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SingaporeMathsSyllabus dot point

Given two points, how do we find the length of the segment joining them and the coordinates of its midpoint?

Calculate the length of a line segment using the distance formula and find the midpoint of a line segment

A focused answer to the N(A)-Level Mathematics outcome on length and midpoint. The distance formula from Pythagoras, the midpoint formula, and applying both to points on a grid.

Generated by Claude Opus 4.88 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
  4. Try this

What this dot point is asking

SEAB wants you to find the length of a line segment joining two points using the distance formula, and to find the coordinates of its midpoint. Both come from simple ideas - Pythagoras for length and averaging for the midpoint - and they appear throughout coordinate geometry.

The answer

The distance formula

The length of the segment joining (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) comes straight from Pythagoras' theorem. The horizontal gap and the vertical gap form the two short sides of a right-angled triangle, and the segment is the hypotenuse:

length=(x2−x1)2+(y2−y1)2\text{length} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Because the differences are squared, it does not matter which point you call first - a negative difference squared becomes positive.

Why it is just Pythagoras

If you plot the two points and draw a horizontal line and a vertical line to form a right-angled triangle, the horizontal side has length x2−x1x_2 - x_1 and the vertical side has length y2−y1y_2 - y_1. The straight-line distance between the points is the hypotenuse, so length2=(horizontal)2+(vertical)2\text{length}^2 = (\text{horizontal})^2 + (\text{vertical})^2.

The midpoint formula

The midpoint is exactly halfway between the two points. You find it by averaging the xx values and averaging the yy values:

midpoint=(x1+x22,y1+y22)\text{midpoint} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)

For (2,4)(2, 4) and (6,10)(6, 10), the midpoint is (2+62,4+102)=(4,7)\left( \dfrac{2 + 6}{2}, \dfrac{4 + 10}{2} \right) = (4, 7).

Horizontal and vertical segments

When the two points share an xx-coordinate, the segment is vertical and its length is just the difference of the yy-values. When they share a yy-coordinate, the segment is horizontal and its length is the difference of the xx-values. In these cases you do not need the full formula, though it still gives the same answer.

Rounding the length

A length is often a surd such as 20\sqrt{20}. Leave it exact if the question allows, or round to the requested accuracy (commonly 33 significant figures) if a decimal is asked for. Keep the value exact through the working and round only at the very end, so rounding does not build up errors.

Examples in context

Example 1. Checking a shape. To test whether three points form an isosceles triangle, find the length of each side with the distance formula and compare. If two sides come out equal, the triangle is isosceles. The distance formula turns a geometry question into pure arithmetic.

Example 2. The centre of a diameter. If a circle has a diameter from (1,2)(1, 2) to (7,8)(7, 8), its centre is the midpoint of that diameter: (1+72,2+82)=(4,5)\left( \dfrac{1 + 7}{2}, \dfrac{2 + 8}{2} \right) = (4, 5). The midpoint formula locates the centre directly, a useful link to circle work.

Try this

  • Cue. Find the length from (0,0)(0, 0) to (6,8)(6, 8). Compute 62+82=100=10\sqrt{6^2 + 8^2} = \sqrt{100} = 10.
  • Cue. Find the midpoint of (1,3)(1, 3) and (5,11)(5, 11). Average: (1+52,3+112)=(3,7)\left( \dfrac{1 + 5}{2}, \dfrac{3 + 11}{2} \right) = (3, 7).
  • Cue. Find the length from (2,1)(2, 1) to (2,6)(2, 6). The points share an xx, so the length is just 6−1=56 - 1 = 5.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original3 marksFind the length of the line segment joining A(1,2)A(1, 2) and B(4,6)B(4, 6).
Show worked answer →

Use the distance formula (from Pythagoras): length =(x2−x1)2+(y2−y1)2= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.

Differences: x2−x1=4−1=3x_2 - x_1 = 4 - 1 = 3 and y2−y1=6−2=4y_2 - y_1 = 6 - 2 = 4.

Length =32+42=9+16=25=5= \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5.

What markers reward: the correct distance formula, the horizontal and vertical differences squared, and the final length. This is a 33-44-55 triangle, so the answer is exactly 55.

Original2 marksFind the midpoint of the line segment joining P(2,7)P(2, 7) and Q(8,3)Q(8, 3).
Show worked answer →

The midpoint is the average of the coordinates:

midpoint =(x1+x22,y1+y22)=(2+82,7+32)=(5,5)= \left( \dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2} \right) = \left( \dfrac{2 + 8}{2}, \dfrac{7 + 3}{2} \right) = (5, 5).

What markers reward: averaging the xx values and the yy values separately, and giving the answer as a coordinate pair. A common slip is to subtract instead of add, which gives the wrong point.

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