IGCSE · Practice questions · Parallel lines

Keep the gradient.

Six original Cambridge-style questions on parallel lines: equal gradients, equations through a point and parallel lines in context.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Same \(m\), new \(c\).

Gradient: write the given line as \(y = mx + c\) and read \(m\).
Parallel: the new line has the same gradient \(m\).
Intercept: substitute the given point into \(y = mx + c\) to find the new \(c\).
Shortcut (Extended): parallel to \(ax + by = c\) is \(ax + by = k\); put the point in to find \(k\).
Proof: equal gradients show that two sides are parallel.

Recall

[2 marks]

Which of these lines are parallel to each other?

A: \(y = 2x + 3\) B: \(y = 3 - 2x\) C: \(y = 2x - 7\) D (Extended): \(2y = 4x + 1\)

Gradients: A is 2, B is \(-2\), C is 2, and D is \(y = 2x + 0.5\), gradient 2. ✓
A, C and D are parallel. ✓

Calculation

[2 marks]

Find the equation of the line parallel to \(y = -3x + 2\) that passes through \((2, 5)\).

Gradient \(-3\): \(5 = -3 \times 2 + c = -6 + c\), so \(c = 11\). ✓
\(y = -3x + 11\) ✓

Calculation

[3 marks]

(Extended) Find the equation of the line parallel to \(5x + 2y = 8\) that passes through \((-2, 3)\). Give your answer in the form \(ax + by = c\).

Parallel line: \(5x + 2y = k\). ✓
Substitute \((-2, 3)\): \(k = 5 \times (-2) + 2 \times 3 = -10 + 6 = -4\). ✓
\(5x + 2y = -4\) ✓

Proof

[3 marks]

The vertices of quadrilateral \(ABCD\) are \(A(0, 0)\), \(B(6, 2)\), \(C(5, 5)\) and \(D(2, 4)\).

Show that \(ABCD\) is a trapezium.

Gradient \(AB = \dfrac{2}{6} = \dfrac{1}{3}\); gradient \(DC = \dfrac{5 - 4}{5 - 2} = \dfrac{1}{3}\). So \(AB \parallel DC\). ✓
Gradient \(AD = \dfrac{4}{2} = 2\); gradient \(BC = \dfrac{5 - 2}{5 - 6} = -3\). So \(AD\) and \(BC\) are not parallel. ✓
Exactly one pair of parallel sides, so \(ABCD\) is a trapezium. ✓

Analysis

[3 marks]

A line \(L\) drawn on a grid passes through \((0, 4)\) and \((2, 0)\).

Find the equation of the line parallel to \(L\) that passes through \((3, 1)\).

Gradient of \(L = \dfrac{0 - 4}{2 - 0} = -2\). ✓
\(1 = -2 \times 3 + c\), so \(c = 7\). ✓
\(y = -2x + 7\) ✓

Analysis

[4 marks]

Cyclist A's distance from a town after \(t\) hours is \(d = 15t + 6\) km. Cyclist B leaves the town at \(t = 0\) and rides at the same speed as A.

(a) Write down the equation for B's distance.

(b) Explain, using the graphs, why B never catches A.

(a) Same speed means the same gradient 15, starting from 0: \(d = 15t\). ✓
(b) The lines \(d = 15t + 6\) and \(d = 15t\) have equal gradients, so they are parallel and never meet: B is always 6 km behind A. ✓
(c) \(18t = 15t + 6\), so \(3t = 6\) and \(t = 2\) hours. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.