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Question
Find the image of the point (2, 1) with respect to the line mirror x + y − 5 = 0.
Solution
Let the image of A (2, 1) be B (a, b). Let M be the midpoint of AB.
\[\therefore \text { Coordinates of M are } \equiv \left( \frac{2 + a}{2}, \frac{1 + b}{2} \right)\]
The point M lies on the line x + y − 5 = 0
\[\therefore \frac{2 + a}{2} + \frac{1 + b}{2} - 5 = 0\]
\[\Rightarrow a + b = 7\] ... (1)
Now, the lines x + y − 5 = 0 and AB are perpendicular.
∴ Slope of AB \[\times\] Slope of CD = −1
\[\Rightarrow \frac{b - 1}{a - 2} \times \left( - 1 \right) = - 1\]
\[ \Rightarrow a - 2 = b - 1\]
⇒ \[a - b = 1\] ... (2)
Adding eq (1) and eq (2):
\[2a = 8\]
\[ \Rightarrow a = 4\]
Now, from equation (1):
\[4 + b = 7\]
\[ \Rightarrow b = 3\]
Hence, the image of the point (2, 1) with respect to the line mirror x + y − 5 = 0 is (4, 3).
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