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 $P'(x_1, y_1)$.Since $Q$ is the midpoint of $PP'$,its coordinates are $Q = \left(\frac{x_1 + 4}{2}, \fra

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$(-1, -14)$

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(A) Let the image of the point $P(4, -13)$ be $P'(x_1, y_1)$.Since $Q$ is the midpoint of $PP'$,its coordinates are $Q = \left(\frac{x_1 + 4}{2}, \frac{y_1 - 13}{2}\right)$.Since $Q$ lies on the line $5x + y + 6 = 0$,we have:$5\left(\frac{x_1 + 4}{2}\right) + \left(\frac{y_1 - 13}{2}\right) + 6 = 0$$5x_1 + 20 + y_1 - 13 + 12 = 0$$5x_1 + y_1 + 19 = 0$ $\ldots$ $(i)$Since the line $PP'$ is perpendicular to the line $5x + y + 6 = 0$ (which has slope $-5$),the slope of $PP'$ is $\frac{1}{5}$.Thus,$\frac{y_1 - (-13)}{x_1 - 4} = \frac{1}{5}$$5(y_1 + 13) = x_1 - 4$$x_1 - 5y_1 - 69 = 0$ $\ldots$ (ii)Solving equations $(i)$ and (ii):Multiply $(i)$ by $5$: $25x_1 + 5y_1 + 95 = 0$Add to (ii): $(25x_1 + 5y_1 + 95) + (x_1 - 5y_1 - 69) = 0$$26x_1 + 26 = 0 \Rightarrow x_1 = -1$Substitute $x_1 = -1$ into $(i)$: $5(-1) + y_1 + 19 = 0 \Rightarrow y_1 = -14$.Thus,the image is $(-1, -14)$.

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

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