Question

Find the equation of a line passing through the point of intersection of the lines $x+2y+3=0$ and $3x+4y+7=0$ and perpendicular to the straight line $y-x=8$. Justify whether it bisects the angle between them.

Answer 1

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

$3x+4y+7=\mathrm{0.....}\left(ii\right)$

Equation of line passing through their point of intersection is

$x+2y+3+\lambda (3x+4y+7)=0(1+3\lambda )x+(2+4\lambda )y+3+7\lambda =\mathrm{0......}\left(i\right)$

Slope of line $=m=-\frac{a}{b}=-\frac{1+3\lambda }{2+4\lambda }$

$y-x=\mathrm{8......}\left(ii\right)$

Slope of $\left(ii\right)$ $=m^{\prime }=-\frac{a}{b}=-\frac{-1}{1}=1$

As the lines are perpendicular

$\Rightarrow m{m}^{\prime }=-1-\frac{1+3\lambda }{2+4\lambda }\times 1=-11+3\lambda =2+4\lambda \Rightarrow \lambda =-1$

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

$(1+3(-1\left)\right)x+(2+4(-1\left)\right)y+3+7(-1)=0-2x-2y-4=0x+y+2=0$

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

$\frac{x+2y+3}{\sqrt{1^2+2^2}}=\pm \frac{3x+4y+7}{\sqrt{3^2+4^2}}\frac{x+2y+3}{\sqrt{5}}=\pm \frac{3x+4y+7}{5}\sqrt{5}(x+2y+3)=\pm (3x+4y+7)$

Equation of angle bisector is different

So $x+y+2$ do not bisect the angle between the lines