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Question Number 77514 by liki last updated on 07/Jan/20

	Answered by jagoll last updated on 07/Jan/20

	$L_{1} : y = - \frac{3}{4} (x - 2) + (- 1)$ $L_{1} : y = - \frac{3}{4} x + \frac{1}{2} \Rightarrow a = \frac{1}{2}$ $l e t p o i n C (x, y), C A p e r p e n d i c u l a r$ $t o L_{1} \Rightarrow y - \frac{1}{2} = \frac{4}{3} x$ $y = \frac{4}{3} x + \frac{1}{2}$
	Commented by liki last updated on 07/Jan/20

	$. T h a n k s a l o t s i r .$
	Answered by Kunal12588 last updated on 07/Jan/20

	$l e t c o o r d i n a t e s o f p o i n t C b e (h, k)$ $L_{1} : 3 x + 4 y = 2$ $∵ A (0, a) s a t i s f y L_{1}$ $\Rightarrow a = \frac{1}{2}$ $C A : 6 y - 8 x = 3$ $L_{2} : x = 2$ $∵ C (h, k) s a t i s f y C A a n d L_{2}$ $∴ 6 k - 8 h = 3 a n d h = 2$ $\Rightarrow k = \frac{19}{6}$ $p o i n t C (2, \frac{19}{6})$
	Commented by liki last updated on 07/Jan/20

	$. T h a n k s s o m u c h s i r .$
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