Question

Find the equation of a straight line through the point of intersection of the lines 4x − 3y = 0 and 2x − 5y + 3 = 0 and parallel to 4x + 5y + 6 = 0.

Answer 1

The equation of the straight line passing through the points of intersection of 4x − 3y = 0 and 2x − 5y + 3 = 0 is given below:

4x − 3y + λ (2x − 5y + 3) = 0

$\Rightarrow$(4 + 2λ)x + (−3 − 5λ)y + 3λ = 0

$\Rightarrow y=\frac{\left(4+2\lambda \right)}{\left(3+5\lambda \right)}x+\frac{3\lambda }{\left(3+5\lambda \right)}$

The required line is parallel to 4x + 5y + 6 = 0 or, $y=-\frac{4}{5}x-\frac{6}{5}$
$$\begin{gathered} \therefore \frac{\left(4+2\lambda \right)}{3+5\lambda }=-\frac{4}{5} \\ \Rightarrow \lambda =-\frac{16}{15} \end{gathered}$$

Hence, the required equation is


$$\begin{gathered} \left(4-\frac{32}{15}\right)x-\left(3-\frac{80}{15}\right)y-\frac{48}{15}=0 \\ \Rightarrow 28x+35y-48=0 \end{gathered}$$