A ray of light travels along the line `2x-3y+5=0` and strikes a plane mirror lying along the line `x+y=2`. The equation of the straight line containing the reflected ray is

Options

A. `2x-3y+3=0`
B. `3x-2y+3=0`
C. `21x-7y+1=0`
D. `21x+7y-1=0`

Answer

To solve the problem of finding the equation of the reflected ray after a ray of light strikes a plane mirror, we can follow these steps:

### Step 1: Find the intersection point of the incident ray and the mirror
We need to solve the equations of the incident ray and the mirror to find their intersection point.

1. The equation of the incident ray is:
\[ 2x - 3y + 5 = 0 \]
Rearranging gives:
\[ 3y = 2x + 5 \quad \Rightarrow \quad y = \frac{2}{3}x + \frac{5}{3} \]

2. The equation of the mirror is:
\[ x + y - 2 = 0 \quad \Rightarrow \quad y = -x + 2 \]

3. To find the intersection point, we set the two equations for \(y\) equal:
\[ \frac{2}{3}x + \frac{5}{3} = -x + 2 \]

4. Multiplying through by 3 to eliminate the fraction:
\[ 2x + 5 = -3x + 6 \]

5. Rearranging gives:
\[ 5x = 1 \quad \Rightarrow \quad x = \frac{1}{5} \]

6. Substituting \(x = \frac{1}{5}\) back into the equation of the mirror to find \(y\):
\[ y = -\frac{1}{5} + 2 = \frac{9}{5} \]

Thus, the intersection point \(A\) is:
\[ A\left(\frac{1}{5}, \frac{9}{5}\right) \]

### Step 2: Find the slope of the incident ray
The slope \(m_1\) of the incident ray can be derived from its equation:
\[ 2x - 3y + 5 = 0 \quad \Rightarrow \quad 3y = 2x + 5 \quad \Rightarrow \quad y = \frac{2}{3}x + \frac{5}{3} \]
Thus, the slope \(m_1 = \frac{2}{3}\).

### Step 3: Find the slope of the mirror
The slope \(m_2\) of the mirror can be derived from its equation:
\[ x + y - 2 = 0 \quad \Rightarrow \quad y = -x + 2 \]
Thus, the slope \(m_2 = -1\).

### Step 4: Use the reflection formula to find the slope of the reflected ray
Using the formula for the slopes of the incident ray and reflected ray:
\[ \tan(\theta_i) = \tan(\theta_r) \]
Where:
\[ \tan(\theta_i) = \frac{m_1 - m_2}{1 + m_1 m_2} \]
Let \(m\) be the slope of the reflected ray. Then:
\[ \tan(\theta_r) = \frac{m - m_2}{1 + m m_2} \]

Setting the two expressions equal:
\[ \frac{\frac{2}{3} - (-1)}{1 + \frac{2}{3}(-1)} = \frac{m - (-1)}{1 + m(-1)} \]

Calculating the left-hand side:
\[ \frac{\frac{2}{3} + 1}{1 - \frac{2}{3}} = \frac{\frac{5}{3}}{\frac{1}{3}} = 5 \]

Setting this equal to the right-hand side:
\[ 5 = \frac{m + 1}{1 - m} \]

Cross-multiplying gives:
\[ 5(1 - m) = m + 1 \quad \Rightarrow \quad 5 - 5m = m + 1 \]
\[ 5 - 1 = 6m \quad \Rightarrow \quad 4 = 6m \quad \Rightarrow \quad m = \frac{2}{3} \]

### Step 5: Find the equation of the reflected ray
Now that we have the slope of the reflected ray \(m = \frac{3}{2}\) and the point \(A\left(\frac{1}{5}, \frac{9}{5}\right)\), we can use the point-slope form of the line equation:
\[ y - y_1 = m(x - x_1) \]
Substituting the values:
\[ y - \frac{9}{5} = \frac{3}{2}\left(x - \frac{1}{5}\right) \]

Multiplying through by 10 to eliminate fractions:
\[ 10y - 18 = 15x - 3 \quad \Rightarrow \quad 15x - 10y + 15 = 0 \]

Thus, the equation of the reflected ray is:
\[ 3x - 2y + 3 = 0 \]

### Final Answer
The equation of the straight line containing the reflected ray is:
\[ 3x - 2y + 3 = 0 \]