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Find the equations to the straight lines...

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 cutting off a given distance a from the axis of y

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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 cutting off a distance \(a\) from the y-axis, we can follow these steps: ### Step-by-Step Solution: 1. **Identify the Point of Intersection:** The point of intersection of the two lines can be found by solving the equations simultaneously. Let’s denote the point of intersection as \(P(x_0, y_0)\). 2. **Form the General Equation of the Line:** The general form of a line passing through the point of intersection can be expressed as: \[ A x + B y + C + \lambda (A' x + B' y + C') = 0 \] where \(\lambda\) is a parameter that will help us find the specific line we need. 3. **Determine the Intercept on the Y-Axis:** Since the line cuts the y-axis at a distance \(a\), it means that when \(x = 0\), \(y = a\). Therefore, the point \((0, a)\) must satisfy the line equation. 4. **Substitute the Point into the Line Equation:** Substitute \(x = 0\) and \(y = a\) into the line equation: \[ A(0) + B(a) + C + \lambda (A'(0) + B'(a) + C') = 0 \] This simplifies to: \[ Ba + C + \lambda B'a + \lambda C' = 0 \] 5. **Solve for \(\lambda\):** Rearranging the equation gives: \[ \lambda (B'a + C') = - (Ba + C) \] Thus, we can express \(\lambda\) as: \[ \lambda = -\frac{Ba + C}{B'a + C'} \] 6. **Substitute \(\lambda\) Back into the General Equation:** Substitute the value of \(\lambda\) back into the general equation: \[ A x + B y + C - \frac{(Ba + C)}{(B'a + C')} (A' x + B' y + C') = 0 \] 7. **Simplify the Equation:** After substituting, we can multiply through by \(B'a + C'\) to eliminate the fraction and simplify the equation to find the specific line. 8. **Final Form of the Equation:** The final equation will be in the form: \[ (A B' - A' B) x + (B C' - B' C) y + (C B' - C' B) = 0 \] This represents the required straight lines.
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