Question

Find the equation of the straight line which passes through the point intersection of the lines $3x-y=5$ and $x+3y=1$ and makes equal and positive intercepts on the axes.

Answer 1

The equation of the straight line passing through the point of intersection of $3x-y=5$ and $x+3y=1$ is

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

$\Rightarrow (3+\lambda )x+(-1+3\lambda )y-5\lambda =0$ ...(i)

$\Rightarrow y=-\left(\frac{3+\lambda }{-1+\lambda }\right)x+\frac{5+\lambda }{-1+\lambda }$

The slope of the line that makes equal and positive intercepts on the axis is -1 From equation (i), we have

$-\frac{3+\lambda }{-1+3\lambda }=-1$

$\Rightarrow \lambda =2$

Substitute the value of $\lambda$ in (i), we get the equation of the required line,

$\Rightarrow (3+2)x+(-1+6)y-5-2=0$

$\Rightarrow 5x+5y-7=0$