Question:

Step 1: Identify the nature of the equation. The given equation $3y^2 - 4xy + 4x + 2y - 5 = 0$ represents a pair of straight lines, which correspond to the incident and reflected rays. Step 2: Express the equation as a quadratic in $y$. Rearranging the terms, we obtain a quadratic equation in $y$: $$3y^2 + (-4x + 2)y + (4x - 5) = 0$$ This is in the form $ay^2 + by + c = 0$, where $a=3$, $b=(-4x+2)$, and $c=(4x-5)$. Step 3: Apply the quadratic formula to solve for $y$. Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $$y = \frac{-( -4x + 2) \pm \sqrt{(-4x + 2)^2 - 4(3)(4x - 5)}}{2(3)}$$ $$y = \frac{(4x - 2) \pm \sqrt{(16x^2 - 16x + 4) - (48x - 60)}}{6}$$ $$y = \frac{(4x - 2) \pm \sqrt{16x^2 - 16x + 4 - 48x + 60}}{6}$$ $$y = \frac{(4x - 2) \pm \sqrt{16x^2 - 64x + 64}}{6}$$ $$y = \frac{(4x - 2) \pm \sqrt{16(x^2 - 4x + 4)}}{6}$$ $$y = \frac{(4x - 2) \pm 4\sqrt{(x - 2)^2}}{6}$$ $$y = \frac{(4x - 2) \pm 4(x - 2)}{6}$$ Step 4: Determine the equations of the two lines and their slopes. We consider the two cases from the $\pm$ sign: Case 1: Using the positive sign $$y = \frac{(4x - 2) + 4(x - 2)}{6} = \frac{4x - 2 + 4x - 8}{6} = \frac{8x - 10}{6} = \frac{4x - 5}{3}$$ This line has a slope $m_1 = \frac{4}{3}$. Case 2: Using the negative sign $$y = \frac{(4x - 2) - 4(x - 2)}{6} = \frac{4x - 2 - 4x + 8}{6} = \frac{6}{6} = 1$$ This line has a slope $m_2 = 0$. Step 5: Identify the reflected ray and its slope. The two lines represented by the equation are $y = \frac{4x-5}{3}$ and $y=1$. The problem states that the ray is reflected on the line $y=1$. The line $y=1$ itself represents the mirror. A ray traveling along the mirror is considered a trivial case. Therefore, the reflected ray is the other line, $y = \frac{4x-5}{3}$. The slope of the reflected ray is $\frac{4}{3}$. When the slope is expressed as a simplified fraction $\frac{p}{q}$, we have $p=4$ and $q=3$. The required value is $p+q = 4+3=7$.