Sheet № 145 · Higher only · AQA · Edexcel · OCR
Parallel and Perpendicular Lines –
Understanding the gradient relationships for parallel and perpendicular lines is a key Higher topic. If two lines have the same gradient they are parallel; if the product of their gradients is −1 they are perpendicular. These rules let you find equations of lines in many exam questions. This page covers the theory, the method for finding
§Key definitions
Question:
Line A has equation y = 3x + 2. Line B has equation y = 3x − 5. Are they parallel?
Answer:
Yes, they are parallel (both have gradient 3).
Q1 (Foundation):
Are the lines y = 4x + 1 and y = 4x − 7 parallel, perpendicular, or neither?
Q2 (Higher):
Find the equation of the line parallel to y = −2x + 3 passing through (5, 1).
Q3 (Higher):
Find the equation of the line perpendicular to y = (3/4)x − 2 passing through (6, 1).
§Formulas to memorise
Parallel lines have equal gradients: m₁ = m₂
Perpendicular lines satisfy m₁ × m₂ = −1, so m₂ = −1/m₁
Equation of a straight line: y − y₁ = m(x − x₁)
Find the gradient of the given line. — If the equation is y = mx + c, the gradient is m. If it is in the form ax + by = c, rearrange to y = mx + c.
For a parallel line — , use the same gradient m.
For a perpendicular line — , use the negative reciprocal: −1/m.
Substitute the gradient and the given point — into y − y₁ = m(x − x₁).
Simplify — to get y = mx + c.
Worked example
Line A has equation y = 3x + 2. Line B has equation y = 3x − 5. Are they parallel?
Working:
⚠ Common mistakes
- ✗Using m instead of −1/m for perpendicular lines. The perpendicular gradient is the negative reciprocal, not just the negative. For example, if m = 2, the perpendicular gradient is −1/2, not −2.
- ✗Forgetting to flip the fraction. If the gradient is 3/4, the perpendicular gradient is −4/3, not −3/4.
- ✗Not rearranging to find the gradient. If the line is given as 2x + 3y = 12, you must rearrange to y = −2x/3 + 4 to see the gradient is −2/3.
✦ Exam tips
- →State clearly whether lines are parallel or perpendicular and justify with the gradient rule.
- →If you are finding the equation of a line, the point-gradient form y − y₁ = m(x − x₁) is the most efficient starting point.
- →Questions often combine this topic with midpoints — find the midpoint first, then use it as the point for the perpendicular bisector.
- →Double-check by substituting the given point into your final equation.