Question:

Step 1: Determine the image of point $P$ reflected across the x-axis. When a ray of light from point $P(x_1, y_1)$ reflects off the x-axis (the line $y=0$), the reflected ray appears to come from the image of $P$ with respect to the x-axis. Let $P'$ be the image of $P(1,2)$. The coordinates of the image of a point $(x,y)$ reflected across the x-axis are $(x, -y)$. $$P'(1, -2)$$ Step 2: Find the equation of the line passing through the image point $P'$ and point $R$. The reflected ray passes through $Q$ on the x-axis and then through $R(4,3)$. This means $P'$, $Q$, and $R$ are collinear. We find the equation of the line passing through $P'(1,-2)$ and $R(4,3)$. The slope $m$ of the line is given by: $$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - (-2)}{4 - 1} = \frac{5}{3}$$ Using the point-slope form $y - y_1 = m(x - x_1)$ with $P'(1,-2)$: $$y - (-2) = \frac{5}{3}(x - 1)$$ $$3(y + 2) = 5(x - 1)$$ $$3y + 6 = 5x - 5$$ $$5x - 3y - 11 = 0$$ Step 3: Determine the coordinates of point $Q$. Point $Q$ is the intersection of the line $P'R$ and the x-axis. On the x-axis, $y=0$. Substitute $y=0$ into the equation of the line $P'R$: $$5x - 3(0) - 11 = 0$$ $$5x - 11 = 0$$ $$x = \frac{11}{5}$$ Thus, the coordinates of point $Q$ are $\left(\frac{11}{5}, 0\right)$. Step 4: Determine the coordinates of point $S(h,k)$ using the parallelogram property. For a parallelogram $PQRS$, the diagonals bisect each other. This means the midpoint of $PR$ must be the same as the midpoint of $QS$. Given points: $P(1,2)$, $Q\left(\frac{11}{5}, 0\right)$, $R(4,3)$, $S(h,k)$. Midpoint of $PR$: $$M_{PR} = \left(\frac{1+4}{2}, \frac{2+3}{2}\right) = \left(\frac{5}{2}, \frac{5}{2}\right)$$ Midpoint of $QS$: $$M_{QS} = \left(\frac{\frac{11}{5} + h}{2}, \frac{0 + k}{2}\right) = \left(\frac{11+5h}{10}, \frac{k}{2}\right)$$ Equating the coordinates of the midpoints: $$\frac{11+5h}{10} = \frac{5}{2} \implies 11+5h = 25 \implies 5h = 14 \implies h = \frac{14}{5}$$ $$\frac{k}{2} = \frac{5}{2} \implies k = 5$$ So, the coordinates of point $S$ are $\left(\frac{14}{5}, 5\right)$. Step 5: Calculate the value of $hk^2$. Substitute the values of $h = \frac{14}{5}$ and $k = 5$ into the expression $hk^2$: $$hk^2 = \left(\frac{14}{5}\right) (5^2)$$ $$hk^2 = \left(\frac{14}{5}\right) (25)$$ $$hk^2 = 14 \times 5$$ $$hk^2 = 70$$ The final answer is $\boxed{\text{70}}$.