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# A ray of light incident along x-2y-3=0 and strikes a line mirror. If equation of normal on line mirror at the point of incidence is 2x+3y+1=0 then find the equation of reflected ray.

2075 Points
one year ago
hello pooja, we find the intersection of incident ray x-2y-3=0 and normal 2x+3y+1=0, which comes out to be
(1, – 1).
now, the reflected ray surely passes through (1, – 1). note that (3, 0) lies on x-2y-3=0. since the normal line bisects the angle b/w incident and reflected rays, so the image/reflection of (3, 0) about the normal shall lie on reflected ray.
let the image of (3, 0) about normal 2x+3y+1=0 be (m, n).
then, mid point of (3, 0) and (m, n) shall lie on normal, or (3+m)/2, (0+n)/2 or [(3+m)/2, n/2] lies on 2x+3y+1=0
so, 3+m+3n/2+1= 0
or 4+m+3n/2= 0
also, they are perpendicular so pdt of slopes= – 1
or (n – 0)/(m – 3)* (– 2/3)= – 1
or 2n= 3(m – 3)
solving we get.
m= 11/13, n= – 42/13
so, by point form of line we can write eqn of line passing thru (m, n) and (1, – 1) as
2y+31= 29x
kindly approve :)