Question

A ray of light coming from the point $P(1, 2)$ gets reflected from the point $Q$ on the x-axis and then passes through the point $R(4, 3)$. If the point $S(h, k)$ is such that $PQRS$ is a parallelogram, then $hk^2$ is equal to:

• A. 80
• B. 90
• C. 60
• D. 70

Answer 1

Answer: (D)

70

Answer 2

Step 1: Find the reflection point Q The ray reflects at point Q on the x -axis. Let the coordinates of Q be $(a, 0)$. Since the ray is reflected, Q lies on the x -axis, and the slope of $PQ$ is equal to the negative of the slope of $QR$.

The slope of $PQ$ is: $\text{slope of } PQ = \frac{2 - 0}{1 - a} = \frac{2}{1 - a}.$

The slope of $QR$ is: $\text{slope of } QR = \frac{3 - 0}{4 - a} = \frac{3}{4 - a}.$

By the law of reflection: $\frac{2}{1 - a} = -\frac{3}{4 - a}.$

Cross-multiply to solve for $a$: $2(4 - a) = -3(1 - a).$ $8 - 2a = -3 + 3a.$ $8 + 3 = 5a.$ $a = \frac{11}{5}.$

Thus, $Q$ is $\left(\frac{11}{5}, 0\right)$.

Step 2: Find the coordinates of $S$ The points $P(1, 2)$, $Q \left(\frac{11}{5}, 0\right)$, $R(4, 3)$, and $S(h, k)$ form a parallelogram. The diagonals of a parallelogram bisect each other, so the midpoint of $PR$ must equal the midpoint of $QS$.

The midpoint of $PR$ is: $\text{Midpoint of } PR = \left(\frac{1 + 4}{2}, \frac{2 + 3}{2}\right) = \left(\frac{5}{2}, \frac{5}{2}\right).$

The midpoint of $QS$ is: $\text{Midpoint of } QS = \left(\frac{\frac{11}{5} + h}{2}, \frac{0 + k}{2}\right).$

Equating the midpoints: $\frac{\frac{11}{5} + h}{2} = \frac{5}{2}, \quad \frac{k}{2} = \frac{5}{2}.$

Solve for $h$ and $k$: $\frac{11}{5} + h = 5 \implies h = 5 - \frac{11}{5} = \frac{25}{5} - \frac{11}{5} = \frac{14}{5}.$ $\frac{k}{2} = \frac{5}{2} \implies k = 5.$

Thus, $S$ is $\left(\frac{14}{5}, 5\right)$.

Step 3: Calculate $hk^2$ $hk^2 = \left(\frac{14}{5}\right)(5^2) = \frac{14}{5}(25) = 70.$

Final Answer: Option (4).