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.

Answer 1

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).