What are straight line graphs?
A straight line graph shows a linear relationship between two variables. At GCSE, the equation of a straight line is written in the form:
y = mx + c
where m is the gradient (steepness) and c is the y-intercept (where the line crosses the y-axis).
This topic appears on every GCSE paper. You need to be able to:
- Identify the gradient and y-intercept from an equation or graph
- Plot a straight line from its equation
- Find the equation of a line from two points or from a graph
- Calculate midpoints and lengths of line segments
- Recognise and find equations of parallel and perpendicular lines
Gradient and y-intercept
The gradient tells you how steep the line is. A positive gradient slopes upward (left to right); a negative gradient slopes downward. Calculate it using:
m = (y₂ − y₁) / (x₂ − x₁)
The y-intercept is the value of y when x = 0. If the equation is already in y = mx + c form, you can read off both values directly.
Worked Example 1 — Reading from an equation
State the gradient and y-intercept of: y = 3x − 5
1
The equation is already in y = mx + c form. Compare: y = 3x + (−5)
2
Gradient m = 3; y-intercept c = −5 (the line crosses the y-axis at (0, −5))
AnswerGradient = 3, y-intercept = −5
Worked Example 2 — Calculating the gradient from two points
Find the gradient of the line passing through (1, 4) and (5, 12)
1
Label the points: (x₁, y₁) = (1, 4) and (x₂, y₂) = (5, 12)
2
Substitute into the formula: m = (12 − 4) / (5 − 1) = 8 / 4 = 2
AnswerGradient = 2
Finding the equation of a line
To find the equation of a line, you need the gradient and one point on the line. Substitute into y − y₁ = m(x − x₁) and rearrange into y = mx + c.
Worked Example 3 — Equation from a point and gradient
Find the equation of the line with gradient 4 passing through (2, 3)
1
Use y − y₁ = m(x − x₁): y − 3 = 4(x − 2)
Answery = 4x − 5
Parallel and perpendicular lines
Parallel lines have the same gradient. If one line has gradient m, any parallel line also has gradient m.
Perpendicular lines meet at a right angle. The gradients of two perpendicular lines multiply to −1. So if one line has gradient m, the perpendicular gradient is −1/m (the negative reciprocal).
Worked Example 4 — Perpendicular line
Find the equation of the line perpendicular to y = 2x + 1 that passes through (4, 5)
1
Gradient of given line: m = 2. Perpendicular gradient: −1/2
2
Use y − y₁ = m(x − x₁): y − 5 = −½(x − 4)
Answery = −½x + 7
Midpoint and length of a line segment
The midpoint of a line segment joining (x₁, y₁) and (x₂, y₂) is: ((x₁ + x₂)/2, (y₁ + y₂)/2)
The length of the segment uses Pythagoras: √((x₂ − x₁)² + (y₂ − y₁)²)
Worked Example 5 — Finding the midpoint
Find the midpoint of the line segment joining A(3, 7) and B(9, 1)
1
Average the x-coordinates: (3 + 9) / 2 = 6
2
Average the y-coordinates: (7 + 1) / 2 = 4
AnswerMidpoint = (6, 4)
Worked Example 6 — Finding the length
Find the exact length of the line segment joining P(1, 3) and Q(7, 11)
1
Find the horizontal distance: 7 − 1 = 6
2
Find the vertical distance: 11 − 3 = 8
3
Apply Pythagoras: √(6² + 8²) = √(36 + 64) = √100 = 10
AnswerLength = 10
Practice questions
Work through each question before checking the answers. Difficulty is shown for each question.
Foundation (Grade 3–5)
Q1State the gradient and y-intercept of: y = 5x + 2Foundation
Q2State the gradient and y-intercept of: y = −3x − 1Foundation
Q3Find the gradient of the line passing through (0, 2) and (4, 10)Foundation
Q4A line has gradient 2 and passes through (0, −3). Write its equation.Foundation
Q5Find the midpoint of the line segment joining (2, 4) and (8, 10)Foundation
Higher (Grade 5–7)
Q6Rearrange into y = mx + c form and state the gradient: 2y − 6x = 8Higher
Q7Find the equation of the line passing through (1, 5) and (3, 11)Higher
Q8Find the equation of the line parallel to y = 4x − 1 passing through (2, 9)Higher
Q9Find the length of the line segment joining (1, 2) and (7, 10)Higher
Q10Find the gradient of a line perpendicular to y = ⅓x + 5Higher
Higher — Hard (Grade 8–9)
Q11Find the equation of the perpendicular bisector of the line segment joining A(2, 1) and B(8, 5)Grade 8–9
Q12The line L passes through (−1, 7) and is perpendicular to 3y − x = 9. Find the equation of L.Grade 8–9
Q13Show that the points (0, −1), (3, 5) and (5, 9) are collinear (lie on the same straight line)Grade 8–9
Q14Two lines y = 3x − 2 and y = −x + 6 intersect. Find the coordinates of the intersection point.Grade 8–9
Q15A line passes through P(−2, 5) and Q(4, k). If the gradient is 2, find k and the equation of the line.Grade 8–9
Answers
Foundation (Q1–Q5)
Q1Gradient = 5, y-intercept = 2
Q2Gradient = −3, y-intercept = −1
Q3Gradient = 2((10 − 2) / (4 − 0) = 8/4)
Q4y = 2x − 3(m = 2, c = −3; passes through (0, −3) so c is the y-intercept directly)
Q5(5, 7)((2+8)/2 = 5; (4+10)/2 = 7)
Higher (Q6–Q10)
Q6y = 3x + 4; gradient = 3(divide by 2: y = 3x + 4)
Q7y = 3x + 2(m = (11−5)/(3−1) = 3; y − 5 = 3(x − 1) → y = 3x + 2)
Q8y = 4x + 1(same gradient 4; y − 9 = 4(x − 2) → y = 4x + 1)
Q910(√((7−1)² + (10−2)²) = √(36 + 64) = √100 = 10)
Q10Gradient = −3(negative reciprocal of ⅓ is −3)
Higher — Hard (Q11–Q15)
Q11y = −(3/2)x + 21/2(midpoint = (5,3); gradient AB = (5−1)/(8−2) = 2/3; perp. gradient = −3/2; y − 3 = −3/2(x − 5) → y = −3/2x + 15/2 + 3 = −3/2x + 21/2)
Q12y = −3x + 4(3y = x + 9 → gradient = 1/3; perp. = −3; y − 7 = −3(x + 1) → y = −3x + 4)
Q13Both gradients = 2, so collinear ✓(m between (0,−1)&(3,5) = 6/3 = 2; m between (3,5)&(5,9) = 4/2 = 2)
Q14(2, 4)(3x − 2 = −x + 6 → 4x = 8 → x = 2; y = 4)
Q15k = 17; y = 2x + 9(gradient: (k−5)/(4−(−2)) = 2 → k − 5 = 12 → k = 17; y − 5 = 2(x + 2) → y = 2x + 9)
Common mistakes
These are the errors Alamin sees most frequently with straight line graphs at GCSE. Recognising them now will save you marks in the exam.
Common Mistake 1
Confusing gradient and y-intercept
In y = 3x + 5, students sometimes say the gradient is 5 and the y-intercept is 3. Always match to y = mx + c: the gradient m is the coefficient of x, and c is the constant.
Common Mistake 2
Getting the gradient fraction the wrong way round
The formula is m = (y₂ − y₁) / (x₂ − x₁) — that is, change in y over change in x, not the other way round. Flipping the fraction gives a completely different gradient.
Common Mistake 3
Parallel lines — changing the gradient
Parallel lines have the exact same gradient. A common error is to change the gradient slightly when writing a parallel line. Only c (the y-intercept) changes — m stays identical.
Common Mistake 4
Perpendicular gradient — forgetting to flip AND negate
If the gradient is 2, the perpendicular gradient is −½, not −2 or ½. You must take the reciprocal (flip) AND change the sign. Both steps are required.
Common Mistake 5
Not rearranging before reading off the gradient
From 3y = 6x + 9, students read gradient = 6. This is wrong. Divide by 3 first: y = 2x + 3. The gradient is 2. Always get y on its own before comparing to y = mx + c.
Common Mistake 6
Midpoint — averaging incorrectly
For points (1, 3) and (5, 7), students sometimes write the midpoint as (4, 4) — subtracting instead of averaging. The midpoint is ((1+5)/2, (3+7)/2) = (3, 5).
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