Find the image of the point (1,2) in the line x-3y+4 =0 - Maths - Straight Lines

Question

Find the image of the point (1,2) in the line x-3y+4 =0

Aakash Sharma · Accepted Answer

The equation of the given line is x â€“ 3y + 4 = 0 (1) Let Q (a, b) be the image of the point (1, 2) in the mirror line (1) Then PQ is perpendicularly bisected at same point M Now Slope of line (1) i e AB is given as $M_{1} = \frac{- 1}{- 3} = \frac{1}{3}$ Since PQ âŠ¥ AB âˆ´ Slope of the PQ $(M_{2}) = \frac{- 1}{\frac{1}{3}} = - 3$ $\Rightarrow \frac{b - 2}{a - 1} = - \frac{3}{1} \Rightarrow b - 2 = - 3 (a - 1) = - 3 a + 3$ â‡’ 3a + b â€“ 5 = 0 (2) Now M is the mid point of PQ $∴ Coordinates of M = (\frac{a + 1}{2}, \frac{b + 2}{2})$ Putting this value in (1) we get $\frac{a + 1}{2} - 3 (\frac{b + 2}{2}) + 4 = 0 \Rightarrow \frac{a + 1 - 3 b - 6 + 8}{2} = 0$ â‡’ a â€“ 3b + 3 = 0 (3) Solving (2) and (3) we get $a = \frac{6}{5} and b = \frac{7}{5}$ Hence the required image of point (1, 2) is $(\frac{6}{5}, \frac{7}{5})$