AP ICET 2014

Question 140

A (4, 2), B (6, 5) and C (1, 4) are the vertices of a $$\triangle$$ABC. The median from A meets the side BC at D. Then $$2AD^2 =$$

Solution

D is the midpoint of BC. Hence coordinates of D is (3.5,4.5)

Now applying distance formula , AD is $$\sqrt{\ \left(0.5\right)^2+\left(2.5\right)^2}=\sqrt{\ 0.25+6.25}=\sqrt{\ 6.5}$$

$$2AD^2=2\times\ 6.5=13$$


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