Question

Find the equation of the line passing through origin and the point of intersection of the lines $\frac{x}{a}+\frac{y}{b}=1$ and $\frac{x}{b}+\frac{y}{a}=1$ proving that it bisects the angle between them.

Answer 1

$\frac{x}{a}+\frac{y}{b}-1=\mathrm{0......}\left(i\right)\frac{x}{b}+\frac{y}{a}-1=\mathrm{0......}\left(ii\right)$

Equation of line passing through point of intersection of $\left(i\right)$ and $\left(ii\right)$ is

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

It passes through $(0,0)$

$0+0-1+\lambda (0+0+-1)=0-1-\lambda =0\Rightarrow \lambda =-1$

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

$\frac{x}{a}+\frac{y}{b}-1+(-1)(\frac{x}{b}+\frac{y}{a}-1)\frac{x}{a}+\frac{y}{b}-1-\frac{x}{b}-\frac{y}{a}+1=0x(\frac{1}{a}-\frac{1}{b})-y(\frac{1}{a}-\frac{1}{b})=0y-x=0y=x$

Now any point on $y=x$ is of the form $P(h,h)$

Let perpendicular distance of $P$ from $\left(i\right)$ $=p_1$

$p_1=\frac{|\frac{h}{a}+\frac{h}{b}-1|}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}$

Perpendicular distance of $P$ from $\left(ii\right)$ $=p_2$

$p_2=\frac{|\frac{h}{b}+\frac{h}{a}-1|}{\sqrt{\frac{1}{b^2}+\frac{1}{a^2}}}$

Clearly $p_2=p_1$

If perpendicular distance of any point from the line passing through the point of intersection of two lines are equal, then the line is a bisector of the angle between the other two.

Hence, $y=x$ also bisects the angle between the given lines.