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

To solve this problem, we understand that a ray of light from point \( P(1, 2) \) is reflected at point \( Q \) on the x-axis and then passes through point \( R(4, 3) \). We need to find point \( S(h, k) \) such that \( PQRS \) forms a parallelogram, and subsequently calculate \( hk^2 \).

1. Since point \( Q \) lies on the x-axis, let \( Q \) be \((x, 0)\).
2. The slope of line \( PR \) is calculated as: 
   \(\text{slope of } PR = \frac{3 - 2}{4 - 1} = \frac{1}{3}\)
3. Using the concept of reflection, the angle of incidence is equal to the angle of reflection. Hence, the slope of line \( PQ \) is negative reciprocal of the slope of \( QR \).
4. The slope of \( QR \) is: 
   \(\text{slope of } QR = \frac{3 - 0}{4 - x}\)
5. Since \( PQ \) and \( QR \) form equal angles at \( Q \), we have: 
   \(\frac{2 - 0}{1 - x} = -\left(\frac{3 - 0}{4 - x}\right)\)
6. Cross-multiplying gives: 
   \(2(4 - x) = -3(1 - x)\) 
   \(8 - 2x = -3 + 3x\) 
   Solving for \( x \): \(5x = 11 \Rightarrow x = \frac{11}{5}\)
7. Thus, \( Q \) is \(\left(\frac{11}{5}, 0\right)\).
8. For \( PQRS \) to be a parallelogram, vector \( PS \) should be equal to vector \( QR \).
9. Vector \( QR = (4 - \frac{11}{5}, 3 - 0) = \left(\frac{9}{5}, 3\right)\).
10. \( PS \) should also be \(\left(\frac{9}{5}, 3\right)\), thus \( S(h, k) \) is given by: 
    \(h = 1 + \frac{9}{5} = \frac{14}{5},\) 
    \(k = 2 + 3 = 5\)
11. Finally, calculate \( hk^2 \): 
    \(hk^2 = \left(\frac{14}{5}\right) \times 5^2 = \frac{14}{5} \times 25 = 70\)

Thus, the value of \( hk^2 \) is 70, which corresponds to the correct option.

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).