Find the equation to the straight line passing through the intersection of the lines $2x - 5y + 1 = 0$ and $x + 3y - 4 = 0$ and cutting off equal intercepts from the axes.
Equation of a line passing through the intersection of the lines
$2x - 5y + 1 = 0$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $x + 3y - 4 = 0$ $\;\;\; \cdots \; (2)$ is
$\left(2x - 5y + 1\right) + \lambda \left(x + 3y - 4\right) = 0$ $\;\;\; \cdots \; (3)$ $\;$ where $\lambda$ is a scalar.
i.e. $\;$ $\left(2 + \lambda\right)x + \left(-5 + 3 \lambda\right)y + \left(1 - 4 \lambda\right) = 0$ $\;\;\; \cdots \; (4)$
i.e. $\;$ $\left(2 + \lambda\right)x + \left(-5 + 3 \lambda\right)y = \left(4 \lambda - 1\right)$
i.e. $\;$ $\dfrac{\left(2 + \lambda\right) x}{4 \lambda -1} + \dfrac{\left(3 \lambda - 5\right) y}{4 \lambda - 1} = 1$
i.e. $\;$ $\dfrac{x}{\dfrac{4\lambda - 1}{2 + \lambda}} + \dfrac{y}{\dfrac{4 \lambda - 1}{3 \lambda - 5}} = 1$
$\therefore \;$ $X$ intercept of the required line $= a = \dfrac{4 \lambda - 1}{2 + \lambda}$
$Y$ intercept of the required line $= b = \dfrac{4 \lambda - 1}{3 \lambda - 5}$
Given: $\;$ $a = b$
i.e. $\;$ $\dfrac{4 \lambda - 1}{2 + \lambda} = \dfrac{4 \lambda - 1}{3 \lambda - 5}$
i.e. $\;$ $3 \lambda - 5 = 2 + \lambda$
i.e. $\;$ $2 \lambda = 7$ $\implies$ $\lambda = \dfrac{7}{2}$
Substituting the value of $\lambda$ in equation $(4)$ gives the required equation of line as
$\dfrac{11}{2} x + \dfrac{11}{2} y - 13 = 0$
i.e. $\;$ $11 x + 11y - 26 = 0$