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 intercepes on the axis.

Answer 1

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

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

Or $x+y=a$ ___(i)

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

$\therefore (\frac{1}{13},\frac{5}{13})$ is the point of the 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 $13x+13y=16$.