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).