Question

Find the equation of the stright line passing through the point of intersection of $2x+3y+1=0$ and $3x-5y-5=0$and equally inclined to the axes.

Answer 1

The required line is

$(2x+3y-1)+\lambda (3x-5y-5)=0$

or $x(2+3\lambda )+y(3-5\lambda )-1-5\lambda =0$

Since this lines is equally inclined to both the axes,it slope should be 1 or -1

$\therefore \frac{-2-3\lambda }{3-5\lambda }=1$ or, $\therefore \frac{-2-3\lambda }{3-5\lambda }=-1$

$\Rightarrow 3-5\lambda =-2-3\lambda$ or , $\Rightarrow -2-3\lambda =-3\lambda =-3+5\lambda$

$\Rightarrow 5=2\lambda$ or, $\Rightarrow 1=8\lambda$

$\Rightarrow \lambda =\frac{5}{2}or,\Rightarrow \lambda =\frac{1}{8}$

$\therefore$ The required line is

$2x+3y+1+\frac{5}{2}(3x-5y-5)=0$

$4x+6y+2+15x-25y-25=0$

$19x-19y-23=0$

or

$(2x+3y+1)+\frac{1}{8}(3x-5y-5)=0$

$16x+24y+8+3x-5y-5=0$

$19x+19y+3=0$

$\therefore$ The two possible equation are

$19x-19y-23=0$ or

$19x+19y+3=0$