Find the equations to the straight lines passing through the point of intersection of the straight lines
`Ax + By + C= 0 and A'x + B'y + C'= 0` and passing through the origin

Answer

To find the equations of the straight lines passing through the point of intersection of the lines \(Ax + By + C = 0\) and \(A'x + B'y + C' = 0\) and passing through the origin, we can follow these steps:

### Step 1: Identify the equations of the lines
We have two lines given by:
1. \(Ax + By + C = 0\)
2. \(A'x + B'y + C' = 0\)

### Step 2: Find the point of intersection
To find the point of intersection \((x_1, y_1)\) of the two lines, we can solve the equations simultaneously.

From the first equation, we can express \(y\) in terms of \(x\):
\[ By = -Ax - C \implies y = -\frac{A}{B}x - \frac{C}{B} \]
Substituting this expression for \(y\) into the second equation:
\[ A'x + B'\left(-\frac{A}{B}x - \frac{C}{B}\right) + C' = 0 \]
This simplifies to:
\[ A'x - \frac{B'A}{B}x - \frac{B'C'}{B} + C' = 0 \]
Combining the \(x\) terms:
\[ \left(A' - \frac{B'A}{B}\right)x + C' - \frac{B'C}{B} = 0 \]
From here, we can solve for \(x_1\):
\[ x_1 = \frac{\frac{B'C}{B} - C'}{A' - \frac{B'A}{B}} \]

Substituting \(x_1\) back into the equation for \(y\):
\[ y_1 = -\frac{A}{B}x_1 - \frac{C}{B} \]

### Step 3: Find the slope \(m\)
The line that passes through the origin and the point of intersection \((x_1, y_1)\) can be expressed in slope-intercept form \(y = mx\). The slope \(m\) can be calculated as:
\[ m = \frac{y_1}{x_1} \]

### Step 4: Substitute \(y_1\) and \(x_1\) into the slope formula
Substituting the expressions for \(y_1\) and \(x_1\) into the slope formula gives:
\[ m = \frac{-\frac{A}{B}x_1 - \frac{C}{B}}{x_1} \]
This simplifies to:
\[ m = -\frac{A}{B} - \frac{C}{Bx_1} \]

### Step 5: Write the equation of the line
The equation of the line passing through the origin with slope \(m\) is:
\[ y = mx \]
Substituting the value of \(m\) gives:
\[ y = \left(-\frac{A}{B} - \frac{C}{Bx_1}\right)x \]

### Final Step: General form of the line
The general form of the line can be expressed as:
\[ By = -Ax - C \]
This represents the family of lines passing through the point of intersection of the two given lines and through the origin.