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 \[\sqrt{5}\].

\[\left| \frac{1 - 9\lambda}{\sqrt{\left( 1 + 2\lambda \right)^2 + \left( 5\lambda - 3 \right)^2}} \right| = \sqrt{5}\]

\[ \Rightarrow 1 + 81 \lambda^2 - 18\lambda = 145 \lambda^2 - 130\lambda + 50\]

\[ \Rightarrow 64 \lambda^2 - 112\lambda + 49 = 0\]

\[ \Rightarrow \left( 8\lambda - 7 \right)^2 = 0\]

\[ \Rightarrow \lambda = \frac{7}{8}\]

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

\[\left( 1 + \frac{14}{8} \right)x + \left( - 3 + \frac{35}{8} \right)y + 1 - \frac{63}{8} = 0\]

\[ \Rightarrow 22x + 11y - 55 = 0\]

\[ \Rightarrow 2x + y - 5 = 0\]