From the point (-1, -1), two rays are sent making angles of 45° with the line x + y = 0. These rays get reflected from the mirror x + 2y = 1. If the…

The given line is $x + y = 0$x+y=0, which has a slope of $m_1 = -1$m1​=−1.

Let the slope of an incident ray be $m$m. Since it makes an angle of $45^{\circ}$45∘ with the line $x + y = 0$x+y=0, we have:

$$\left| \dfrac{m - (-1)}{1 + m(-1)} \right| = \tan 45^{\circ} = 1$$​1+m(−1)m−(−1)​​=tan45∘=1

$$\Rightarrow \dfrac{m + 1}{1 - m} = \pm 1$$⇒1−mm+1​=±1

Taking $+1$+1: $m + 1 = 1 - m \Rightarrow 2m = 0 \Rightarrow m = 0$m+1=1−m⇒2m=0⇒m=0

Taking $-1$−1: $m + 1 = -1 + m \Rightarrow 1 = -1$m+1=−1+m⇒1=−1 (which is not possible, implying $m \to \infty$m→∞).

The incident rays pass through $P(-1, -1)$P(−1,−1). Thus, their equations are:

Ray 1: $y = -1$y=−1

Ray 2: $x = -1$x=−1

Next, we find the points of intersection of these incident rays with the mirror $x + 2y = 1$x+2y=1.

For Ray 1 ($y = -1$y=−1): $x + 2(-1) = 1 \Rightarrow x = 3$x+2(−1)=1⇒x=3. The point of incidence is $A(3, -1)$A(3,−1).

For Ray 2 ($x = -1$x=−1): $-1 + 2y = 1 \Rightarrow 2y = 2 \Rightarrow y = 1$−1+2y=1⇒2y=2⇒y=1. The point of incidence is $B(-1, 1)$B(−1,1).

The reflected rays will appear to originate from the image of $P(-1, -1)$P(−1,−1) in the mirror $x + 2y - 1 = 0$x+2y−1=0. Let the image be $P'(x', y')$P′(x′,y′).

$$\dfrac{x' - (-1)}{1} = \dfrac{y' - (-1)}{2} = -2 \dfrac{1(-1) + 2(-1) - 1}{1^2 + 2^2}$$1x′−(−1)​=2y′−(−1)​=−212+221(−1)+2(−1)−1​

$$\dfrac{x' + 1}{1} = \dfrac{y' + 1}{2} = -2 \left(\dfrac{-4}{5}\right) = \dfrac{8}{5}$$1x′+1​=2y′+1​=−2(5−4​)=58​

$\Rightarrow x' = \dfrac{8}{5} - 1 = \dfrac{3}{5}$⇒x′=58​−1=53​ and $y' = \dfrac{16}{5} - 1 = \dfrac{11}{5}$y′=516​−1=511​

So, $P' = \left(\dfrac{3}{5}, \dfrac{11}{5}\right)$P′=(53​,511​).

The reflected rays pass through $P'$P′ and their respective points of incidence.

Equation of Reflected Ray 1 (passing through $A$A and $P'$P′):

Slope $m_1' = \dfrac{\dfrac{11}{5} - (-1)}{\dfrac{3}{5} - 3} = \dfrac{\dfrac{16}{5}}{-\dfrac{12}{5}} = -\dfrac{4}{3}$m1′​=53​−3511​−(−1)​=−512​516​​=−34​

$$y - (-1) = -\dfrac{4}{3}(x - 3) \Rightarrow 3y + 3 = -4x + 12 \Rightarrow 4x + 3y = 9$$y−(−1)=−34​(x−3)⇒3y+3=−4x+12⇒4x+3y=9

Comparing with $ax + by = 9$ax+by=9, we get $a = 4, b = 3$a=4,b=3.

Equation of Reflected Ray 2 (passing through $B$B and $P'$P′):

Slope $m_2' = \dfrac{\dfrac{11}{5} - 1}{\dfrac{3}{5} - (-1)} = \dfrac{\dfrac{6}{5}}{\dfrac{8}{5}} = \dfrac{3}{4}$m2′​=53​−(−1)511​−1​=58​56​​=43​

$$y - 1 = \dfrac{3}{4}(x - (-1)) \Rightarrow 4y - 4 = 3x + 3 \Rightarrow 3x - 4y = -7 \Rightarrow -3x + 4y = 7$$y−1=43​(x−(−1))⇒4y−4=3x+3⇒3x−4y=−7⇒−3x+4y=7

Comparing with $cx + dy = 7$cx+dy=7, we get $c = -3, d = 4$c=−3,d=4.

Finally, we calculate $ad + bc$ad+bc:

$$ad + bc = (4)(4) + (3)(-3) = 16 - 9 = 7$$ad+bc=(4)(4)+(3)(−3)=16−9=7

Answer: $7$7