The given line is x+y=0, which has a slope of m1=−1.
Let the slope of an incident ray be m. Since it makes an angle of 45∘ with the line x+y=0, we have:
1+m(−1)m−(−1)=tan45∘=1
⇒1−mm+1=±1
Taking +1: m+1=1−m⇒2m=0⇒m=0
Taking −1: m+1=−1+m⇒1=−1 (which is not possible, implying m→∞).
The incident rays pass through P(−1,−1). Thus, their equations are:
Ray 1: y=−1
Ray 2: x=−1
Next, we find the points of intersection of these incident rays with the mirror x+2y=1.
For Ray 1 (y=−1): x+2(−1)=1⇒x=3. The point of incidence is A(3,−1).
For Ray 2 (x=−1): −1+2y=1⇒2y=2⇒y=1. The point of incidence is B(−1,1).
The reflected rays will appear to originate from the image of P(−1,−1) in the mirror x+2y−1=0. Let the image be P′(x′,y′).
1x′−(−1)=2y′−(−1)=−212+221(−1)+2(−1)−1
1x′+1=2y′+1=−2(5−4)=58
⇒x′=58−1=53 and y′=516−1=511
So, P′=(53,511).
The reflected rays pass through P′ and their respective points of incidence.
Equation of Reflected Ray 1 (passing through A and P′):
Slope m1′=53−3511−(−1)=−512516=−34
y−(−1)=−34(x−3)⇒3y+3=−4x+12⇒4x+3y=9
Comparing with ax+by=9, we get a=4,b=3.
Equation of Reflected Ray 2 (passing through B and P′):
Slope m2′=53−(−1)511−1=5856=43
y−1=43(x−(−1))⇒4y−4=3x+3⇒3x−4y=−7⇒−3x+4y=7
Comparing with cx+dy=7, we get c=−3,d=4.
Finally, we calculate ad+bc:
ad+bc=(4)(4)+(3)(−3)=16−9=7
Answer: 7