Find the equation of the straight line which passes through
the origin and the point of intersection of the lines `ax+by+c=0 and a'x+b'y+c'=0`.

Answer

To find the equation of the straight line that passes through the origin and the point of intersection of the lines \( ax + by + c = 0 \) and \( a'x + b'y + c' = 0 \), we can follow these steps:

### Step 1: Find the point of intersection of the two lines.
To find the point of intersection, we can use the method of elimination. We will manipulate the two equations:

1. \( ax + by + c = 0 \) (Equation 1)
2. \( a'x + b'y + c' = 0 \) (Equation 2)

We can multiply Equation 1 by \( b' \) and Equation 2 by \( b \):

- From Equation 1:
\[ b' (ax + by + c) = 0 \implies ab'x + b'y + b'c = 0 \]
- From Equation 2:
\[ b (a'x + b'y + c') = 0 \implies ba'x + b^2y + bc' = 0 \]

### Step 2: Subtract the modified equations.
Now, we subtract the two modified equations:

\[ (ab'x + b'y + b'c) - (ba'x + b^2y + bc') = 0 \]

This simplifies to:

\[ (ab' - ba')x + (b' - b^2)y + (b'c - bc') = 0 \]

### Step 3: Solve for \( x \).
From the simplified equation, we can isolate \( x \):

\[ (ab' - ba')x = - (b'c - bc') \]
\[ x = \frac{b'c - bc'}{ab' - ba'} \]

### Step 4: Substitute \( x \) back to find \( y \).
Now we substitute the value of \( x \) back into one of the original equations to find \( y \). We use Equation 1:

\[ a\left(\frac{b'c - bc'}{ab' - ba'}\right) + by + c = 0 \]

Rearranging gives:

\[ by = -a\left(\frac{b'c - bc'}{ab' - ba'}\right) - c \]

### Step 5: Solve for \( y \).
This can be simplified to:

\[ y = \frac{-a(b'c - bc') - c(ab' - ba')}{b(ab' - ba')} \]

### Step 6: Find the slope of the line through the origin and the intersection point.
The slope \( m \) of the line passing through the origin (0, 0) and the point of intersection \( (x, y) \) is given by:

\[ m = \frac{y}{x} = \frac{\frac{-a(b'c - bc') - c(ab' - ba')}{b(ab' - ba')}}{\frac{b'c - bc'}{ab' - ba'}} \]

### Step 7: Write the equation of the line.
Using the point-slope form of the equation of a line:

\[ y - 0 = m(x - 0) \implies y = mx \]

Substituting the value of \( m \):

\[ y = \frac{-a(b'c - bc') - c(ab' - ba')}{b(ab' - ba')} x \]

This gives us the required equation of the line.

### Final Equation
Thus, the equation of the line is:

\[ y = \frac{a'c - ac'}{bc' - b'c} x \]