IGCSE · Practice questions · Midpoints, lengths and perpendicular bisectors

Equal distances, exact answers.

Six original Cambridge-style questions on midpoints, lengths and perpendicular bisectors: exact surd lengths, endpoints from midpoints, bisector equations, equidistant points, areas from coordinates, a rhombus and a circle centre, all without a calculator.

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

Midpoint, flip the gradient, then use it.

Midpoint: add the coordinates and halve; for a missing end, \(B = 2M - A\).
Length: \(\sqrt{(\Delta x)^{2} + (\Delta y)^{2}}\), simplified to an exact surd.
Bisector: through the midpoint, gradient \(-\dfrac{1}{m_{AB}}\).
Equidistant from \(A\) and \(B\) means on the bisector: solve it with the other given line.
Area: \(\tfrac{1}{2} d_1 d_2\) for a rhombus or kite, \(\tfrac{1}{2}\) base \(\times\) perpendicular height, or the shoelace formula.

Recall

[3 marks]

\(P\) is the point \((-3, -1)\) and \(Q\) is the point \((5, 3)\). Find the coordinates of the midpoint of \(PQ\) and the length of \(PQ\), giving your answer in the form \(k\sqrt{5}\).

Midpoint: \(\left(\dfrac{-3 + 5}{2}, \dfrac{-1 + 3}{2}\right) = (1, 1)\). ✓
\(PQ = \sqrt{8^{2} + 4^{2}} = \sqrt{80}\). ✓
\(\sqrt{80} = \sqrt{16 \times 5} = 4\sqrt{5}\). ✓

Calculation

[2 marks]

The midpoint of \(AB\) is \(M(4, -1)\) and \(A\) is the point \((7, 3)\). Find the coordinates of \(B\).

\(B = (2 \times 4 - 7,\ 2 \times (-1) - 3)\). ✓
\(B = (1, -5)\). Check: \(\dfrac{7 + 1}{2} = 4\) and \(\dfrac{3 + (-5)}{2} = -1\). ✓

Calculation

[4 marks]

\(A\) is the point \((-4, 3)\) and \(B\) is the point \((2, -1)\).

(a) Find the equation of the perpendicular bisector of \(AB\), giving your answer in the form \(ax + by + c = 0\) where \(a\), \(b\) and \(c\) are integers.

(b) Find the coordinates of the point where this bisector crosses the \(y\)-axis.

(a) Midpoint \(\left(\dfrac{-4 + 2}{2}, \dfrac{3 - 1}{2}\right) = (-1, 1)\). ✓
Gradient of \(AB\): \(\dfrac{-1 - 3}{2 - (-4)} = -\dfrac{2}{3}\), so the perpendicular gradient is \(\dfrac{3}{2}\). ✓
\(y - 1 = \dfrac{3}{2}(x + 1)\), so \(2y - 2 = 3x + 3\), giving \(3x - 2y + 5 = 0\). ✓
(b) \(x = 0\): \(-2y + 5 = 0\), so the point is \(\left(0, \tfrac{5}{2}\right)\). ✓

Analysis

[6 marks]

The points \(A(1, 6)\) and \(B(7, 2)\) are given. The point \(C\) lies on the line \(y = x + 1\) and \(AC = BC\).

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

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

(a) \(C\) lies on the perpendicular bisector of \(AB\). Midpoint \(M(4, 4)\); \(m_{AB} = -\dfrac{4}{6} = -\dfrac{2}{3}\), so the bisector has gradient \(\dfrac{3}{2}\). ✓
\(y - 4 = \dfrac{3}{2}(x - 4)\), so \(3x - 2y = 4\). ✓
Substitute \(y = x + 1\): \(3x - 2x - 2 = 4\), so \(x = 6\), \(y = 7\): \(C(6, 7)\). Check: \(CA^{2} = 25 + 1 = 26 = 1 + 25 = CB^{2}\). ✓
(b) \(CM\) is perpendicular to \(AB\), so it is the height. \(AB = \sqrt{6^{2} + 4^{2}} = \sqrt{52} = 2\sqrt{13}\) and \(CM = \sqrt{2^{2} + 3^{2}} = \sqrt{13}\). ✓
Area \(= \tfrac{1}{2} \times 2\sqrt{13} \times \sqrt{13} = 13\) square units (the shoelace method also gives \(\tfrac{1}{2}\left|-5 + 7 + 24\right| = 13\)). ✓

Analysis

[6 marks]

\(ABCD\) is a rhombus. \(A\) is the point \((2, -1)\) and \(C\) is the point \((6, 7)\). The vertex \(B\) lies on the \(x\)-axis.

(a) Find the equation of the diagonal \(BD\).

(b) Find the coordinates of \(B\) and of \(D\).

(a) \(BD\) is the perpendicular bisector of \(AC\). Midpoint \(M(4, 3)\); \(m_{AC} = \dfrac{8}{4} = 2\), so \(BD\) has gradient \(-\dfrac{1}{2}\). ✓
\(y - 3 = -\dfrac{1}{2}(x - 4)\), so \(2y - 6 = -x + 4\), giving \(x + 2y = 10\). ✓
(b) \(y = 0\): \(x = 10\), so \(B(10, 0)\). Then \(D = (2 \times 4 - 10,\ 2 \times 3 - 0) = (-2, 6)\). ✓
(c) \(AC = \sqrt{4^{2} + 8^{2}} = \sqrt{80} = 4\sqrt{5}\) and \(BD = \sqrt{12^{2} + 6^{2}} = \sqrt{180} = 6\sqrt{5}\). ✓
Area \(= \tfrac{1}{2} \times 4\sqrt{5} \times 6\sqrt{5} = \tfrac{1}{2} \times 120 = 60\) square units. ✓

Analysis

[6 marks]

A circle passes through the points \(A(-2, 4)\), \(B(4, 4)\) and \(C(6, 0)\).

(a) Explain why the centre of the circle lies on the line \(x = 1\).

(b) Find the equation of the perpendicular bisector of \(BC\).

(d) Show that the point \(D(4, -4)\) also lies on the circle.

(a) The centre is equidistant from \(A\) and \(B\), so it lies on the perpendicular bisector of \(AB\). \(AB\) is horizontal with midpoint \((1, 4)\), so that bisector is the vertical line \(x = 1\). ✓
(b) Midpoint of \(BC\): \((5, 2)\). \(m_{BC} = \dfrac{0 - 4}{6 - 4} = -2\), so the bisector has gradient \(\dfrac{1}{2}\): \(y - 2 = \dfrac{1}{2}(x - 5)\), or \(x - 2y = 1\). ✓
(c) At \(x = 1\): \(1 - 2y = 1\), so \(y = 0\). The centre is \((1, 0)\). ✓
Radius: distance to \(A\) is \(\sqrt{3^{2} + 4^{2}} = 5\) (to \(C\) it is \(6 - 1 = 5\) as well). ✓
(d) Distance from \((1, 0)\) to \(D(4, -4)\) is \(\sqrt{3^{2} + 4^{2}} = 5\), equal to the radius, so \(D\) is on the circle. ✓

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.