Equations of Straight Lines
The algebra of straight lines is a foundational skill for representing 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.