Parallel lines have equal gradients, and perpendicular lines have gradients whose product is −1. To write the equation, find the gradient you need, then use the point the line passes through.
This lesson is part of coordinate geometry. It uses gradients from using gradient, midpoint and distance together.
How do you write the equation?
Use y − y₁ = m(x − x₁) with the gradient m and a point (x₁, y₁) on the line. Then tidy it to the form the question asks for.
Worked example: one line, three answers
The line L is 2x + 3y = 12. Rearranged, y = −2x/3 + 4, so its gradient is −2/3.
Parallel through A(2, 5). The gradient is −2/3. y − 5 = −2/3(x − 2), so 3y − 15 = −2x + 4, which gives 2x + 3y = 19. Check: 2(2) + 3(5) = 19.
Perpendicular through B(−1, 4). The gradient is 3/2. y − 4 = 3/2(x + 1), so 2y − 8 = 3x + 3, which gives 3x − 2y + 11 = 0. Check: 3(−1) − 2(4) + 11 = 0.
Perpendicular bisector of P(1, 2) and Q(7, 6). Midpoint = (4, 4). Gradient of PQ = 4/6 = 2/3, so the perpendicular gradient is −3/2.
y − 4 = −3/2(x − 4), so 2y − 8 = −3x + 12, which gives 3x + 2y = 20. Check: 3(4) + 2(4) = 20.
The mistake that costs marks
A common slip is to change only the sign, or only flip the fraction. Take a gradient of −2/3.
| Attempt | Gradient | Product with −2/3 |
|---|---|---|
| Negative only | 2/3 | −4/9 |
| Reciprocal only | −3/2 | 1 |
| Negative reciprocal | 3/2 | −1 |
Only the negative reciprocal gives a product of −1. The check takes five seconds: multiply the two gradients before you use the result.
Check yourself
Find the equation of the line through (0, 3) that is perpendicular to x − 2y = 6.
Answer
Rearrange: −2y = −x + 6, so y = x/2 − 3. The gradient is 1/2.
The perpendicular gradient is −2. The line passes through (0, 3), so the intercept is 3: y = −2x + 3, or 2x + y = 3.
Check: the gradients 1/2 and −2 multiply to −1, and (0, 3) satisfies 2(0) + 3 = 3.
What to study next
Go on to calculating areas from coordinates, then the coordinate geometry practice set.
The word-problem structure worksheet helps you list the points and conditions. A teacher can work through longer questions with you in online one-to-one Additional Mathematics tuition.