IGCSE · Practice questions · Length and midpoint of a line segment

Distance and middle.

Extended. Six original Cambridge-style questions on the length and midpoint of a line segment, ending with a map 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

Differences for length, sums for midpoint.

Steps: work out the change in \(x\) and the change in \(y\), with brackets for negatives.
Length: \(\sqrt{(\text{change in } x)^2 + (\text{change in } y)^2}\).
Midpoint: \(\left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)\).
Missing end point: repeat the step from the known end to the midpoint.
Accuracy: exact surd or 3 significant figures.

Recall

[2 marks]

Find the midpoint of the line segment joining \((-4, 3)\) and \((6, -9)\).

\(x = \dfrac{-4 + 6}{2} = 1\), \(y = \dfrac{3 + (-9)}{2} = -3\). ✓
Midpoint \((1, -3)\). ✓

Calculation

[2 marks]

Calculate the length of the line segment joining \((2, -3)\) and \((-4, 5)\).

Change in \(x = -6\), change in \(y = 8\). ✓
Length \(= \sqrt{36 + 64} = \sqrt{100} = 10\) units. ✓

Calculation

[2 marks]

Calculate the length of the line segment joining \((1, 7)\) and \((6, -1)\). Give your answer correct to 3 significant figures.

Change in \(x = 5\), change in \(y = -8\). ✓
Length \(= \sqrt{25 + 64} = \sqrt{89} = 9.43\) units (3 s.f.). ✓

Analysis

[4 marks]

\(M(4, -1)\) is the midpoint of \(AB\), where \(A\) is \((-2, 5)\).

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

(b) Calculate the length of \(AB\), correct to 3 significant figures.

(a) From \(A\) to \(M\): \(+6\) across, \(-6\) up. So \(B = (4 + 6, -1 - 6) = (10, -7)\). ✓
(b) Change in \(x = 12\), change in \(y = -12\). ✓
\(AB = \sqrt{144 + 144} = \sqrt{288} = 16.97\ldots = 17.0\) units (3 s.f.). ✓

Proof

[3 marks]

The vertices of triangle \(ABC\) are \(A(1, 1)\), \(B(7, 3)\) and \(C(3, 7)\).

Show that triangle \(ABC\) is isosceles.

\(AB = \sqrt{6^2 + 2^2} = \sqrt{40}\). ✓
\(AC = \sqrt{2^2 + 6^2} = \sqrt{40}\). ✓
\(BC = \sqrt{(-4)^2 + 4^2} = \sqrt{32}\). ✓
\(AB = AC\), so two sides are equal and the triangle is isosceles. ✓

Analysis

[4 marks]

On a map, one unit on the grid represents 1 km. Town \(P\) is at \((-6, 2)\) and town \(Q\) is at \((10, 14)\). A straight road joins them, and a service station is built exactly halfway along it.

(a) Find the coordinates of the service station.

(b) A driver travels from \(P\) to the service station at an average speed of 60 km/h. How many minutes does this take?

(a) \(\left(\dfrac{-6 + 10}{2}, \dfrac{2 + 14}{2}\right) = (2, 8)\). ✓
(b) \(PQ = \sqrt{16^2 + 12^2} = \sqrt{400} = 20\) km, so \(P\) to the station is 10 km. ✓
Time \(= 10 \div 60 = \dfrac{1}{6}\) hour \(= 10\) minutes. ✓

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.