Question

Find the equation of the line passing through the point of intersection of the lines $3x-4y+1=0$ and $5x+y-1=0$ and cutting off equal intercepts from the axes.

Answer 1

$3x-4y+1=\mathrm{0.......}\left(i\right)5x+y-1=\mathrm{0........}\left(ii\right)$

Equation of line passing through the point of intersection of $\left(i\right)$ and $\left(ii\right)$ is

$3x-4y+1+\lambda (5x+y-1)=0(3+5\lambda )x+(\lambda -4)y+1-\lambda =\mathrm{0.......}\left(i\right)x=0\Rightarrow y=\frac{\lambda -1}{\lambda -4}y=0\Rightarrow x=\frac{\lambda -1}{5\lambda +3}$

As the intercepts are given equal

$\therefore \frac{\lambda -1}{\lambda -4}=\frac{\lambda -1}{5\lambda +3}5\lambda +3=\lambda -44\lambda =-7\lambda =-\frac{7}{4}$

substituting $\lambda$ in $\left(i\right)$

$(3+5(-\frac{7}{4}))x+(-\frac{7}{4}-4)y+1+\frac{7}{4}=0-23x-23y+11=023x+23y=11$