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:

Options

A. 90
B. 80
C. 70
D. 60

Answer

To solve the problem step by step, we will follow the geometric properties of the points involved and the properties of a parallelogram.

### Step 1: Identify Points and Reflection
We have the point \( P(1, 2) \) from which a ray of light is coming. It reflects off a point \( Q \) on the x-axis and then passes through the point \( R(4, 3) \). The coordinates of point \( Q \) can be represented as \( Q(x, 0) \).

### Step 2: Find the Reflection Point
The reflection of point \( P \) across the x-axis will be \( P'(1, -2) \). This is because the x-coordinate remains the same while the y-coordinate changes sign.

### Step 3: Find the Slope of Line \( P'R \)
Next, we need to find the slope of the line connecting \( P' \) and \( R \):
- Coordinates of \( P' \) are \( (1, -2) \)
- Coordinates of \( R \) are \( (4, 3) \)

The slope \( m \) of line \( P'R \) is calculated as:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - (-2)}{4 - 1} = \frac{5}{3} \]

### Step 4: Equation of Line \( P'R \)
Using the point-slope form of the equation of a line, we can write the equation of line \( P'R \):
\[ y - (-2) = \frac{5}{3}(x - 1) \]
This simplifies to:
\[ y + 2 = \frac{5}{3}x - \frac{5}{3} \]
\[ y = \frac{5}{3}x - \frac{5}{3} - 2 \]
\[ y = \frac{5}{3}x - \frac{11}{3} \]

### Step 5: Find Intersection with the x-axis
To find point \( Q \), we set \( y = 0 \) in the equation of line \( P'R \):
\[ 0 = \frac{5}{3}x - \frac{11}{3} \]
Solving for \( x \):
\[ \frac{5}{3}x = \frac{11}{3} \]
\[ x = \frac{11}{5} \]
Thus, the coordinates of point \( Q \) are \( Q\left(\frac{11}{5}, 0\right) \).

### Step 6: Use Parallelogram Properties
Since \( PQRS \) is a parallelogram, the midpoints of the diagonals \( PR \) and \( QS \) must be the same.

The midpoint of \( PR \) is:
\[ \left(\frac{1 + 4}{2}, \frac{2 + 3}{2}\right) = \left(\frac{5}{2}, \frac{5}{2}\right) \]

Let the coordinates of point \( S \) be \( S(h, k) \). The midpoint of \( QS \) is:
\[ \left(\frac{\frac{11}{5} + h}{2}, \frac{0 + k}{2}\right) \]

### Step 7: Set Midpoints Equal
Setting the midpoints equal gives us the following equations:
1. \( \frac{\frac{11}{5} + h}{2} = \frac{5}{2} \)
2. \( \frac{k}{2} = \frac{5}{2} \)

From the second equation:
\[ k = 5 \]

From the first equation:
\[ \frac{11}{5} + h = 5 \implies h = 5 - \frac{11}{5} = \frac{25}{5} - \frac{11}{5} = \frac{14}{5} \]

### Step 8: Calculate \( hk^2 \)
Now we have \( h = \frac{14}{5} \) and \( k = 5 \). We need to find \( hk^2 \):
\[ hk^2 = \left(\frac{14}{5}\right)(5^2) = \left(\frac{14}{5}\right)(25) = \frac{14 \times 25}{5} = 14 \times 5 = 70 \]

### Final Answer
Thus, the value of \( hk^2 \) is \( \boxed{70} \).