The image of the point $P(3,8)$ with respect to the line $x+3 y=7$, assuming the line to be a plane mirror, is equal to .........

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Aicharya · Accepted Answer

Let image is $P^{\prime}(h, k)$ Then $P P^{\prime}$ is perpendicular to given line and mid point of $P P^{\prime}$ lies over line. So, we have $\begin{aligned} & & \left(\frac{8-k}{3-h}\right) \times \frac{-1}{3} & =-1 \ \Rightarrow & & 8-k & =3(3-h) \quad \ldots (i) \ & \text {And, } \quad & \frac{3+h}{2}+3\left(\frac{8+k}{2}\right) & =7 \ \Rightarrow & & 3+h+24+3 k & =14 \quad \ldots (ii) \ \Rightarrow & & h+3 k+13 & =0 \end{aligned}$ $\Rightarrow$ $h=-1 \text { and } k=-4$ So image is, $(-1,-4)$.

The image of the point \(P(3,8)\) with respect to the line \(x+3 y=7\), assuming the line to be a plane…

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