IGCSE · Practice questions · Perpendicular lines and bisectors

Right angles by algebra.

Extended. Six original Cambridge-style questions on perpendicular lines and perpendicular bisectors, ending with a shortest-distance problem.

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

Flip, change the sign, substitute.

Gradient: write the given line as \(y = mx + c\) and read \(m\).
Perpendicular: new gradient \(-\dfrac{1}{m}\); check the product is \(-1\).
Through a point: substitute it into \(y = mx + c\) to find \(c\).
Bisector of \(AB\): midpoint of \(AB\) plus the gradient perpendicular to \(AB\).
Tidy: clear fractions if asked for the form \(ax + by = c\).

Recall

[2 marks]

Write down the gradient of a line perpendicular to

(a) \(y = 5x - 1\) (b) \(y = 3 - \dfrac{3}{4}x\).

(a) \(-\dfrac{1}{5}\) ✓
(b) the gradient is \(-\dfrac{3}{4}\), so the perpendicular gradient is \(\dfrac{4}{3}\). ✓

Proof

[2 marks]

Show that the lines \(y = 4x - 3\) and \(x + 4y = 12\) are perpendicular.

\(x + 4y = 12\) gives \(y = -\dfrac{1}{4}x + 3\), gradient \(-\dfrac{1}{4}\). ✓
\(4 \times \left(-\dfrac{1}{4}\right) = -1\), so the lines are perpendicular. ✓

Calculation

[3 marks]

Find the equation of the line perpendicular to \(x + 3y = 6\) that passes through \((2, -1)\).

\(y = -\dfrac{1}{3}x + 2\), gradient \(-\dfrac{1}{3}\), so the perpendicular gradient is 3. ✓
\(-1 = 3 \times 2 + c\), so \(c = -7\). ✓
\(y = 3x - 7\) ✓

Proof

[3 marks]

The vertices of triangle \(PQR\) are \(P(-1, 1)\), \(Q(3, 3)\) and \(R(1, 7)\).

Show that angle \(PQR\) is a right angle.

Gradient \(PQ = \dfrac{3 - 1}{3 - (-1)} = \dfrac{2}{4} = \dfrac{1}{2}\). ✓
Gradient \(QR = \dfrac{7 - 3}{1 - 3} = \dfrac{4}{-2} = -2\). ✓
\(\dfrac{1}{2} \times (-2) = -1\), so \(PQ \perp QR\) and angle \(PQR = 90^\circ\). ✓

Analysis

[4 marks]

\(A\) is the point \((1, 7)\) and \(B\) is the point \((7, -1)\).

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

(b) Show that the perpendicular bisector passes through the origin.

(a) Midpoint \(= (4, 3)\). Gradient \(AB = \dfrac{-1 - 7}{7 - 1} = -\dfrac{4}{3}\), so the perpendicular gradient is \(\dfrac{3}{4}\). ✓
\(3 = \dfrac{3}{4} \times 4 + c = 3 + c\), so \(c = 0\): \(y = \dfrac{3}{4}x\) (or \(3x - 4y = 0\)). ✓
(b) \(c = 0\), so when \(x = 0\), \(y = 0\): the line passes through \((0, 0)\). ✓

Analysis

[5 marks]

On a map with a 1 km grid, a straight road follows the line \(y = 2x + 1\). A house is at \(H(7, 0)\). A path is to be built from \(H\) to the road, as short as possible, so it must meet the road at right angles.

(a) Find the equation of the path.

(b) Find the point where the path meets the road.

(a) Perpendicular gradient \(-\dfrac{1}{2}\). \(0 = -\dfrac{1}{2} \times 7 + c\), so \(c = 3.5\): \(y = -\dfrac{1}{2}x + 3.5\). ✓
(b) \(2x + 1 = -\dfrac{1}{2}x + 3.5\), so \(2.5x = 2.5\) and \(x = 1\); \(y = 2 \times 1 + 1 = 3\). The point is \((1, 3)\). ✓
(c) Length \(= \sqrt{(7 - 1)^2 + (0 - 3)^2} = \sqrt{36 + 9} = \sqrt{45} = 6.71\) km (3 s.f.). ✓

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.