If P^{prime}(a, b) is the image of the point | vedclass

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(C) The image $(h, k)$ of a point $P(x_1, y_1)$ with respect to the line $ax+by+c=0$ is given by the formula: $\frac{h-x_1}{a} = \frac{k-y_1}{b} = -2

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$\frac{7}{\sqrt{5}}$

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(C) The image $(h, k)$ of a point $P(x_1, y_1)$ with respect to the line $ax+by+c=0$ is given by the formula: $\frac{h-x_1}{a} = \frac{k-y_1}{b} = -2 \frac{ax_1+by_1+c}{a^2+b^2}$.Substituting the values $P(-1, 2)$ and the line $x-2y+3=0$ into the formula:$\frac{a-(-1)}{1} = \frac{b-2}{-2} = -2 \frac{1(-1) - 2(2) + 3}{1^2 + (-2)^2} = -2 \frac{-1-4+3}{5} = -2 \frac{-2}{5} = \frac{4}{5}$.Thus,$a+1 = \frac{4}{5} \implies a = -\frac{1}{5}$ and $\frac{b-2}{-2} = \frac{4}{5} \implies b-2 = -\frac{8}{5} \implies b = \frac{2}{5}$.The point $P^{\prime}$ is $(-\frac{1}{5}, \frac{2}{5})$.The length of the perpendicular from $P^{\prime}$ to the line $2x+y-7=0$ is given by $d = \frac{|2(-\frac{1}{5}) + \frac{2}{5} - 7|}{\sqrt{2^2 + 1^2}} = \frac{|-\frac{2}{5} + \frac{2}{5} - 7|}{\sqrt{5}} = \frac{|-7|}{\sqrt{5}} = \frac{7}{\sqrt{5}}$.

If $P^{\prime}(a, b)$ is the image of the point $P(-1, 2)$ with respect to the line $x-2y+3=0$,then the length of the perpendicular from $P^{\prime}$ onto the line $2x+y-7=0$ is

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