Find the image of the point (1,2) in the line x-3y+4 =0
The equation of the given line is
x – 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
Since PQ ⊥ AB
∴ Slope of the PQ
⇒ 3a + b – 5 = 0 ......... (2)
Now M is the mid point of PQ
Putting this value in (1) we get
⇒ a – 3b + 3 = 0 ....... (3)
Solving (2) and (3) we get
Hence the required image of point (1, 2) is