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

The equation of the given line is

*x* – 3*y* + 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

⇒ 3*a* + *b* – 5 = 0 ......... (2)

Now M is the mid point of PQ

Putting this value in (1) we get

⇒ *a* – 3*b* + 3 = 0 ....... (3)

Solving (2) and (3) we get

Hence the required image of point (1, 2) is

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