
My = (0 + 12) ÷ 2 = 6
Answer: The midpoint is (2.5, 6).
Example 4: Calculate the midpoint between points (-6, -4) and (-2, 2).
Given: (x₁, y₁) = (-6, -4), (x₂, y₂) = (-2, 2)
Mx = (-6 + (-2)) ÷ 2 = -8 ÷ 2 = -4
My = (-4 + 2) ÷ 2 = -2 ÷ 2 = -1
Answer: The midpoint is (-4, -1).
Example 5: Find the midpoint of weld points at (3, -5) and (9, 1).
Given: (x₁, y₁) = (3, -5), (x₂, y₂) = (9, 1)
Mx = (3 + 9) ÷ 2 = 12 ÷ 2 = 6
My = (-5 + 1) ÷ 2 = -4 ÷ 2 = -2
Answer: The midpoint is (6, -2).
The gradient of a line measures how steep it is and in which direction it slopes. It is the vertical change (rise) divided by the horizontal change (run) between any two points on the line. In a welding workshop the gradient describes the slope of an inclined brace, a loading ramp or a roof truss rafter, so a fabricator can check that a member is cut and set at the intended angle. The symbol for gradient is m, and it is a plain number with no units because it is a ratio of two lengths.
m = rise ÷ run = (y₂ - y₁) ÷ (x₂ - x₁)
Procedure for finding a gradient from two points:
The two subtractions must follow the same order. Reversing both gives the same answer, but reversing only one changes the sign wrongly.
A line has a positive gradient when it rises as you move from left to right. Both the rise and the run have the same sign, so m > 0. The larger the value of m, the steeper the line. A brace welded from the bottom left of a frame up to the top right has a positive gradient.
Example: A brace runs from (1, 2) to (5, 10) on a fabrication drawing, with units in cm. Find its gradient.
Given: (x₁, y₁) = (1, 2), (x₂, y₂) = (5, 10)
m = (y₂ - y₁) ÷ (x₂ - x₁)
m = (10 - 2) ÷ (5 - 1)
m = 8 ÷ 4
m = 2
Answer: The gradient is 2, a positive gradient. The brace rises 2 cm for every 1 cm along.
A line has a negative gradient when it falls as you move from left to right. The rise and the run have opposite signs, so m < 0. A support strut that starts high on the left and slopes down to the right has a negative gradient. The size of m, ignoring the sign, still tells how steep the line is.
Example: A strut runs from (0, 9) to (6, 3), with units in cm. Find its gradient.
Given: (x₁, y₁) = (0, 9), (x₂, y₂) = (6, 3)
m = (y₂ - y₁) ÷ (x₂ - x₁)
m = (3 - 9) ÷ (6 - 0)
m = -6 ÷ 6
m = -1
Answer: The gradient is -1, a negative gradient. The strut falls 1 cm for every 1 cm along.
A horizontal line has a gradient of zero. All its points have the same y value, so the rise is 0 and m = 0 ÷ run = 0. A weld bead run along the level top edge of a plate, or a horizontal cross-bar, has zero gradient. Its equation has the form y = constant.
Example: A cross-bar runs from (2, 4) to (9, 4), with units in cm. Find its gradient.
Given: (x₁, y₁) = (2, 4), (x₂, y₂) = (9, 4)
m = (y₂ - y₁) ÷ (x₂ - x₁)
m = (4 - 4) ÷ (9 - 2)
m = 0 ÷ 7
m = 0
Answer: The gradient is 0, so the bar is horizontal and lies on the line y = 4.
A vertical line has an infinite gradient, also described as undefined. All its points have the same x value, so the run is 0 and the division rise ÷ 0 cannot be completed. A vertical post or column on a welded frame is an example. Its equation has the form x = constant.
Example: A post runs from (3, 1) to (3, 8), with units in cm. Find its gradient.
Given: (x₁, y₁) = (3, 1), (x₂, y₂) = (3, 8)
m = (y₂ - y₁) ÷ (x₂ - x₁)
m = (8 - 1) ÷ (3 - 3)
m = 7 ÷ 0
Division by zero is not possible.
Answer: The gradient is infinite (undefined), so the post is vertical and lies on the line x = 3.
Parallel lines never meet because they slope in exactly the same way. Two lines are parallel if and only if their gradients are equal: m₁ = m₂. This is how a fabricator checks that two stiffeners or rails on a drawing are parallel without measuring the angle. Two vertical lines are also parallel, even though their gradients are infinite.
Example: Stiffener A passes through (0, 1) and (4, 9). Stiffener B passes through (2, 0) and (5, 6). Are they parallel?
Given: A: (0, 1), (4, 9); B: (2, 0), (5, 6)
mA = (9 - 1) ÷ (4 - 0) = 8 ÷ 4 = 2
mB = (6 - 0) ÷ (5 - 2) = 6 ÷ 3 = 2
Since mA = mB, the lines are parallel.
Answer: The stiffeners are parallel, each with gradient 2.
Two lines are perpendicular when they meet at a right angle. Their gradients multiply to give -1: m₁ × m₂ = -1. This means one gradient is the negative reciprocal of the other: m₂ = -1 ÷ m₁. A horizontal line (m = 0) and a vertical line (infinite gradient) are also perpendicular, but they are a special case because the product rule cannot be applied to an infinite gradient. A welder uses this idea to check that a gusset or brace meets a member at a right angle.
Example 1: Line P passes through (0, 0) and (4, 3). Line Q passes through (0, 0) and (3, -4). Are they perpendicular?
Given: P: (0, 0), (4, 3); Q: (0, 0), (3, -4)
mP = (3 - 0) ÷ (4 - 0) = 3 ÷ 4
mQ = (-4 - 0) ÷ (3 - 0) = -4 ÷ 3
mP × mQ = (3 ÷ 4) × (-4 ÷ 3) = -1
Answer: The product is -1, so the lines are perpendicular.
Example 2: A line has gradient 2. Find the gradient of a line perpendicular to it.
Given: m₁ = 2
m₂ = -1 ÷ m₁
m₂ = -1 ÷ 2
m₂ = -0.5
Answer: The perpendicular gradient is -0.5. A line with gradient 0.5 would not be perpendicular, because 2 × 0.5 = 1, not -1.
The y-intercept is the point where a line crosses the y-axis. At that point x = 0, so the intercept is written (0, c), and the number c is often simply called the y-intercept. On a drawing it shows where a line starts on the vertical datum, for example the height at which an inclined brace meets the left edge of a plate. When the gradient m and the intercept c are known, the line can be written as y = mx + c.
Methods for finding the y-intercept:
Procedure for finding the y-intercept from two points:
Example 1: A brace has gradient 2 and passes through (2, 7), with units in cm. Find its y-intercept.
Given: m = 2, (x, y) = (2, 7)
c = y - mx
c = 7 - 2 × 2
c = 7 - 4
c = 3
Answer: The y-intercept is 3, so the brace meets the y-axis at (0, 3).
Example 2: A strut passes through (1, 5) and (3, 1), with units in cm. Find its y-intercept.
Given: (x₁, y₁) = (1, 5), (x₂, y₂) = (3, 1)
m = (1 - 5) ÷ (3 - 1) = -4 ÷ 2 = -2
c = y₁ - m × x₁
c = 5 - (-2) × 1
c = 5 + 2
c = 7
Check with the second point: y = -2 × 3 + 7 = 1, which matches.
Answer: The y-intercept is 7, so the strut meets the y-axis at (0, 7).
Example 3: Find the y-intercept of the horizontal bar through (2, 4) and (9, 4).
Given: the bar is horizontal, so its equation is y = 4
At x = 0, y = 4
Answer: The y-intercept is 4, at the point (0, 4).
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Create a free accountThis chapter covered the fundamentals of coordinate geometry beginning with the Cartesian plane, where the x and y axes define the coordinate system and points are identified using positive and negative coordinates. The concept of gradient was explored in detail, including positive, negative, zero, and infinite gradients, as well as the relationships between gradients of parallel and perpendicular lines. The y-intercept was also introduced as a key feature of linear equations. Linear equations were examined through the properties of straight lines, including how to identify parallel and perpendicular lines algebraically. Graphs of linear equations were presented with a focus on plotting straight lines accurately. The chapter then introduced polar geometry, explaining the magnitude and direction that define points in the polar coordinate system and how to graph these points. Conversions between linear and polar coordinates were demonstrated to facilitate switching between the two systems. Finally, methods for solving polar equations were discussed, completing the comprehensive treatment of coordinate geometry in engineering mathematics.
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 \(A(2, 4)\) and \(B(6, 10)\). (3 marks)
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