Question

Find the equation of the line passing through the point of intersection of the lines $x-2y-a=0$ and $x+3y-2a=0$ and parallel to the straight line $3x+4y=0$ .

Answer 1

$x-2y-a=\mathrm{0.....}\left(i\right)$

$x+3y-2a=\mathrm{0....}\left(ii\right)$

equation of line passing through point of intersection of given lines is

$x-2y-a+\lambda (x+3y-2a)=0(1+\lambda )x+(3\lambda -2)y-(1+2\lambda )a=\mathrm{0......}\left(i\right)$

Slope of line $=m=-\frac{a}{b}=-\frac{1+\lambda }{3\lambda -2}$

$3x+4y=\mathrm{0......}\left(ii\right)$

Slope of $\left(ii\right)$ $=m^{\prime }=-\frac{a}{b}=-\frac{3}{4}$

Both the lines are parallel

$\Rightarrow m=m^{\prime }-\frac{1+\lambda }{3\lambda -2}=-\frac{3}{4}4+4\lambda =9\lambda -610=5\lambda \Rightarrow \lambda =2$

Substituting $\lambda$ in $\left(i\right)$

$(1+2)x+\left(3\right(2)-2)y-(1+2(2\left)\right)a=03x+4y-5a=03x+4y=5a$

Equation of angle bisector of $\left(i\right)$ and $\left(ii\right)$ is

$\frac{x-2y-a}{\sqrt{1^2+2^2}}=\pm \frac{x+3y-2a}{\sqrt{1^2+3^2}}\frac{x-2y-a}{\sqrt{5}}=\pm \frac{x+3y-2a}{\sqrt{10}}\sqrt{2}(x-2y-a)=\pm (x+3y-2a)$

Angle bisector of line is different.