Home / IGCSE Add Maths 0606 / Straight-line graphs / Midpoints, lengths and perpendicular bisectors
Cambridge IGCSE · Topic 7.3
IGCSE Add Maths 0606 / Topic 7 / IGCSE

Halfway, and at right angles.

Two points give you a midpoint, a length and a gradient. Put the midpoint and the perpendicular gradient together and you get the perpendicular bisector: the line of every point the same distance from both ends. It finds equidistant points, centres of circles, missing vertices of a rhombus and exact areas, all without a calculator.

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The key idea

For \(A(x_1, y_1)\) and \(B(x_2, y_2)\): the midpoint is \(M\left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)\) and the length is \[AB = \sqrt{(x_2 - x_1)^{2} + (y_2 - y_1)^{2}}\], left as an exact surd. The perpendicular bisector of \(AB\) passes through \(M\) with gradient \(-\dfrac{1}{m_{AB}}\); every point on it is equidistant from \(A\) and \(B\).

x y O −2 2 4 6 2 4 6 A(−2, 1) B(4, 5) M(1, 3) P(3, 0) AB, gradient 2/3 3x + 2y = 9 gradient −3/2 M = midpoint of AB AM = MB = √13 bisector ⊥ AB at M P is on the bisector, so PA = PB = √26
Fig. 1: the segment from A(-2, 1) to B(4, 5) has gradient 2/3 and midpoint M(1, 3), so AM = MB = √13. The perpendicular bisector passes through M with gradient -3/2, giving 3x + 2y = 9. Any point on it, such as P(3, 0), is the same distance from A and B: PA = PB = √26
Section 01

Midpoint and length.

The midpoint of \(AB\) is the average of the two \(x\)-coordinates and the average of the two \(y\)-coordinates. The length comes from Pythagoras: the horizontal step \(x_2 - x_1\) and the vertical step \(y_2 - y_1\) are the two shorter sides of a right-angled triangle with \(AB\) as its hypotenuse. On Paper 1 you have no calculator, so leave a length as a simplified surd such as \(3\sqrt{5}\). In the lab, drag the two points and watch the midpoint and the right-angled triangle behind the length formula change.

Worked example Midpoint and length together
\(A\) is the point \((-3, 4)\) and \(B\) is the point \((5, -2)\). Find the midpoint of \(AB\) and the length of \(AB\).
  1. Midpoint: \(\left(\dfrac{-3 + 5}{2}, \dfrac{4 + (-2)}{2}\right) = (1, 1)\).
  2. Steps: \(x_2 - x_1 = 5 - (-3) = 8\) and \(y_2 - y_1 = -2 - 4 = -6\).
  3. \(AB = \sqrt{8^{2} + (-6)^{2}} = \sqrt{64 + 36} = \sqrt{100}\).
Answer: Midpoint \((1, 1)\), \(AB = 10\)
Worked example Finding the other end
The midpoint of \(AB\) is \(M(2, -1)\) and \(A\) is \((-1, 3)\). Find the coordinates of \(B\).
  1. From \(A\) to \(M\): \(x\) goes up by \(3\) and \(y\) goes down by \(4\).
  2. Repeat the same step from \(M\): \(x = 2 + 3 = 5\), \(y = -1 - 4 = -5\).
  3. Check with the formula: \(\dfrac{-1 + 5}{2} = 2\) and \(\dfrac{3 + (-5)}{2} = -1\).
Answer: \(B(5, -5)\)
Your turn
Find the midpoint and the exact length of \(PQ\), where \(P\) is \((1, -2)\) and \(Q\) is \((7, 1)\).
Examiner trap

The midpoint ADDS the coordinates; it never subtracts them. And to find a missing endpoint, do not just average \(A\) and \(M\): that gives the point a quarter of the way along. With \(M(2, -1)\) and \(A(-1, 3)\), \(B\) is \((5, -5)\), not \(\left(\tfrac{1}{2}, 1\right)\). Bracket negatives in the length formula: \(5 - (-3) = 8\), not \(2\).

Section 02

The perpendicular bisector.

The perpendicular bisector of \(AB\) cuts \(AB\) in half (it passes through the midpoint \(M\)) and meets it at right angles (its gradient is the negative reciprocal of the gradient of \(AB\)). It is also the set of all points that are the same distance from \(A\) as from \(B\), so whenever a question says a point is equidistant from \(A\) and \(B\), or \(PA = PB\), that point lies on the perpendicular bisector (Fig. 1).

Worked example The bisector in Fig. 1
Find the equation of the perpendicular bisector of the line joining \(A(-2, 1)\) and \(B(4, 5)\), in the form \(ax + by = c\).
  1. Midpoint: \(M = \left(\dfrac{-2 + 4}{2}, \dfrac{1 + 5}{2}\right) = (1, 3)\).
  2. Gradient of \(AB\): \(\dfrac{5 - 1}{4 - (-2)} = \dfrac{4}{6} = \dfrac{2}{3}\), so the perpendicular gradient is \(-\dfrac{3}{2}\).
  3. \(y - 3 = -\dfrac{3}{2}(x - 1)\). Multiply by 2: \(2y - 6 = -3x + 3\).
  4. Rearrange: \(3x + 2y = 9\). Check \(M\): \(3 + 6 = 9\).
Answer: \(3x + 2y = 9\)
Worked example A point on a given line, equidistant from A and B
The point \(C\) lies on the line \(y = 2x - 6\) and is equidistant from \(A(1, 5)\) and \(B(7, 1)\). Find the coordinates of \(C\).
  1. \(C\) is equidistant from \(A\) and \(B\), so it lies on their perpendicular bisector.
  2. Midpoint \((4, 3)\); gradient of \(AB\) is \(\dfrac{1 - 5}{7 - 1} = -\dfrac{2}{3}\), so the perpendicular gradient is \(\dfrac{3}{2}\).
  3. \(y - 3 = \dfrac{3}{2}(x - 4)\), so \(2y - 6 = 3x - 12\), giving \(3x - 2y = 6\).
  4. Substitute \(y = 2x - 6\): \(3x - 4x + 12 = 6\), so \(x = 6\) and \(y = 6\).
  5. Check: \(CA^{2} = 5^{2} + 1^{2} = 26\) and \(CB^{2} = 1^{2} + 5^{2} = 26\).
Answer: \(C(6, 6)\)
Your turn
Using Fig. 1, find the point on the \(x\)-axis that is equidistant from \(A(-2, 1)\) and \(B(4, 5)\).
Examiner trap

The bisector passes through the MIDPOINT, not through \(A\) or \(B\), and its gradient is the NEGATIVE reciprocal. With \(m_{AB} = \dfrac{2}{3}\), the perpendicular gradient is \(-\dfrac{3}{2}\), not \(\dfrac{3}{2}\) (sign forgotten), not \(-\dfrac{2}{3}\) (not flipped) and not \(\dfrac{2}{3}\) (that is \(AB\) itself).

Stage 1 · Learn

Check your understanding

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Find the midpoint of the line joining \((-4, 7)\) and \((6, -1)\).

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Find the length of the line joining \((2, -1)\) and \((5, 3)\).

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\(A\) is \((1, 2)\) and \(B\) is \((5, 10)\). What is the gradient of the perpendicular bisector of \(AB\)?

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\(M(3, -2)\) is the midpoint of \(AB\), where \(A\) is \((-1, 4)\). Find \(B\).

Section 03

Rhombuses, areas and centres.

The diagonals of a rhombus bisect each other at right angles, so each diagonal lies on the perpendicular bisector of the other. In a kite only one diagonal does this: the line of symmetry is the perpendicular bisector of the other diagonal. For an area, use \(\tfrac{1}{2} \times d_1 \times d_2\) for a rhombus or kite, \(\tfrac{1}{2} \times \text{base} \times \text{height}\) when you know a perpendicular height as a length, or the shoelace method for any triangle or polygon. Finally, the centre of a circle is equidistant from every point on it, so it lies on the perpendicular bisector of every chord: two bisectors meet at the centre (circles are covered fully in topic 8).

Worked example Completing a rhombus
\(ABCD\) is a rhombus with \(A(1, 1)\) and \(C(7, 5)\). The vertex \(B\) lies on the line \(y = -3\). Find the coordinates of \(B\) and \(D\) and the area of the rhombus.
  1. \(BD\) is the perpendicular bisector of \(AC\). Midpoint of \(AC\): \(M(4, 3)\); gradient of \(AC\) is \(\dfrac{4}{6} = \dfrac{2}{3}\), so \(BD\) has gradient \(-\dfrac{3}{2}\).
  2. \(y - 3 = -\dfrac{3}{2}(x - 4)\), so \(3x + 2y = 18\). With \(y = -3\): \(3x = 24\), \(x = 8\), so \(B(8, -3)\).
  3. \(M\) is also the midpoint of \(BD\): \(D = (2 \times 4 - 8,\ 2 \times 3 - (-3)) = (0, 9)\).
  4. \(AC = \sqrt{6^{2} + 4^{2}} = \sqrt{52}\) and \(BD = \sqrt{8^{2} + 12^{2}} = \sqrt{208}\).
  5. Area \(= \tfrac{1}{2}\sqrt{52}\sqrt{208} = \tfrac{1}{2}\sqrt{10816} = \tfrac{1}{2} \times 104 = 52\). (Or \(\sqrt{52} = 2\sqrt{13}\), \(\sqrt{208} = 4\sqrt{13}\), so \(\tfrac{1}{2} \times 2\sqrt{13} \times 4\sqrt{13} = 52\).)
Answer: \(B(8, -3)\), \(D(0, 9)\), area \(52\) square units
Worked example Centre of a circle through three points (preview)
A circle passes through \(A(-1, 3)\), \(B(5, 3)\) and \(C(3, -1)\). Find its centre and its radius.
  1. \(AB\) is horizontal with midpoint \((2, 3)\), so its perpendicular bisector is \(x = 2\).
  2. \(BC\): midpoint \((4, 1)\), gradient \(\dfrac{-1 - 3}{3 - 5} = 2\), so the bisector has gradient \(-\dfrac{1}{2}\): \(y - 1 = -\dfrac{1}{2}(x - 4)\).
  3. At \(x = 2\): \(y = 1 - \dfrac{1}{2}(-2) = 2\). The centre is \((2, 2)\).
  4. Radius: distance to \(A\) is \(\sqrt{3^{2} + 1^{2}} = \sqrt{10}\); to \(B\) and \(C\) it is also \(\sqrt{10}\).
Answer: Centre \((2, 2)\), radius \(\sqrt{10}\)
Your turn
Find the area of the triangle with vertices \((1, 2)\), \((7, 4)\) and \((3, 8)\).
Examiner trap

Do not forget the \(\tfrac{1}{2}\): the rhombus area is HALF the product of the diagonals, and the shoelace total must be halved too. In base \(\times\) height, the height must be PERPENDICULAR to the base: a slanted side is not a height. And a kite is not a rhombus: only its symmetry diagonal bisects the other one.

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Find the equation of the perpendicular bisector of the line joining \(P(-1, 5)\) and \(Q(5, 1)\).

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Find the point on the \(y\)-axis that is equidistant from \(A(2, 1)\) and \(B(6, 5)\).

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The points \(A(1, 1)\), \(B(5, 3)\) and \(C(3, 7)\) form a triangle with a right angle at \(B\). Find its area.

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\(ABCD\) is a rhombus with \(A(-1, 2)\) and \(C(5, 6)\). Find the equation of the diagonal \(BD\).

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Six original Cambridge-style questions
Midpoint and exact length, a missing endpoint, a perpendicular bisector and its intercept, an equidistant point with an exact area, a rhombus from two vertices, and the centre of a circle through three points.