IGCSE · Practice questions · Equations of lines: parallel and perpendicular

Find the gradient, then the line.

Six original Cambridge-style questions on equations of straight lines: integer forms, parallel and perpendicular lines, intersections, collinear points and unknown constants, all with exact fractions as on Paper 1.

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

Gradient, point, equation, check.

Gradient: \(\dfrac{y_{2} - y_{1}}{x_{2} - x_{1}}\) from two points, or make \(y\) the subject of a given line.
Parallel: same gradient. Perpendicular: negative reciprocal, \(m_{1}m_{2} = -1\).
Equation: \(y - y_{1} = m(x - x_{1})\), then clear fractions and rearrange to the form asked.
Intersection: solve the two equations simultaneously.
Check: substitute a known point into your final equation.

Recall

[2 marks]

Find the gradient of the line \(5x + 2y - 6 = 0\) and the coordinates of the point where it crosses the \(y\)-axis.

\(2y = -5x + 6\), so \(y = -\dfrac{5}{2}x + 3\). ✓
Gradient \(-\dfrac{5}{2}\); it crosses the \(y\)-axis at \((0, 3)\). ✓

Calculation

[3 marks]

Find the equation of the line through \(A(-3, 4)\) and \(B(5, -2)\), giving your answer in the form \(ax + by + c = 0\), where \(a\), \(b\) and \(c\) are integers.

\(m = \dfrac{-2 - 4}{5 - (-3)} = \dfrac{-6}{8} = -\dfrac{3}{4}\). ✓
\(y - 4 = -\dfrac{3}{4}(x + 3)\), so \(4y - 16 = -3x - 9\). ✓
\(3x + 4y - 7 = 0\). Check with \(B\): \(15 - 8 - 7 = 0\). ✓

Calculation

[3 marks]

Find the equation of the line through \((6, -1)\) that is perpendicular to the line \(2y = 3x + 5\). Give your answer in the form \(ax + by = c\), where \(a\), \(b\) and \(c\) are integers.

\(y = \dfrac{3}{2}x + \dfrac{5}{2}\) has gradient \(\dfrac{3}{2}\), so the perpendicular gradient is \(-\dfrac{2}{3}\). ✓
\(y + 1 = -\dfrac{2}{3}(x - 6)\), so \(3y + 3 = -2x + 12\). ✓
\(2x + 3y = 9\). ✓

Proof

[4 marks]

The points \(A(-2, -1)\), \(B(1, 5)\) and \(C(3, 9)\) are given.

(a) Show that \(A\), \(B\) and \(C\) are collinear.

(b) Find the equation of the line through them in the form \(y = mx + c\).

(a) \(m_{AB} = \dfrac{5 - (-1)}{1 - (-2)} = \dfrac{6}{3} = 2\) and \(m_{BC} = \dfrac{9 - 5}{3 - 1} = \dfrac{4}{2} = 2\). ✓
The gradients are equal and \(B\) is common to both, so the points are collinear. ✓
(b) \(y - 5 = 2(x - 1)\), so \(y = 2x + 3\). ✓
Check: \(A\) gives \(-4 + 3 = -1\) and \(C\) gives \(6 + 3 = 9\). ✓

Analysis

[5 marks]

The line \(l_{1}\) passes through \(P(1, 7)\) and \(Q(5, -1)\). The line \(l_{2}\) passes through \(R(4, 6)\) and is perpendicular to \(l_{1}\).

(a) Find the equation of \(l_{1}\).

(b) Find the equation of \(l_{2}\).

(a) \(m = \dfrac{-1 - 7}{5 - 1} = -2\), so \(y - 7 = -2(x - 1)\), giving \(y = -2x + 9\). ✓
(b) Perpendicular gradient \(\dfrac{1}{2}\): \(y - 6 = \dfrac{1}{2}(x - 4)\), giving \(y = \dfrac{1}{2}x + 4\). ✓
(c) \(-2x + 9 = \dfrac{1}{2}x + 4\), so \(5 = \dfrac{5}{2}x\) and \(x = 2\). ✓
\(y = -2(2) + 9 = 5\), so they meet at \((2, 5)\). ✓
Check in \(l_{2}\): \(\dfrac{1}{2}(2) + 4 = 5\). ✓

Analysis

[4 marks]

The line \(kx + 4y = 3\) is perpendicular to the line through \((2, 1)\) and \((5, 7)\).

(a) Find the value of \(k\).

(b) The point \((h, -2)\) lies on the line \(kx + 4y = 3\). Find the value of \(h\).

(a) The line through \((2, 1)\) and \((5, 7)\) has gradient \(\dfrac{7 - 1}{5 - 2} = 2\), so the perpendicular gradient is \(-\dfrac{1}{2}\). ✓
\(kx + 4y = 3\) gives \(y = -\dfrac{k}{4}x + \dfrac{3}{4}\), so \(-\dfrac{k}{4} = -\dfrac{1}{2}\) and \(k = 2\). ✓
(b) \(2h + 4(-2) = 3\), so \(2h = 11\). ✓
\(h = \dfrac{11}{2}\). ✓

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.