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Question
Find the image of the point (3, 8) with respect to the line x + 3y = 7 assuming the line to be a plane mirror.
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Solution
Let the image of A (3,8) be B (a,b). Also, let M be the midpoint of AB.
\[\therefore\text { Coordinates of M } = \left( \frac{3 + a}{2}, \frac{8 + b}{2} \right)\]
Point M lies on the line x + 3y = 7
\[\therefore \frac{3 + a}{2} + 3 \times \left( \frac{8 + b}{2} \right) = 7\]
\[\Rightarrow a + 3b + 13 = 0\] ... (1)
Lines CD and AB are perpendicular.
∴ Slope of AB \[\times\] Slope of CD = −1
\[\Rightarrow \frac{b - 8}{a - 3} \times \left( - \frac{1}{3} \right) = - 1\]
\[ \Rightarrow b - 8 = 3a - 9\]
\[\Rightarrow 3a - b - 1 = 0\] ... (2)
Solving (1) and (2) by cross multiplication, we get:
\[\frac{a}{- 3 + 13} = \frac{b}{39 + 1} = \frac{1}{- 1 - 9}\]
\[ \Rightarrow a = - 1, b = - 4\]
Hence, the image of the point (3, 8) with respect to the line mirror x + 3y = 7 is (−1, −4).
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