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 equations $4x + 7y \- 3 = 0$ and $2x \- 3y + 1 = 0$, we obtain

$x =\frac{ 1}{13}$ and $y =\frac{ 5}{13}$

$∴ (\frac{1}{13}, \frac{5}{13})$ is the point of intersection of two given lines.
Since equation (1) passes through point $(\frac{1}{13},\frac{ 5}{13})$
$\frac{1}{13} +\frac{ 5}{13} = a$

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

∴ Equation (1) becomes
$x + y =\frac{ 6}{13}$
$13x + 13y = 6$

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