The equation of the line passing through the point of intersection of the lines $2 x+y-4=0, x-3 y+5=0$ and…
The equation of the line passing through the point of intersection of the lines $2 x+y-4=0, x-3 y+5=0$ and lying at a distance of $\sqrt{5}$ units from the origin, is
$x-2 y-5=0$
$x+2 y-5=0$
$x+2 y+5=0$
$x-2 y+5=0$
Solution
The equation of a line passing through the
in ter section of 2x + y − 4 = 0 and x − 3y + 5 = 0 is
(2x + y − 4) + λ(x − 3y + 5) = 0 …(i)
⇒ x(2 + λ) + y(1 − 3λ) + 5λ − 4 = 0
This is at a distance of $\sqrt{5}$ units from the origin.
$\begin{array}{ll}
\therefore & \left|\frac{5 \lambda-4}{\sqrt{(2+\lambda)^2+(1-3 \lambda)^2}}\right|=\sqrt{5} \\
\Rightarrow & \frac{(5 \lambda-4)^2}{4+\lambda^2+4 \lambda+1+9 \lambda^2-6 \lambda}=5 \\
\Rightarrow & \frac{(5 \lambda-4)^2}{10 \lambda^2-2 \lambda+5}=5 \\
\Rightarrow & 25 \lambda^2+16-40 \lambda=50 \lambda^2-10 \lambda+25 \\
\Rightarrow & 25 \lambda^2+30 \lambda+9=0
\end{array}$
By solving, we get $\lambda=-\frac{3}{5}$
Putting the value of $\lambda=-\frac{3}{5}$ in Eq. (i), we get
$\begin{array}{rrrl}
\Rightarrow (2 x+y-4)-\frac{3}{5}(x-3 y+5) =0 \\
\Rightarrow 10 x+5 y-20-3 x+9 y-15 =0 \\
\Rightarrow 7 x+14 y-35 =0 \\
\Rightarrow x+2 y-5 =0
\end{array}$