Question

Find the equations of the lines through the point of intersection of the lines x − 3y + 1 = 0 and 2x + 5y − 9 = 0 and whose distance from the origin is $\sqrt{5}$.

Answer 1

The equation of the straight line passing through the point of intersection of x − 3y + 1 = 0 and 2x + 5y − 9 = 0 is given below:

x − 3y + 1 + λ(2x + 5y − 9) = 0

$\Rightarrow$(1 + 2λ)x + (−3 + 5λ)y + 1 − 9λ = 0 ... (1)

The distance of this line from the origin is $\sqrt{5}$.

$$\begin{gathered} \left|\frac{1-9\lambda }{\sqrt{{\left(1+2\lambda \right)}^2+{\left(5\lambda -3\right)}^2}}\right|=\sqrt{5} \\ \Rightarrow 1+81{\lambda }^2-18\lambda =145{\lambda }^2-130\lambda +50 \\ \Rightarrow 64{\lambda }^2-112\lambda +49=0 \\ \Rightarrow {\left(8\lambda -7\right)}^2=0 \\ \Rightarrow \lambda =\frac{7}{8} \end{gathered}$$

Substituting the value of λ in (1), we get the equation of the required line.

$$\begin{gathered} \left(1+\frac{14}{8}\right)x+\left(-3+\frac{35}{8}\right)y+1-\frac{63}{8}=0 \\ \Rightarrow 22x+11y-55=0 \\ \Rightarrow 2x+y-5=0 \end{gathered}$$