Assuming that straight lines work as the plane mirror for a point, find the image of the point (1, 2) in the line `x -3y + 4 = 0`.

Explanation

Let `D(h,k)` be the image of point `C(1,2)` in line `x-3y+4=0`
Let `x-3y+4=0` be the equation of line `AB`.
Slope of `AB=-(1)/(-3)=(1)/(3)`
`:.` Slope of `CD=-(1)/((1)/(3))=-3`

and equation of `Cd`
`y-2=-3(x-1)`
`implies y-2=-3x+3`
`implies3x+y=5`
On solving equation of `AB`, `x-3y+4=0` and equation of `CD`, `3x+y=5`, the point of intersection `M((11)/(10), (17)/(10))`
Now, `M` is the mid-point of `CD`
`:. ((h+1)/(2),(k+2)/(2))=((11)/(10),(17)/(10))`
`(h+1)/(2)=(11)/(10)impliesh=6//5`
and `(k+2)/(2)=(17)/(10)`
`implies k=(7)/(5)`
Now, the image of point `(1,2)` in the straight line is `((6)/(5),(7)/(5))`