If Q is the image of the point P(1, 1) with respect to the straight line x + y + 1 = 0, then the length of the perpendicular drawn from Q to the line 3x-4y + 3 =0 is

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If Q is the image of the point P(1, 1) with respect to the straight line $x + y + 1 = 0$ , then the length of the perpendicular drawn from Q to the line $3 x - 4 y + 3 = 0$ is

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$\frac{5}{2}$

$\frac{1}{2}$

answer is C.

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Detailed Solution

Let Q (h.k) be the image of P(1, 1) $\Rightarrow \frac{h -1}{1} = \frac{k-1}{1} = \frac{-2 (1+1+1)}{1+1}$

$∴ Q = (h, k) = (-2, -2)$

$⊥^{r}$ distance from (-2, -2) to 3x – 4y + 3 = 0 is $\frac{|-6 + 8 + 3|}{\sqrt{9 + 16}}$ = 1

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