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$(2 + k) x + (1 + k) y = 5 + 7 k$ for different values of $k$ pass through a fixed point

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$(2, 8)$

$(- 2, 9)$

$(3, 4)$

$(7, - 2)$

answer is B.

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Detailed Solution

$(2 + k) x + (1 + k) y = 5 + 7 k$

$2 x + k x + y + k y - 5 - 7 k = 0$

$2 x + y - 5 + k (x + y - 7) = 0$

$x = - 2$ sub in above any equation

$\begin{array}{l} 2 (- 2) + y - 5 = 0 \\ - 4 + y - 5 = 0 \\ y - 9 = 0 \\ y = 9 \end{array}$

$∴$ Fixed point $= (- 2, 9)$

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