# Q: 12     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.

Point of intersection of the lines  $4x+7y-3=0$  and  $2x-3y+1=0$ is  $\left ( \frac{1}{13},\frac{5}{13} \right )$
We know that the intercept form of the line is
$\frac{x}{a}+\frac{y}{b}= 1$
It is given that line make equal intercepts on x and  y axis
Therefore,
a = b
Now, the equation reduces to
$x+y = a$         -(i)
It passes through point  $\left ( \frac{1}{13},\frac{5}{13} \right )$
Therefore,
$a = \frac{1}{13}+\frac{5}{13}= \frac{6}{13}$
Put the value of a in equation (i)
we will get
$13x+13y=6$
Therefore, equation of line is   $13x+13y=6$

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