The image of A(1, -2) with respect to the str | vedclass

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(C) For point $B$,the image of $A(1, -2)$ with respect to $2x - 3y + 5 = 0$ is given by $\frac{x - 1}{2} = \frac{y + 2}{-3} = -2 \frac{2(1) - 3(-2) +

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$(-1, 1)$

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(C) For point $B$,the image of $A(1, -2)$ with respect to $2x - 3y + 5 = 0$ is given by $\frac{x - 1}{2} = \frac{y + 2}{-3} = -2 \frac{2(1) - 3(-2) + 5}{2^2 + (-3)^2} = -2 \frac{2 + 6 + 5}{4 + 9} = -2 \frac{13}{13} = -2$. Thus,$x - 1 = -4 \Rightarrow x = -3$ and $y + 2 = 6 \Rightarrow y = 4$. So,$B \equiv (-3, 4)$. The equation of line $AB$ passing through $A(1, -2)$ and $B(-3, 4)$ is $y - (-2) = \frac{4 - (-2)}{-3 - 1}(x - 1)$ $\Rightarrow y + 2 = \frac{6}{-4}(x - 1)$ $\Rightarrow 2(y + 2) = -3(x - 1)$ $\Rightarrow 3x + 2y + 1 = 0$. The foot of the perpendicular from $P(-4, -1)$ onto $3x + 2y + 1 = 0$ is $R(x, y)$,given by $\frac{x - (-4)}{3} = \frac{y - (-1)}{2} = - \frac{3(-4) + 2(-1) + 1}{3^2 + 2^2} = - \frac{-12 - 2 + 1}{13} = - \frac{-13}{13} = 1$. Thus,$x + 4 = 3 \Rightarrow x = -1$ and $y + 1 = 2 \Rightarrow y = 1$. Therefore,$R \equiv (-1, 1)$.

The image of $A(1, -2)$ with respect to the straight line $L \equiv 2x - 3y + 5 = 0$ is $B$. The foot of the perpendicular from $P(-4, -1)$ onto the line joining $AB$ is:

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