AS & A Level · Practice questions · Equations of straight lines

Gradient, point, equation.

Six original Cambridge-style questions on straight lines: equations in all three forms, parallel and perpendicular lines, lengths, midpoints, intersections and a shortest-distance problem.

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

The method

Find a gradient and a point, then write the line.

Gradient from two points, or by rearranging a given line to \(y = mx + c\).
Parallel: same gradient. Perpendicular: \(-\dfrac{1}{m}\).
Equation: \(y - y_{1} = m(x - x_{1})\), then rearrange to the form asked for.
Length and midpoint from the formulae; intersections by solving simultaneously.
Sketch the points to check that your answers are sensible.

Recall

[2 marks]

Find the gradient and the \(y\)-intercept of the line \(5x - 2y + 8 = 0\).

\(2y = 5x + 8 \Rightarrow y = \dfrac{5}{2}x + 4\). ✓
Gradient \(\dfrac{5}{2}\), \(y\)-intercept \(4\). ✓

Calculation

[3 marks]

Find the equation of the line through \((4, -3)\) and \((-2, 6)\), giving your answer in the form \(ax + by + c = 0\) where \(a\), \(b\) and \(c\) are integers.

\(m = \dfrac{6 - (-3)}{-2 - 4} = \dfrac{9}{-6} = -\dfrac{3}{2}\). ✓
\(y + 3 = -\dfrac{3}{2}(x - 4) \Rightarrow 2y + 6 = -3x + 12\). ✓
\(3x + 2y - 6 = 0\) (check \((-2, 6)\): \(-6 + 12 - 6 = 0\)). ✓

Calculation

[4 marks]

The points \(A(-2, 1)\) and \(B(4, 5)\) are given.

(a) Find the equation of the perpendicular bisector of \(AB\).

(b) Find the point where the perpendicular bisector crosses the \(x\)-axis.

(a) Midpoint \((1, 3)\). Gradient \(AB = \dfrac{4}{6} = \dfrac{2}{3}\), so the perpendicular gradient is \(-\dfrac{3}{2}\). ✓
\(y - 3 = -\dfrac{3}{2}(x - 1) \Rightarrow 2y - 6 = -3x + 3 \Rightarrow 3x + 2y - 9 = 0\). ✓
(b) \(y = 0\): \(3x = 9\), so \((3,\ 0)\). ✓

Analysis

[5 marks]

The points \(A(0, 2)\) and \(B(6, 5)\) are given. The line through \(B\) perpendicular to \(AB\) meets the \(x\)-axis at \(C\).

(a) Find the coordinates of \(C\).

(b) Find the exact area of triangle \(ABC\).

(a) Gradient \(AB = \dfrac{1}{2}\); perpendicular gradient \(-2\): \(y = -2x + 17\). ✓
\(y = 0 \Rightarrow x = \dfrac{17}{2}\), so \(C\left(\dfrac{17}{2}, 0\right)\). ✓
(b) \(AB = \sqrt{45} = 3\sqrt{5}\), \(BC = \sqrt{\left(\dfrac{5}{2}\right)^{2} + 5^{2}} = \dfrac{5\sqrt{5}}{2}\). ✓
Angle \(ABC = 90^{\circ}\), so area \(= \dfrac{1}{2} \times 3\sqrt{5} \times \dfrac{5\sqrt{5}}{2} = \dfrac{75}{4}\). ✓

Proof

[3 marks]

The points \(P(1, 2)\), \(Q(4, 6)\) and \(R(8, 3)\) are given. Show that triangle \(PQR\) is right-angled and isosceles.

Gradient \(PQ = \dfrac{4}{3}\), gradient \(QR = \dfrac{-3}{4}\); \(\dfrac{4}{3} \times \left(-\dfrac{3}{4}\right) = -1\), so angle \(PQR = 90^{\circ}\). ✓
\(PQ = \sqrt{3^{2} + 4^{2}} = 5\) and \(QR = \sqrt{4^{2} + 3^{2}} = 5\). ✓
Two equal sides meeting at a right angle: the triangle is right-angled and isosceles. ✓

Analysis

[5 marks]

Coordinates are in kilometres. A ship sails in a straight line from \(A(2, 1)\) to \(B(14, 10)\). A lighthouse is at \(L(10, 1)\).

(a) Find the equation of the ship's path \(AB\).

(b) Find the point on \(AB\) where the ship is closest to the lighthouse, and the shortest distance.

(a) Gradient \(\dfrac{9}{12} = \dfrac{3}{4}\): \(y - 1 = \dfrac{3}{4}(x - 2) \Rightarrow 3x - 4y - 2 = 0\). ✓
(b) The closest point is where the perpendicular from \(L\) meets \(AB\). Perpendicular gradient \(-\dfrac{4}{3}\): \(y - 1 = -\dfrac{4}{3}(x - 10) \Rightarrow 4x + 3y - 43 = 0\). ✓
Solve \(3x - 4y = 2\) and \(4x + 3y = 43\): \(x = \dfrac{178}{25} = 7.12\), \(y = \dfrac{121}{25} = 4.84\). ✓
Distance \(= \sqrt{(10 - 7.12)^{2} + (1 - 4.84)^{2}} = \sqrt{8.2944 + 14.7456} = \sqrt{23.04}\). ✓
Shortest distance \(= 4.8\) km, at \((7.12,\ 4.84)\). ✓

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.