Graphs
Grade 5-8Gradient, Parallel & Perpendicular Lines
Gradient is the single idea that ties together parallel and perpendicular lines: parallel lines share the same gradient, and perpendicular lines have gradients that multiply together to give -1. This lesson covers finding the gradient between two coordinates, identifying parallel lines, and finding the equation of a line parallel or perpendicular to a given line through a point.
Written by Asad, Co-Founder of Teachably
Video walkthrough coming soon
The written lesson below covers everything you need in the meantime.
Finding the gradient between two points
The gradient between two points is the change in y divided by the change in x, calculated as (y₂ - y₁) ÷ (x₂ - x₁). Keep the points in the same order in the numerator and denominator to avoid a sign error.
The gradient of the line joining A(1, 2) and B(5, 10): (10 - 2) ÷ (5 - 1) = 8 ÷ 4 = 2.
Identifying parallel lines
Parallel lines have exactly the same gradient, regardless of their y-intercept. Compare the number in front of x in each equation to check.
y = 3x + 2 and y = 3x - 5 both have a gradient of 3, so the lines are parallel, even though their y-intercepts are different.
Finding the equation of a parallel line
A line parallel to a given line has the same gradient. Use that gradient with the given point, substitute both into y = mx + c, and solve for c to find the full equation.
The line parallel to y = 2x + 3 through the point (1, 9): the gradient is 2, so 9 = 2(1) + c, giving c = 7, and the equation y = 2x + 7.
Perpendicular gradients and finding a perpendicular line
Perpendicular lines meet at a right angle, and their gradients are negative reciprocals of each other, meaning you flip the fraction and change the sign. Once the perpendicular gradient is found, use the same method as for parallel lines to find the full equation.
A line perpendicular to y = 2x + 1 through the point (4, 3): the perpendicular gradient is -1/2, so 3 = -1/2(4) + c, giving c = 5, and the equation y = -1/2x + 5.
Worked Examples
Three exam-style questions, fully solved.
Find the gradient of the line joining A(1, 2) and B(5, 10).
Easy- 1.Find the change in y: 10 - 2 = 8
- 2.Find the change in x: 5 - 1 = 4
- 3.Divide the change in y by the change in x: 8 ÷ 4
Answer: 2
Find the equation of the line parallel to y = 2x + 3 that passes through the point (1, 9).
Medium- 1.A parallel line has the same gradient: 2
- 2.Substitute the point (1, 9) into y = 2x + c: 9 = 2(1) + c
- 3.Solve for c: c = 7
Answer: y = 2x + 7
Find the equation of the line perpendicular to y = 2x + 1 that passes through the point (4, 3).
Hard- 1.Find the perpendicular gradient by flipping and negating 2: -1/2
- 2.Substitute the point (4, 3) into y = -1/2x + c: 3 = -1/2(4) + c
- 3.Solve for c: 3 = -2 + c, so c = 5
Answer: y = -1/2x + 5
Avoid These
The most common mistakes students make.
Subtracting the coordinates in the wrong order when finding the gradient, which flips the sign of the answer.
Assuming two lines are parallel because their equations look similar, instead of checking that the gradients are exactly equal.
Using the negative of the original gradient for a perpendicular line, instead of the negative reciprocal (flipping the fraction as well as changing the sign).
Substituting the given point into the original line's equation instead of the new parallel or perpendicular line's equation when finding c.
Not fully flipping a fractional gradient when finding a perpendicular gradient, for example turning 3/5 into -3/5 instead of -5/3.
FAQ
Questions parents and students ask.
Before this topic, make sure you know
Want a plan built around your child specifically?
Get our free 8-video course, or book a free Roadmap Call for a personalised plan.