CAIE IGCSE Additional Math 7.3 Solve Midpoint Length and Bisector Problems Questions

Practise Cambridge IGCSE Additional Mathematics by using midpoint, distance and perpendicular-bisector methods to solve exact coordinate geometry problems.

Exam points

Use midpoint and distance formulas to analyse points and line segments.
Find perpendicular bisectors, circle diameters and intersections using coordinate geometry.
Solve for exact lengths, coordinates and equations while checking all geometric conditions.

[Maximum number: 7]

4 The coordinates of points A, B, C and D are as follows.

A(-4,3)\quad B(6,-9)\quad C(15,10)\quad D(14,-1)

The line L has equation y=11 x-75 .

The perpendicular bisector of the line A B meets L at the point E .

Find the area of triangle C D E .

$m_{A B}=\frac{-9-3}{6--4}$ oe isw or $-\frac{6}{5}$ $\left[m_{\perp}=\right] \frac{5}{6}$ or $\frac{-1}{\text { their } m_{A B}}$
FT their $m_{A B}$ providing that
$\frac{\text { difference in } y}{\text { difference in } x}$ attempted
[Midpoint] $\left(\frac{-4+6}{2}, \frac{3-9}{2}\right)$ oe isw or (1,-3)
Finds equation of perpendicular bisector:
$y \text {-their }(-3)=-\frac{1}{\text { their }\left(-\frac{6}{5}\right)}(x-\text { their } 1) \mathrm{oe}$
FT their perpendicular gradient providing gradient is $\frac{-1}{\text { their } m_{A B}}$ and using their midpoint

Their midpoint must not be any of A, B, C or D $11 x-75=\frac{5}{6} x-\frac{23}{6}$
Eliminates one variable
dep on previous $\mathbf{M}$ mark
11 x-75= their perp bisector equation
E(7,2)
[Area C D E= ] 40
Dependent on all previous marks

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CAIE IGCSE Additional Math 7.3 Solve Midpoint Length and Bisector Problems Questions question 1