From the point $$(-1, -1)$$, two rays are sent making angle  of $$45°$$ with the line $$x + y = 0$$. The rays get reflected from the mirror $$x + 2y = 1$$. If the equations of the reflected rays are $$ax + by = 9$$ and $$cx + dy = 7$$,$$a,b,c,d \in \mathbb{Z}$$ thenthe value of  $$ad + bc$$ is :

Let $$P' (x_i, y_i)$$ be the image of $$P(-1, -1)$$ with respect to the mirror $$x + 2y - 1 = 0$$:

$$\frac{x_i - (-1)}{1} = \frac{y_i - (-1)}{2} = -2\frac{1(-1) + 2(-1) - 1}{1^2 + 2^2}$$

$$\frac{x_i + 1}{1} = \frac{y_i + 1}{2} = -2\frac{-4}{5} = \frac{8}{5}$$

So, the image point is $$P'\left(\frac{3}{5}, \frac{11}{5}\right)$$. Both reflected rays must pass through $$P'$$

For Ray 1 ($$x = -1$$): $$-1 + 2y = 1 \implies y = 1$$. Point of incidence $$Q_1 = (-1, 1)$$.

For Ray 2 ($$y = -1$$): $$x + 2(-1) = 1 \implies x = 3$$. Point of incidence $$Q_2 = (3, -1)$$.

Reflected Ray 1 (through $$Q_1(-1, 1)$$ and $$P'$$):

$$\text{Slope } m_1 = \frac{\frac{11}{5} - 1}{\frac{3}{5} - (-1)} = \frac{\frac{6}{5}}{\frac{8}{5}} = \frac{3}{4}$$

$$-3x + 4y = 7$$

$$c = -3, d = 4$$

Reflected Ray 2 (through $$Q_2(3, -1)$$ and $$P'$$):

$$\text{Slope } m_2 = \frac{\frac{11}{5} - (-1)}{\frac{3}{5} - 3} = \frac{\frac{16}{5}}{-\frac{12}{5}} = -\frac{4}{3}$$

$$\implies 4x + 3y = 9$$

$$a = 4, b = 3$$

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