The image of the point (4, -13) with respect | vedclass

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(A) Let the image of the point $P(4, -13)$ be $P^{\prime}(x_1, y_1)$.The line $AB$ is $5x + y + 6 = 0$.The formula for the image $(x_1, y_1)$ of a poi

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$(-1, -14)$

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(A) Let the image of the point $P(4, -13)$ be $P^{\prime}(x_1, y_1)$.The line $AB$ is $5x + y + 6 = 0$.The formula for the image $(x_1, y_1)$ of a point $(x_0, y_0)$ with respect to the line $ax + by + c = 0$ is given by:$\frac{x_1 - x_0}{a} = \frac{y_1 - y_0}{b} = -2 \frac{ax_0 + by_0 + c}{a^2 + b^2}$Substituting the values $x_0 = 4, y_0 = -13, a = 5, b = 1, c = 6$:$\frac{x_1 - 4}{5} = \frac{y_1 - (-13)}{1} = -2 \frac{5(4) + 1(-13) + 6}{5^2 + 1^2}$$\frac{x_1 - 4}{5} = \frac{y_1 + 13}{1} = -2 \frac{20 - 13 + 6}{25 + 1}$$\frac{x_1 - 4}{5} = \frac{y_1 + 13}{1} = -2 \frac{13}{26}$$\frac{x_1 - 4}{5} = \frac{y_1 + 13}{1} = -1$Now,solving for $x_1$ and $y_1$:$\frac{x_1 - 4}{5} = -1$ $\Rightarrow x_1 - 4 = -5$ $\Rightarrow x_1 = -1$$\frac{y_1 + 13}{1} = -1$ $\Rightarrow y_1 + 13 = -1$ $\Rightarrow y_1 = -14$Thus,the image of the point is $(-1, -14)$.

The image of the point $(4, -13)$ with respect to the line $5x + y + 6 = 0$ is

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