# Find the equation of the line passing through the point of intersection of the lines

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.

Solution:

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

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

Or $x+y=a$  $\ldots(1)$

On solving equations $4 x+7 y-3=0$ and $2 x-3 y+1=0$, we obtain $x=\frac{1}{13}$ and $y=\frac{5}{13}$.

$\therefore\left(\frac{1}{13}, \frac{5}{13}\right)$ is the point of intersection of the two given lines.

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

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

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

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

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