Key Concepts and Formulas

• Reflection across the x-axis: The reflection of a point $(x, y)$ across the x-axis is $(x, -y)$.
• Equation of a line passing through two points: The equation of a line passing through points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\frac{y - y_1}{x - x_1} = \frac{y_2 - y_1}{x_2 - x_1}$.
• Midpoint Formula: The midpoint of a line segment with endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$.
• Properties of a Parallelogram: The diagonals of a parallelogram bisect each other.

Step-by-Step Solution

Step 1: Find the reflection of point P across the x-axis

Since the ray of light reflects off the x-axis at point Q, we can use the reflection principle. The reflection of point $P(1, 2)$ across the x-axis is $P'(1, -2)$.

Step 2: Find the equation of the line passing through P' and R

The points $P'(1, -2)$ and $R(4, 3)$ are collinear with $Q$. The equation of the line passing through $P'$ and $R$ is: $\frac{y - (-2)}{x - 1} = \frac{3 - (-2)}{4 - 1}$ $\frac{y + 2}{x - 1} = \frac{5}{3}$ $3(y + 2) = 5(x - 1)$ $3y + 6 = 5x - 5$ $5x - 3y = 11$

Step 3: Find the coordinates of point Q

Since point $Q$ lies on the x-axis, its y-coordinate is 0. Let $Q$ be $(x_Q, 0)$. Substituting $y = 0$ into the equation of the line $5x - 3y = 11$, we get: $5x_Q - 3(0) = 11$ $5x_Q = 11$ $x_Q = \frac{11}{5}$ Thus, $Q = \left(\frac{11}{5}, 0\right)$.

Step 4: Use the parallelogram property to find the coordinates of S

Since $PQRS$ is a parallelogram, the diagonals $PR$ and $QS$ bisect each other. Let $M$ be the midpoint of $PR$ and $QS$. The coordinates of $M$ are: $M = \left(\frac{1 + 4}{2}, \frac{2 + 3}{2}\right) = \left(\frac{5}{2}, \frac{5}{2}\right)$ Also, $M$ is the midpoint of $QS$. So, $M = \left(\frac{\frac{11}{5} + h}{2}, \frac{0 + k}{2}\right)$ Equating the coordinates of $M$, we have: $\frac{\frac{11}{5} + h}{2} = \frac{5}{2} \quad \text{and} \quad \frac{k}{2} = \frac{5}{2}$ Solving for $h$ and $k$: $\frac{11}{5} + h = 5$ $h = 5 - \frac{11}{5} = \frac{25 - 11}{5} = \frac{14}{5}$ $k = 5$ Thus, $S = (h, k) = \left(\frac{14}{5}, 5\right)$.

Step 5: Calculate hk²

We need to find the value of $hk^2$. $hk^2 = \left(\frac{14}{5}\right)(5^2) = \frac{14}{5} \cdot 25 = 14 \cdot 5 = 70$

Common Mistakes & Tips

• Incorrect Reflection: Ensure you reflect the point $P$ correctly across the x-axis. The y-coordinate changes sign, but the x-coordinate remains the same.
• Parallelogram Properties: Remember that the diagonals of a parallelogram bisect each other, which means they share the same midpoint. This is a key property for solving this problem.
• Simplifying Fractions: Pay close attention to simplifying fractions to avoid errors in the final calculation.

Summary

We used the reflection principle to find the image of point P across the x-axis. Then, we found the equation of the line passing through the reflected point and point R, which allowed us to find the coordinates of point Q. Finally, we used the properties of a parallelogram to find the coordinates of point S and calculated $hk^2$, which equals 70.

Final Answer

The final answer is \boxed{70}, which corresponds to option (B).