$In a triangle ABC, if A (2, - 1) and 7 x - 10 y + 1 = 0 and 3 x - 2 y + 5 = 0 are equations of an$ $a l t i t u d e a n d$
$a n a n g l e b i s e c t o r r e s p e c t i v e l y d r a w n f r o m B, t h e n e q u a t i o n o f B C i s$

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$5 x + y + 17 = 0$

$4 x + 9 y + 30 = 0$

$x - 5 y - 7 = 0$

$x + y + 1 = 0$

answer is B.

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Detailed Solution

$B D a n d B E a r e i n t e r s e c t a t B .$

$∴ C o - o r d i n a t e s o f B a r e$ $(- 3, - 2)$

$A l s o m_{A B} = 1 / 5 ∴ \tan θ = |\frac{\frac{3}{2} - \frac{1}{5}}{1 + \frac{3}{10}}| = |\frac{\frac{3}{2} - m}{1 + \frac{3 m}{2}}|$

$\Rightarrow 1 = |\frac{3 - 2 m}{2 + 3 m}| or \pm 1 = \frac{3 - 2 m}{2 + 3 m}$

$\Rightarrow m = 1 / 5 (rejected) or - 5$

$∴ E q u a t i o n o f B C i s y + 2 = - 5 (x + 3) \Rightarrow 5 x + y + 17 = 0$

$Alternatively: Take image of (2, - 1) in the line BD to get a point on BC .$

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