A ray of light passing through the point A(2, 3) reflected at a point B on line `x + y = 0` and then passes through (5, 3). Then the coordinates of B are

Options

A. `((1)/(3),-(1)/(3))`
B. `((2)/(5),-(2)/(5))`
C. `((1)/(13),-(1)/(13))`
D. none of these

Explanation

To solve the problem, we need to find the coordinates of point B where the ray of light reflects on the line \(x + y = 0\) after passing through point A(2, 3) and then through point (5, 3).

### Step-by-Step Solution:

1. **Identify the points and line**:
- Point A: \(A(2, 3)\)
- Point C: \(C(5, 3)\)
- Line: \(x + y = 0\) (which can also be expressed as \(y = -x\))

2. **Find the image of point A with respect to the line**:
- The line \(x + y = 0\) has a slope of -1. The image of point A across this line can be found using the reflection formula.
- The coordinates of the image \(A'\) can be calculated as follows:
- The perpendicular slope to the line \(x + y = 0\) is 1 (the negative reciprocal of -1).
- The equation of the line through A(2, 3) with slope 1 is:
\[ y - 3 = 1(x - 2) \implies y = x + 1 \]
- To find the intersection of this line with \(x + y = 0\):
\[ x + (x + 1) = 0 \implies 2x + 1 = 0 \implies x = -\frac{1}{2} \]
\[ y = -(-\frac{1}{2}) = \frac{1}{2} \]
- The intersection point is \((-0.5, 0.5)\). Now, using this intersection point, we can find the coordinates of the image \(A'\):
\[ A' = (2 - 2(-0.5), 3 - 2(0.5)) = (2 + 1, 3 - 1) = (3, 2) \]

3. **Find the equation of line A'C**:
- Now we need to find the equation of the line passing through points \(A'(3, 2)\) and \(C(5, 3)\).
- The slope \(m\) of line \(A'C\) is:
\[ m = \frac{3 - 2}{5 - 3} = \frac{1}{2} \]
- Using point-slope form:
\[ y - 2 = \frac{1}{2}(x - 3) \]
\[ y - 2 = \frac{1}{2}x - \frac{3}{2} \implies y = \frac{1}{2}x + \frac{1}{2} \]

4. **Find the intersection of line A'C with line \(x + y = 0\)**:
- Set the equations equal to find point B:
\[ \frac{1}{2}x + \frac{1}{2} + x = 0 \]
\[ \frac{3}{2}x + \frac{1}{2} = 0 \implies 3x + 1 = 0 \implies x = -\frac{1}{3} \]
- Substitute \(x\) back into \(y = -x\):
\[ y = -(-\frac{1}{3}) = \frac{1}{3} \]
- Therefore, the coordinates of point B are:
\[ B\left(-\frac{1}{3}, \frac{1}{3}\right) \]

### Final Answer:
The coordinates of point B are \(\left(-\frac{1}{3}, \frac{1}{3}\right)\).