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In a $△^{l e} A B C x + y + 2 = 0$ is the perpendicular bisector of side AB and it meets AB at $(- 1, - 1)$ . If $x - y - 1 = 0$ is $⊥^{l a r}$ bisector of side AC and it meets AC at (2,1) and P is mid point of BC then distance of P from ortho centre of $△^{l e} A B C$ is $\sqrt{k}$ then k is

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$\sqrt{8}$

$7$

answer is C.

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Detailed Solution

Clearly $x + y + 2 = 0,$

$x - y - 1 = 0$ are perpendicular to each other

$∴ ∠ B A C = 90 °$

Mid point of BC = P = circum centre = $(\frac{- 1}{2}, \frac{- 3}{2})$

$∴ P A^{2} = D E^{2} = \sqrt{9 + 4} = \sqrt{13}$

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