Question

Find the equation of the straight line through the point of intersection of lines $x-3y+1=0$ and $2x+5y-9=0$, and whose distance from the origin is $\sqrt{5}$

Answer 1

The equation of straight line passing through point of intersection of lines $x-3y+1=0$ & $2x+5y-9=0$

is given by:

$x-3y+1+\lambda (2x+5y-9)=0$

$\Rightarrow (1+2\lambda )x+(-3+5\lambda )y+(1-9\lambda )=0$-----(1)

The distance of $=n\left(1\right)$ from origin is $\sqrt{5}units$

$\therefore \left|\frac{1-9\lambda }{\sqrt{{(1+2\lambda )}^2+{(-3+5\lambda )}^2}}\right|=\sqrt{5}[\because d=\left|\frac{c}{\sqrt{A^2+B^2}}\right|]$

$\Rightarrow (1-9\lambda {)}^2=5[(1+2\lambda {)}^2+(-3+5\lambda {)}^2]$

$\Rightarrow 1+81{\lambda }^2-18\lambda =5[1+4{\lambda }^2+4\lambda +9+25{\lambda }^2-30\lambda ]$

$\Rightarrow 81{\lambda }^2-18\lambda +1=5[29{\lambda }^2-96\lambda +10]$

$\Rightarrow 81{\lambda }^2-18\lambda +1=145{\lambda }^2-130\lambda +50$

$\Rightarrow 64{\lambda }^2-112\lambda +49=0$

$\Rightarrow (8\lambda {)}^2-2(8\lambda \left)\right(7)+(7{)}^2=0$

$\Rightarrow (8\lambda -7{)}^2=0$ $[\because a^2-2ab+b^2=(a-b{)}^2]$

$\Rightarrow 8\lambda -7=0$ $\Rightarrow \lambda =7/8$

substituting value of $\lambda$ in $=n\left(1\right)$, we get

$(1+7/4{)}^2+(-3+35/8)y+(1-63/8)=0$

$2x+y-5=0$ is the reqd of straight line