The image of $3 x - 4 y + 2 = 0$ with respect to $21 x + 27 y - 16 = 0$ is

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$4 x + 3 y - 5 = 0$

$5 x + 12 y - 2 = 0$

$12 x - 5 y + 2 = 0$

$3 x + 4 y - 5 = 0$

answer is C.

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Detailed Solution

The image of the line $p x + q y + r = 0$ with respect to the line $a x + b y + c = 0$ is

$(a^{2} + b^{2}) (p x + q y + r) - 2 (a p + b q) (a x + b y + c) = 0$

$p = 3, q = - 4, r = 2 & a = 21, b = 27, c = - 16$

$[{(21)}^{2} + {(27)}^{2}] (3 x - 4 y + 2) - 2 (21 \times 3 + 27 \times (- 4)) (21 x + 27 y - 16) = 0]$

$1170 (3 x - 4 y + 2) - 2 (- 45) + (21 x + 27 y - 16) = 0$

$90 [13 (3 x - 4 y + 2) + (21 x + 27 y - 16)] = 0$

$\begin{array}{l} 39 x - 52 y + 26 + 21 x + 27 y - 16 = 0 \\ 60 x - 25 y + 10 = 0 \\ 12 x - 5 y + 2 = 0 \end{array}$

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