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

The Correct Option is D

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).