The problem asks us to find the value of $\text{ad}+\text{bc}$ based on the equations of two reflected rays.
Initial Ray Determination
The initial point is P = $(-1, -1)$. The line $L$ is $x+y=0$, with slope $m_L = -1$. The rays originate from P and make an angle of $45^{\circ}$ with $L$. The angle $135^{\circ}$ is the angle $L$ makes with the positive x-axis. The angles the rays make with the positive x-axis are $135^{\circ} \pm 45^{\circ}$, which are $180^{\circ}$ and $90^{\circ}$.
- Ray 1: Angle $180^{\circ}$, slope $m_{inc1} = \tan(180^{\circ}) = 0$. The equation passing through $(-1, -1)$ is $y - (-1) = 0(x - (-1))$, which simplifies to $y+1=0$.
- Ray 2: Angle $90^{\circ}$, slope is undefined (vertical line). The equation passing through $(-1, -1)$ is $x = -1$.
Mirror and Reflection Calculation
The mirror line is $M: x+2y=1$. Its slope is $m_m = -1/2$.
Reflection of Ray 1 ($y+1=0$)
The point of incidence $(x_i, y_i)$ lies on $y+1=0$ and $x+2y=1$. Substituting $y_i=-1$ into the mirror equation gives $x_i + 2(-1) = 1$, so $x_i=3$. The point of incidence is $(3, -1)$.
Let $m_{ref1}$ be the slope of the reflected ray. Using the reflection formula for slopes:
$ \frac{m_{ref1} - m_m}{1 + m_{ref1} m_m} = \frac{m_m - m_{inc1}}{1 + m_m m_{inc1}} $
$ \frac{m_{ref1} - (-1/2)}{1 + m_{ref1}(-1/2)} = \frac{-1/2 - 0}{1 + (-1/2)(0)} $
$ \frac{m_{ref1} + 1/2}{1 - m_{ref1}/2} = -1/2 $
$ 2(m_{ref1} + 1/2) = -(1 - m_{ref1}/2) $
$ 2m_{ref1} + 1 = -1 + m_{ref1}/2 $
$ (3/2)m_{ref1} = -2 \implies m_{ref1} = -4/3 $
The equation of the reflected ray passing through $(3, -1)$ is:
$ y - (-1) = -4/3 (x - 3) $
$ y+1 = -4/3 x + 4 $
$ 3(y+1) = -4x + 12 $
$ 4x + 3y = 9 $
Comparing this with $ax+by=9$, we get $a=4$ and $b=3$. These are integers.
Reflection of Ray 2 ($x=-1$)
The point of incidence $(x_i, y_i)$ lies on $x=-1$ and $x+2y=1$. Substituting $x_i=-1$ gives $-1 + 2y_i = 1$, so $y_i=1$. The point of incidence is $(-1, 1)$.
The incident ray is vertical. Let the reflected slope be $m_{ref2}$. We can use the reflection formula involving the normal vector $\vec{n}=(1, 2)$ of the mirror and the incident direction $\vec{d}_{inc}=(0, 1)$ (or $(0, -1)$). Using the formula $\vec{d}_{ref} = \vec{d}_{inc} - 2 \text{proj}_{\vec{n}} \vec{d}_{inc}$, we find the direction vector of the reflected ray is proportional to $(-4/5, -3/5)$ or $(4/5, 3/5)$, giving a slope $m_{ref2} = 3/4$.
The equation of the reflected ray passing through $(-1, 1)$ is:
$ y - 1 = 3/4 (x - (-1)) $
$ y - 1 = 3/4 (x+1) $
$ 4(y-1) = 3(x+1) $
$ 4y - 4 = 3x + 3 $
$ 3x - 4y = -7 $
To match the form $cx+dy=7$, we multiply by $-1$: $-3x + 4y = 7$.
Comparing, we get $c=-3$ and $d=4$. These are integers.
Final Calculation
We have $a=4, b=3, c=-3, d=4$. We need to find $\text{ad}+\text{bc}$.
$ \text{ad}+\text{bc} = (4)(4) + (3)(-3) $
$ \text{ad}+\text{bc} = 16 - 9 $
$ \text{ad}+\text{bc} = 7 $
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