IGCSE Mathematics 0580 · Topic 3.5

IGCSE Mathematics: Parallel and Perpendicular Lines Practice Questions

Parallel lines have equal gradients. Perpendicular lines have gradients whose product is minus 1, so each gradient is the negative reciprocal of the other.

Cambridge IGCSE Mathematics (0580) · Topic 3.5: Parallel and Perpendicular Lines

Topic 3.5 of Cambridge IGCSE Mathematics 0580 is short and highly predictable, which makes it reliable marks. The negative reciprocal is where errors happen, because both the flip and the sign change must be applied. The questions below force both steps.

What you need to know for Parallel and Perpendicular Lines

IGCSE Mathematics Parallel and Perpendicular Lines 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.

Question 1[3 marks]

Find the equation of the line parallel to y = 4x minus 3 that passes through the point (1, 9).

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Answer: y = 4x + 5
  1. Parallel lines have equal gradients, so the new line also has gradient 4.
  2. Write the partial equation: y = 4x + c.
  3. Substitute the point (1, 9): 9 = 4 x 1 + c, so 9 = 4 + c.
  4. c = 5, giving the equation y = 4x + 5.
How the marks are awarded. 1 mark for using a gradient of 4. 1 mark for substituting the point correctly. 1 mark for y = 4x + 5.
Where students lose the mark. Changing the gradient because the line must be different. Parallel lines share a gradient. It is the intercept that differs.
Question 2[4 marks]

Find the equation of the line perpendicular to y = 2x + 1 that passes through the point (4, 3).

Show the worked answer
Answer: y = minus one half x + 5
  1. The gradient of the given line is 2. For a perpendicular line, take the negative reciprocal.
  2. The reciprocal of 2 is one half, and changing the sign gives minus one half. Check: 2 multiplied by minus one half = minus 1.
  3. Write the partial equation: y = minus one half x + c.
  4. Substitute (4, 3): 3 = minus one half times 4 + c, so 3 = minus 2 + c and c = 5. The equation is y = minus one half x + 5.
How the marks are awarded. 1 mark for identifying the gradient as 2. 1 mark for the perpendicular gradient of minus one half. 1 mark for substituting the point. 1 mark for the complete equation.
Where students lose the mark. Using minus 2 as the perpendicular gradient. Changing the sign alone is not enough. The fraction must be inverted as well.
Question 3[3 marks]

Determine whether the lines y = 3x minus 1 and y = minus x divided by 3, plus 4, are perpendicular. Justify your answer.

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Answer: Yes, because the product of the gradients is minus 1.
  1. Read the gradient of the first line: m1 = 3.
  2. Read the gradient of the second line: m2 = minus one third.
  3. Multiply the gradients: 3 multiplied by minus one third = minus 1.
  4. The product is minus 1, so the two lines are perpendicular. The y-intercepts are irrelevant to this test.
How the marks are awarded. 1 mark for both gradients identified. 1 mark for multiplying them. 1 mark for the product of minus 1 and the conclusion that the lines are perpendicular.
Where students lose the mark. Comparing the intercepts or the appearance on a sketch. Only the product of the gradients decides perpendicularity.
Question 4[5 marks]

A is the point (1, 2) and B is the point (7, 10). Find the equation of the perpendicular bisector of AB.

Show the worked answer
Answer: y = minus three quarters x + 9
  1. Find the midpoint of AB: x = (1 + 7) divided by 2 = 4, and y = (2 + 10) divided by 2 = 6. The midpoint is (4, 6).
  2. Find the gradient of AB: (10 minus 2) divided by (7 minus 1) = 8 divided by 6 = four thirds.
  3. The perpendicular gradient is the negative reciprocal of four thirds, which is minus three quarters. Check: four thirds times minus three quarters = minus 1.
  4. Write y = minus three quarters x + c and substitute the midpoint (4, 6): 6 = minus three quarters times 4 + c, so 6 = minus 3 + c.
  5. c = 9, giving the perpendicular bisector y = minus three quarters x + 9.
How the marks are awarded. 1 mark for the midpoint (4, 6). 1 mark for the gradient of AB as four thirds. 1 mark for the perpendicular gradient of minus three quarters. 1 mark for substituting the midpoint. 1 mark for the complete equation.
Where students lose the mark. Using point A or point B instead of the midpoint. A bisector must pass through the middle of the segment, not through either end.

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Parallel and Perpendicular Lines FAQs

What is the rule for perpendicular gradients?

The product of the gradients of two perpendicular lines is minus 1, so each gradient is the negative reciprocal of the other. To find it, write the gradient as a fraction, turn it upside down, and change the sign. Multiplying the two values back together should give exactly minus 1.

How do I find a line parallel to another through a given point?

Use the same gradient as the original line, since parallel lines have equal gradients. Write y equals that gradient times x plus c, substitute the coordinates of the given point, and solve for c to complete the equation.

What is a perpendicular bisector?

A line that crosses a line segment at its midpoint and at right angles to it. Find the midpoint of the segment, calculate the gradient of the segment, take the negative reciprocal of that gradient, then use the midpoint and the new gradient to form the equation.

Why is minus 2 not the perpendicular gradient of a line with gradient 2?

Changing the sign alone is not sufficient, because 2 multiplied by minus 2 gives minus 4 rather than minus 1. The gradient must also be inverted, giving minus one half, and 2 multiplied by minus one half does equal minus 1.

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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.