Equations of Straight Lines
This topic covers the essential algebra of straight lines. It's a foundational skill for understanding how to represent linear relationships graphically and algebraically, which is crucial for interpreting data and solving geometric problems in engineering and science.
Part of the ESAT Mathematics 1 syllabus — revision for the Engineering and Science Admissions Test (ESAT), the UAT-UK admissions test for Cambridge, Imperial, Oxford and UCL.
Key points
- Any straight line can be written as y = mx + c, where 'm' is the gradient and 'c' is the y-intercept.
- Parallel lines never meet and have identical gradients. If line 1 has gradient m1 and line 2 has gradient m2, then m1 = m2.
- Perpendicular lines intersect at a 90-degree angle. The product of their gradients is -1. This means one gradient is the negative reciprocal of the other.m1 × m2 = -1m2 = -1/m1
- To find a line's equation, you need one of two things: a single point on the line and its gradient, or two different points on the line.
- Always rearrange equations given in other forms into the y = mx + c form to correctly identify the gradient and intercept.e.g., ax + by = d
Diagram
Formulae
y = mx + c To represent any straight line and quickly identify its gradient (m) and y-intercept (c).
m = (y2 - y1) / (x2 - x1) To calculate the gradient of a line when you know the coordinates of two points (x1, y1) and (x2, y2) on it.
mperpendicular = -1 / moriginal To find the gradient of a line that is perpendicular to a line with a known gradient.
Definitions
- Gradient (m)
- A measure of a line's steepness and direction. It is calculated as the change in the y-coordinate divided by the change in the x-coordinate ('rise over run').
- y-intercept (c)
- The point where the line crosses the vertical y-axis. Its coordinates are always (0, c).
Worked example
Line A is given by the equation 2y + x = 8. Line B is perpendicular to Line A and passes through the point (-3, 5). What is the y-intercept of Line B?
- 1
First, find the gradient of Line A by rearranging its equation into the form y = mx + c.
- 2 2y + x = 8 → 2y = -x + 8 → y = (-1/2)x + 4
- 3
The gradient of Line A (mA) is -1/2.
- 4
Line B is perpendicular to Line A, so its gradient (mB) is the negative reciprocal of mA.
- 5 mB = -1 / (-1/2) = 2
- 6
Now we know Line B has the equation y = 2x + c.
Use the given point (-3, 5) to find c.
- 7
Substitute x = -3 and y = 5 into the equation:
5 = 2*(-3) + c - 8 5 = -6 + c
- 9
Solve for c:
c = 5 + 6 = 11 - 10
The y-intercept of Line B is 11.
Answer: 11
Common mistakes
- ×Sign errors are extremely common, especially when calculating the gradient from two points with negative coordinates. Always write substitutions in brackets, e.g., (5 - (-1)) / (-3 - 2).
- ×Incorrectly finding the perpendicular gradient. Remember it's the *negative reciprocal* (flip the fraction and change the sign), not just the negative or just the reciprocal.
- ×Reading the gradient directly from an equation that isn't in y = mx + c form. For 3y - 6x = 2, the gradient is NOT -6; you must rearrange to y = 2x + 2/3, so the gradient is 2.
No-calculator tips
- ✓Sketching a quick, unscaled graph of the points or lines can provide a visual check. If a line slopes down from left to right, its gradient must be negative.
- ✓Work with fractions throughout. It is far easier to find the negative reciprocal of a fraction like 3/7 (which is -7/3) than to work with its decimal equivalent.
- ✓When substituting a point (p, q) into y = mx + c to find c, rearrange to c = y - mx before substituting. This can help isolate the calculation and reduce arithmetic slips.