Intersection point of AO and XY : 3(2y+3)−2y−5=0[∵x=2y+3] ⇒6y+9−2y−5=0⇒4y+4=0 ⇒y=−1 and x=2(−1)+3 x=1 ∴ Coordinate of O is (1,−1). Slope of line XY=3∕2 Slope of line BO=−
2
3
[∵BO is perpendicular to XO] Equation of reflected ray. y+1=m(x−1), where m is slope of line y=m(x−1)−1 ∠AOB=∠COB[∵ law of reflection] |
1
2
−(−
2
3
)
1+(
1
2
)+(−
2
3
)
|=|
−
2
3
−m
1+(−
2
3
)m
| ⇒
7
6
×
6
5
=|
−2−3m
3−2m
|⇒
2+3m
3−2m
=±
7
5
If
2+3m
3−2m
=
7
5
⇒10+15m=21−14m ⇒29m=11⇒m=
11
29
and y=
11
29
(x−1)−1 ⇒29y=11x−11−1⇒29y=11x−12 ⇒11x−29y−12=0 If