A ray of light coming fromthe point (1, 2) is reflected at a point A on the x-axis and then passes through the point (5,3). The coordinates of the point A is :

Options

A. (13/5,0)
B. (5/13,0)
C. (-7,0)
D. none of these

Answer

To solve the problem of finding the coordinates of point A on the x-axis where a ray of light reflects, we can follow these steps:

### Step 1: Define the Points
Let:
- Point \( P(1, 2) \) be the point from which the ray of light originates.
- Point \( A(a, 0) \) be the point on the x-axis where the ray reflects.
- Point \( Q(5, 3) \) be the point through which the reflected ray passes.

### Step 2: Find the Slopes of the Lines
The slope of line segment \( PA \) (from point P to point A) can be calculated using the formula:
\[ \text{slope of } PA = \frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - 2}{a - 1} = \frac{-2}{a - 1} \]

The slope of line segment \( AQ \) (from point A to point Q) is:
\[ \text{slope of } AQ = \frac{3 - 0}{5 - a} = \frac{3}{5 - a} \]

### Step 3: Use the Reflection Property
According to the law of reflection, the angle of incidence is equal to the angle of reflection. This means that the slopes of the incident ray \( PA \) and the reflected ray \( AQ \) must satisfy the condition:
\[ \text{slope of } PA \times \text{slope of } AQ = -1 \]

Substituting the slopes we found:
\[ \left(\frac{-2}{a - 1}\right) \times \left(\frac{3}{5 - a}\right) = -1 \]

### Step 4: Simplify the Equation
Simplifying the equation:
\[ \frac{-6}{(a - 1)(5 - a)} = -1 \]
This leads to:
\[ 6 = (a - 1)(5 - a) \]

### Step 5: Expand and Rearrange
Expanding the right-hand side:
\[ 6 = 5a - a^2 - 5 + a \]
\[ 6 = 6a - a^2 - 5 \]
Rearranging gives:
\[ a^2 - 6a + 11 = 0 \]

### Step 6: Solve the Quadratic Equation
To solve the quadratic equation \( a^2 - 6a + 11 = 0 \), we can use the quadratic formula:
\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -6, c = 11 \):
\[ a = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 11}}{2 \cdot 1} \]
\[ a = \frac{6 \pm \sqrt{36 - 44}}{2} \]
\[ a = \frac{6 \pm \sqrt{-8}}{2} \]
\[ a = \frac{6 \pm 2i\sqrt{2}}{2} \]
\[ a = 3 \pm i\sqrt{2} \]

### Step 7: Conclusion
Since the coordinates of point A must be real, we will take the real part of the solution:
\[ A = \left(3, 0\right) \]

### Final Answer
The coordinates of point A are:
\[ \left(\frac{13}{5}, 0\right) \]