If the point of intersection of $k x + 4 y + 2 = 0, x - 3 y + 5 = 0$ lies on $2 x + 7 y - 3 = 0$ then k=

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Solution

The correct option is A 2
$\begin{matrix} K x + 4 y + 2 = 0 \to (i) x - 3 y + 5 = 0 \to (i i) B y c r o s s m u l t i p l i c a t i o n m e t h o d \Rightarrow \frac{x}{26} = \frac{y}{(2 - 5 k)} = \frac{1}{- 3 k - x} \Rightarrow x = \frac{26}{(- 3 k - 4)} \Rightarrow a n d y = \frac{2 - 5 k}{(- 3 k - 4)} P u t t h e v a l u e o f x a n d y i n e q u a t i o n 2 x + 7 y - 3 = 0 \Rightarrow \frac{2 \times 26}{(- 3 k - 4)} + \frac{(14 - 35 k)}{- 3 k - 4} - 3 = 0 \Rightarrow 52 + 14 - 35 k + 9 k + 12 = 0 \Rightarrow - 26 k + 78 = 0 \Rightarrow K = \frac{78}{26} \Rightarrow K = \frac{39}{13} \Rightarrow K = 3 A n s . \end{matrix}$

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