If the line $\alpha (3x-4y+1)+\beta (5x+y-1)=0$ cuts off  equal intercepts on the coordinate axes, then find its equation in general form.

$\therefore \frac{-(3\alpha +5\beta )}{(\beta -4\alpha )}=\pm 1\Rightarrow 3\alpha +5\beta =\beta -4\alpha$ $\Rightarrow$ $3\alpha +5\beta =4\alpha -\beta$

$\Rightarrow 7\alpha =-4\beta ,\alpha =6\beta$ $\Rightarrow$ $\frac{\alpha }{\beta }=\frac{-4}{7}$ or  6

$\therefore$ The line are $\frac{-4}{7}(3x-4y+1)+5x+y-1=0$

        or  $6(3x-4y+1)+5x+y-1=0$

$\Rightarrow$$23x+23y-11=0,23x-23y+5=0.$