Question

Find the equation 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 required is

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

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

Distance from origin of this line is

$\left|\frac{(1+2\lambda )0+(-3+5\lambda )0+1-9\lambda }{\sqrt{(1+2\lambda {)}^2+(5\lambda -3{)}^2}}\right|$

$\left[\begin{array}{c}using\frac{a{x}_1+b{y}_1+c}{\sqrt{a^2+b^2}}\end{array}\right]$

$\sqrt{5}=\left|\frac{1-9\lambda }{\sqrt{1+4{\lambda }^2+4\lambda +25{\lambda }^2+9-30\lambda }}\right|$

$\Rightarrow \sqrt{5}=\left|\frac{1-9\lambda }{\sqrt{10+29{\lambda }^2-26\lambda }}\right|$

$\Rightarrow 5(10+29{\lambda }^2-26\lambda )=(19-9\lambda {)}^2$

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

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

$\Rightarrow (8\lambda -7{)}^2=0or\lambda =\frac{7}{8}$

$\therefore$ Required lineis

$x-3y+1+\frac{7}{8}(2x+5y-9)=0$

$\Rightarrow 8x-24y+8+14x+35y-63=0$

$\Rightarrow 22x+11y-55=0$

$\Rightarrow 2x+y-5=0$