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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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Verified Answer
The correct answer is:
\((-1,-4)\)
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, \((-1,-4)\).
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, \((-1,-4)\).
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