First slide

Cartesian plane

Question

In triangle ABC, angle B is right angle and AC=2. If coordinates of A and B are (2,2) and (1,3), respectively, then the lenght of median AD is

Moderate

A

$\frac{1}{2}$
B

$\sqrt{\frac{5}{2}}$
C

$\frac{5}{\sqrt{2}}$
D

$\frac{1}{\sqrt{2}}$

Solution

$AB = \sqrt{2}, AC = 2, BC = \sqrt{2}$
Using Apollonius Theoren, we get
$2 ({AD}^{2} + {BD}^{2}) = {AC}^{2} + {AB}^{2}$
$\begin{array}{l} ∴ 2 ({AD}^{2} + \frac{1}{2}) = 4 + 2 \\ ∴ {AD}^{2} + \frac{1}{2} = 3 \\ ∴ {AD}^{2} = \frac{5}{2} \\ ∴ AD = \sqrt{\frac{5}{2}} \end{array}$

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