Reflected Rays from a Line Mirror and Value of Coefficients

To find the equations of the reflected rays, we first need to find the equations of the incident rays originating from the point $P(-1, -1)$.

Step 1: Finding the equations of the incident rays

The given line is $x + y = 0$, which has a slope of $m_0 = -1$.
Let the slope of an incident ray be $m$. The angle between the incident ray and the line $x + y = 0$ is $45^\circ$. Using the formula for the angle between two lines: $\tan 45^\circ = \left| \frac{m - (-1)}{1 + m(-1)} \right|$ $1 = \left| \frac{m + 1}{1 - m} \right|$

This gives two possibilities:

1. $\frac{m + 1}{1 - m} = 1 \implies m + 1 = 1 - m \implies 2m = 0 \implies m_1 = 0$
2. $\frac{m + 1}{1 - m} = -1 \implies m + 1 = m - 1 \implies 1 = -1$ (which implies $m_2 = \infty$, i.e., a vertical line)

Since both incident rays pass through $P(-1, -1)$:

• Incident Ray 1: $y = -1$
• Incident Ray 2: $x = -1$

---

Step 2: Finding the points of incidence on the mirror

The equation of the mirror is $x + 2y = 1$.

• For Incident Ray 1 ($y = -1$): $x + 2(-1) = 1 \implies x = 3$ So, the point of incidence is $A(3, -1)$.

• For Incident Ray 2 ($x = -1$): $-1 + 2y = 1 \implies 2y = 2 \implies y = 1$ So, the point of incidence is $B(-1, 1)$.

---

Step 3: Finding the image of point $P$ in the mirror line

By the reflection property of light, the reflected rays produced backwards pass through the image of the point source $P(-1, -1)$ in the mirror line $x + 2y - 1 = 0$.

Let $P'(x', y')$ be the reflection of $P(-1, -1)$ in $x + 2y - 1 = 0$. Using the reflection formula: $\frac{x' - (-1)}{1} = \frac{y' - (-1)}{2} = \frac{-2(1(-1) + 2(-1) - 1)}{1^2 + 2^2}$ $\frac{x' + 1}{1} = \frac{y' + 1}{2} = \frac{-2(-4)}{5} = \frac{8}{5}$

Solving for $x'$ and $y'$: $x' + 1 = \frac{8}{5} \implies x' = \frac{3}{5}$ $y' + 1 = \frac{16}{5} \implies y' = \frac{11}{5}$

Thus, $P' = \left( \frac{3}{5}, \frac{11}{5} \right)$.

---

Step 4: Finding the equations of the reflected rays

1. Reflected Ray 1: Passes through $A(3, -1)$ and $P'\left(\frac{3}{5}, \frac{11}{5}\right)$. The slope is: $m_{R1} = \frac{\frac{11}{5} - (-1)}{\frac{3}{5} - 3} = \frac{\frac{16}{5}}{-\frac{12}{5}} = -\frac{4}{3}$ The equation of the line is: $y - (-1) = -\frac{4}{3}(x - 3) \implies 3(y + 1) = -4(x - 3)$ $4x + 3y = 9$ Comparing with $ax + by = 9$, we get $a = 4$ and $b = 3$.

2. Reflected Ray 2: Passes through $B(-1, 1)$ and $P'\left(\frac{3}{5}, \frac{11}{5}\right)$. The slope is: $m_{R2} = \frac{\frac{11}{5} - 1}{\frac{3}{5} - (-1)} = \frac{\frac{6}{5}}{\frac{8}{5}} = \frac{3}{4}$ The equation of the line is: $y - 1 = \frac{3}{4}(x + 1) \implies 4(y - 1) = 3(x + 1)$ $3x - 4y = -7 \implies -3x + 4y = 7$ Comparing with $cx + dy = 7$, we get $c = -3$ and $d = 4$.

---

Step 5: Calculating the value of $ad + bc$

Given $a = 4, b = 3, c = -3, d = 4$: $ad + bc = (4)(4) + (3)(-3) = 16 - 9 = 7$