POP

E3.4 Length and Midpoint of a Line Segment

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01 — Explanation

How do I find the length of a line segment?

The segment joining $A(x_1,y_1)$ and $B(x_2,y_2)$ is the hypotenuse of a right-angled triangle whose legs are the horizontal and vertical distances between the points. Pythagoras' theorem then gives:

AB = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}

02 — Explanation

How do I find the midpoint of a line segment?

The midpoint is simply the average of the two $x$ -coordinates and the average of the two $y$ -coordinates:

M = \left( \frac{x_1+x_2}{2}, \ \frac{y_1+y_2}{2} \right)

Think of it as "meeting in the middle" on each axis separately.

03 — Explanation

How do I tell these two formulas apart under exam pressure?

Length gives you a single number (a distance, always positive, uses square root and squaring). Midpoint gives you a coordinate pair (an $(x,y)$ point, uses only addition and halving — never a square root). If your midpoint answer has a square root in it, you've used the wrong formula.

04 — Explanation

Should my length answer be a surd, a decimal, or a whole number?

If the value under the square root is a perfect square (e.g. $\sqrt{169}=13$ ), give the exact whole-number answer. Otherwise, unless the question specifically says "leave your answer as a surd" or "in exact form", round your decimal answer to 3 significant figures — this is the standard IGCSE rounding convention for a length that doesn't simplify neatly.

05 — Explanation

How do these formulas apply beyond a single pair of points?

Many exam questions use length and midpoint as building blocks inside a bigger problem — for example, finding the length of every side of a triangle plotted on coordinate axes to test whether it's isosceles, or finding the midpoint of a diagonal to help locate a fourth vertex of a parallelogram (since the diagonals of a parallelogram share the same midpoint). Recognising when a wordier geometry question is secretly asking for a length or a midpoint calculation is a key exam skill in itself.

Exam tip

Sketch the two points on a quick mental (or scrap) grid before you commit to a formula — subtracting the coordinates in the wrong order won't change the length (it gets squared away), but it corrupts a midpoint answer instantly if you're not careful with negative signs.

06 — Worked example

Worked example

$P(-3, 8)$ and $Q(9, -2)$ . Find (a) the length of $PQ$ , and (b) the midpoint of $PQ$ .

(a) Length

PQ = \sqrt{(9-(-3))^2 + (-2-8)^2} = \sqrt{12^2 + (-10)^2} = \sqrt{144+100} = \sqrt{244} \approx 15.6 \text{ (3 s.f.)}

(b) Midpoint

M = \left( \frac{-3+9}{2}, \ \frac{8+(-2)}{2} \right) = \left( \frac{6}{2}, \frac{6}{2} \right) = (3, 3)

07 — Worked example

Worked example: using the midpoint of a diagonal

$ABCD$ is a parallelogram with $A(-1, 2)$ , $B(5, 4)$ and $C(7, -2)$ . Find the coordinates of $D$ .

The diagonals $AC$ and $BD$ of a parallelogram share the same midpoint.
Midpoint of $AC = \left( \dfrac{-1+7}{2}, \dfrac{2+(-2)}{2} \right) = (3, 0)$ .
Let $D = (x, y)$ . The midpoint of $BD$ must also equal $(3, 0)$ : $\left( \dfrac{5+x}{2}, \dfrac{4+y}{2} \right) = (3,0)$ .
Solve each coordinate separately: $5+x = 6 \implies x=1$ ; $4+y = 0 \implies y = -4$ . So $D = (1, -4)$ .