Question

Find the equation of the line passing through the point of intersection of the lines$4x-7y-3=0$ and $2x-3y+1=0$ that has equal intercepts on the axes.

Answer 1

Let the equation of the line having equal intercepts on the axes be

$\frac{x}{a}+\frac{y}{a}=1$

$x+y=a$....(1)
On solving equation $4x+7y-3=0$ and $2x-3y+1=0$ we obtain $x=\frac{1}{13}andy=\frac{5}{13}$
$\therefore \{\frac{1}{13},\frac{5}{13}\}$ is the point of intersection of the two given lines.

Since equation (1) passes through point $\{\frac{1}{13},\frac{5}{13}\}$

$\{\frac{1}{13}+\frac{5}{13}\}=a$

$\Rightarrow a=\frac{6}{13}$

$\therefore$ Equation (1) becomes $x+y=\frac{6}{13},i.e.,13x+13y=6$

Thus the required equation of the line is $13x+13y=6$