Find the image of the point (-1, 3) with resp | vedclass

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(D) Let the point be $P(-1, 3)$ and the line be $L: x - y + 1 = 0$. Let the image of $P$ be $P'(x', y')$.Using the formula for the image of a point $(

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$(2, 0)$

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(D) Let the point be $P(-1, 3)$ and the line be $L: x - y + 1 = 0$. Let the image of $P$ be $P'(x', y')$.Using the formula for the image of a point $(x_1, y_1)$ with respect to the line $ax + by + c = 0$:$\frac{x' - x_1}{a} = \frac{y' - y_1}{b} = -2 \frac{ax_1 + by_1 + c}{a^2 + b^2}$Here,$a = 1, b = -1, c = 1, x_1 = -1, y_1 = 3$.$\frac{x' - (-1)}{1} = \frac{y' - 3}{-1} = -2 \frac{1(-1) + (-1)(3) + 1}{1^2 + (-1)^2}$$\frac{x' + 1}{1} = \frac{y' - 3}{-1} = -2 \frac{-1 - 3 + 1}{1 + 1}$$\frac{x' + 1}{1} = \frac{y' - 3}{-1} = -2 \frac{-3}{2} = 3$Now,$x' + 1 = 3 \implies x' = 2$.And $\frac{y' - 3}{-1} = 3 \implies y' - 3 = -3 \implies y' = 0$.Thus,the image point is $(2, 0)$.

Find the image of the point $(-1, 3)$ with respect to the line $x - y + 1 = 0$.

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