A ray of light is incident along a line which meets another line, `7x-y+1 = 0`, at the point (0, 1). The ray isthen reflected from this point along the line, `y+ 2x=1`. Then the equation of the line of incidence of the ray of light is (A) `41x + 38 y - 38 =0` (B) `41 x - 38 y + 38 = 0` (C) `41x + 25 y - 25 = 0` (D) `41x - 25y + 25 =0 `

Options

A. `41x+38y-38=0`
B. `41x-38y+38=0`
C. `41x+25y-25=0`
D. `41x-25y+25=0`

Answer

To solve the problem step by step, we need to find the equation of the line of incidence of a ray of light that reflects off a given line at a specific point.

### Step 1: Identify the given lines and point of intersection
We have two lines:
1. Line 1: \( 7x - y + 1 = 0 \)
2. Line 2: \( y + 2x = 1 \)

The point of intersection is given as \( (0, 1) \).

### Step 2: Find the slope of Line 1
To find the slope of Line 1, we can rewrite it in slope-intercept form \( y = mx + b \):
\[ y = 7x + 1 \]
From this, we can see that the slope \( m_1 \) of Line 1 is \( 7 \).

### Step 3: Find the slope of Line 2
Next, we rewrite Line 2 in slope-intercept form:
\[ y = -2x + 1 \]
Thus, the slope \( m_2 \) of Line 2 is \( -2 \).

### Step 4: Find the slope of the normal line at the point of incidence
The normal line to Line 1 at point \( (0, 1) \) will have a slope that is the negative reciprocal of the slope of Line 1:
\[ m_{\text{normal}} = -\frac{1}{m_1} = -\frac{1}{7} \]

### Step 5: Use the law of reflection
According to the law of reflection, the angle of incidence equals the angle of reflection. Therefore, the slopes of the incident line and the reflected line can be related through the tangent of the angles formed with the normal line.

Let \( m \) be the slope of the incident line. The relationship can be expressed as:
\[ \tan(\theta) = \frac{m - m_{\text{normal}}}{1 + m \cdot m_{\text{normal}}} \]

### Step 6: Set up the equation using slopes
From the slopes we have:
- \( m_{\text{normal}} = -\frac{1}{7} \)
- \( m_2 = -2 \)

Thus, we can write:
\[ \frac{m - (-\frac{1}{7})}{1 + m \cdot (-\frac{1}{7})} = \frac{-2 - (-\frac{1}{7})}{1 + (-2)(-\frac{1}{7})} \]

### Step 7: Solve for \( m \)
Cross-multiplying and simplifying gives:
\[ \frac{m + \frac{1}{7}}{1 - \frac{m}{7}} = \frac{-2 + \frac{1}{7}}{1 + \frac{2}{7}} \]
Calculating the right side:
\[ -2 + \frac{1}{7} = -\frac{14}{7} + \frac{1}{7} = -\frac{13}{7} \]
\[ 1 + \frac{2}{7} = \frac{7}{7} + \frac{2}{7} = \frac{9}{7} \]
Thus, we have:
\[ \frac{m + \frac{1}{7}}{1 - \frac{m}{7}} = -\frac{13}{9} \]

Cross-multiplying gives:
\[ 9(m + \frac{1}{7}) = -13(1 - \frac{m}{7}) \]

### Step 8: Expand and simplify
Expanding both sides:
\[ 9m + \frac{9}{7} = -13 + \frac{13m}{7} \]

Multiply through by 7 to eliminate the fraction:
\[ 63m + 9 = -91 + 13m \]

### Step 9: Solve for \( m \)
Rearranging gives:
\[ 63m - 13m = -91 - 9 \]
\[ 50m = -100 \implies m = -2 \]

### Step 10: Find the equation of the line of incidence
Using the point-slope form of the line:
\[ y - 1 = -2(x - 0) \]
This simplifies to:
\[ y = -2x + 1 \]

### Step 11: Convert to standard form
Rearranging gives:
\[ 2x + y - 1 = 0 \implies 2x + y = 1 \]

### Final Step: Check against the options
We need to check which of the provided options matches our derived equation.