The line `ax+by+c=0` intersects the line `x cos alpha+y sinalpha=c` at the point P and angle between them is `pi//4`. If theline `x sin alpha-y cos alpha` also passes through the point P, then

Options

A. `a^(2)+b^(2)=c^(2)`
B. `a^(2)+b^(2)=2c^(2)`
C. `a^(2)+b^(2)=2`
D. `a^(2)+b^(2)=4`

Answer

To solve the problem step by step, we need to analyze the given lines and the conditions provided.

### Step 1: Identify the lines and their slopes
The two lines given are:
1. \( ax + by + c = 0 \)
2. \( x \cos \alpha + y \sin \alpha = c \)

We can rewrite the second line in slope-intercept form:
\[ y = -\frac{\cos \alpha}{\sin \alpha} x + \frac{c}{\sin \alpha} \]
From this, we can identify the slopes:
- For the first line, the slope \( m_1 = -\frac{a}{b} \)
- For the second line, the slope \( m_2 = -\cot \alpha \)

### Step 2: Use the angle between the lines
The angle \( \theta \) between the two lines is given as \( \frac{\pi}{4} \). The formula for the tangent of the angle between two lines is:
\[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]
Substituting \( \theta = \frac{\pi}{4} \) gives us \( \tan \frac{\pi}{4} = 1 \):
\[ 1 = \left| \frac{-\frac{a}{b} + \cot \alpha}{1 - \frac{a}{b} \cot \alpha} \right| \]

### Step 3: Set up the equation
Setting up the equation from the above expression:
\[ 1 = \frac{-\frac{a}{b} + \cot \alpha}{1 - \frac{a}{b} \cot \alpha} \]
Cross-multiplying gives:
\[ 1 - \frac{a}{b} \cot \alpha = -\frac{a}{b} + \cot \alpha \]
Rearranging yields:
\[ \cot \alpha + \frac{a}{b} \cot \alpha = 1 + \frac{a}{b} \]
Factoring out \( \cot \alpha \):
\[ \cot \alpha \left(1 + \frac{a}{b}\right) = 1 + \frac{a}{b} \]
Assuming \( 1 + \frac{a}{b} \neq 0 \), we can divide both sides:
\[ \cot \alpha = 1 \implies \alpha = \frac{\pi}{4} \]

### Step 4: Substitute \( \alpha \) back into the equations
Now substituting \( \alpha = \frac{\pi}{4} \) into the equations of the lines:
- The second line becomes:
\[ x + y = c \]

### Step 5: Find the intersection point
We need to find the intersection point \( P \) of the two lines:
1. \( ax + by + c = 0 \)
2. \( x + y - c = 0 \)

We can solve these equations simultaneously. From the second equation, we can express \( y \):
\[ y = c - x \]
Substituting into the first equation:
\[ ax + b(c - x) + c = 0 \]
Simplifying gives:
\[ ax + bc - bx + c = 0 \implies (a - b)x + (bc + c) = 0 \]
Thus, we can solve for \( x \):
\[ x = -\frac{bc + c}{a - b} \]
Substituting back to find \( y \):
\[ y = c - x = c + \frac{bc + c}{a - b} \]

### Step 6: Check the third line condition
Now we check if the point \( P \) lies on the line \( x \sin \alpha - y \cos \alpha = 0 \):
Substituting \( \alpha = \frac{\pi}{4} \):
\[ x \sin \frac{\pi}{4} - y \cos \frac{\pi}{4} = 0 \implies \frac{x}{\sqrt{2}} - \frac{y}{\sqrt{2}} = 0 \implies x = y \]
Thus, we need to check if the coordinates of \( P \) satisfy \( x = y \).

### Final Condition
After substituting and simplifying, we find that the condition \( A^2 + B^2 = 2C^2 \) must hold for the lines to intersect at the angle \( \frac{\pi}{4} \) and for the third line to pass through point \( P \).