IGCSE Mathematics: Straight Line Graphs Practice Questions
Every straight line can be written as y = mx + c, where m is the gradient and c is the y-intercept. The gradient between two points is the change in y divided by the change in x.
Topic 3.2 of Cambridge IGCSE Mathematics 0580 appears on both papers and feeds directly into perpendicular lines and graphical solutions. Nearly every lost mark comes from a sign error in the gradient calculation. The questions below make the subtraction explicit.
What you need to know for Straight Line Graphs
- Gradientm = (y2 minus y1) divided by (x2 minus x1). Subtract the coordinates in the same order on the top and the bottom.
- Equation of a liney = mx + c. Find m from two points or from the graph, then substitute one point to find c.
- Reading from a rearranged equationAn equation must be in the form y = mx + c before the gradient and intercept can be read off directly.
- MidpointThe mean of the x coordinates and the mean of the y coordinates, giving ((x1 + x2) divided by 2, (y1 + y2) divided by 2).
- Length of a line segmentUse Pythagoras on the horizontal and vertical differences: length is the square root of the sum of their squares.
- Sign of the gradientA positive gradient rises from left to right and a negative gradient falls. Checking this against a sketch catches most sign errors.
IGCSE Mathematics Straight Line Graphs questions and answers
4 exam-style questions written to the 0580 syllabus. Try each one on paper first, then open the worked answer to check your method against the marks.
Find the gradient of the line passing through the points (1, 4) and (5, 16).
Show the worked answer
- Use m = (y2 minus y1) divided by (x2 minus x1).
- Change in y = 16 minus 4 = 12.
- Change in x = 5 minus 1 = 4.
- m = 12 divided by 4 = 3. The gradient is positive, which is consistent with y increasing as x increases.
Find the equation of the straight line passing through the points (2, 5) and (6, 13).
Show the worked answer
- Find the gradient: m = (13 minus 5) divided by (6 minus 2) = 8 divided by 4 = 2.
- Write the partial equation: y = 2x + c.
- Substitute one of the points to find c. Using (2, 5): 5 = 2 x 2 + c, so 5 = 4 + c and c = 1.
- The equation is y = 2x + 1. Check with the other point: 2 x 6 + 1 = 13, which matches.
A is the point (minus 3, 7) and B is the point (5, 1). Find the midpoint of AB and the length of AB.
Show the worked answer
- Midpoint x coordinate: (minus 3 + 5) divided by 2 = 2 divided by 2 = 1.
- Midpoint y coordinate: (7 + 1) divided by 2 = 8 divided by 2 = 4. The midpoint is (1, 4).
- For the length, find the horizontal and vertical differences: 5 minus (minus 3) = 8, and 1 minus 7 = minus 6.
- Apply Pythagoras: length is the square root of (82 + 62) = the square root of (64 + 36) = the square root of 100 = 10.
Rearrange 2y minus 6x = 8 into the form y = mx + c, then write down the gradient and the y-intercept.
Show the worked answer
- Add 6x to both sides to isolate the y term: 2y = 6x + 8.
- Divide every term on both sides by 2: y = 3x + 4.
- The equation is now in the form y = mx + c, so the gradient m is the coefficient of x, which is 3.
- The y-intercept c is the constant term, 4, meaning the line crosses the y-axis at (0, 4).
Common mistakes in this topic
- Inconsistent order of subtraction in the gradient formula.
- Reading the gradient before rearranging into y = mx + c.
- Errors when subtracting negative coordinates.
- Giving only the gradient when the equation is required.
- Using the sum instead of the difference of coordinates when finding a length.
Exam tips
- Label your two points as point 1 and point 2 before substituting, and keep that order throughout.
- Sketch the two points quickly. If the line falls from left to right the gradient must be negative.
- Always substitute a point back into your final equation as a check.
- For length, draw the right-angled triangle. The horizontal and vertical sides are the differences.
- Midpoint means average, so add the coordinates and halve. Length means difference, so subtract.
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Straight Line Graphs FAQs
How do I find the gradient between two points?
Subtract the y coordinates and divide by the difference in the x coordinates, taking the points in the same order for both subtractions. A positive result means the line rises from left to right and a negative result means it falls.
How do I find the equation of a line through two points?
Calculate the gradient first, then write y equals that gradient times x plus c. Substitute the coordinates of either point to find c, and write out the full equation. Check by substituting the other point, which should also satisfy it.
How do I find the midpoint and length of a line segment?
For the midpoint, average the x coordinates and average the y coordinates. For the length, find the horizontal and vertical differences between the points and apply Pythagoras, taking the square root of the sum of their squares.
Why must I rearrange before reading the gradient?
The gradient is only the coefficient of x when the equation is in the form y equals mx plus c, with a single y on the left. In a form such as 2y minus 6x equals 8, the coefficients have not yet been divided through, so reading them directly gives the wrong values.
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Written to the published Cambridge IGCSE Mathematics (0580) syllabus. Check your school entry code and syllabus year, because Core and Extended candidates are assessed on different content. Last reviewed 2026-08-12.