Question

Find the equation of the straight line joining the point $(a,b)$ to the point of intersection of the lines
$\frac{x}{a}+\frac{y}{b}=1$ and $\frac{x}{b}+\frac{y}{a}=1$
Also prove that sum of intercepts made by this
line on coordination axes is $\frac{(b-a)(a^2-b^2)}{ab}$.

Answer 1

Solving the equation by subtracting, we get $x=y$ and putting in any, the point of intersection is $\left(\frac{ab}{a+b}\frac{ab}{a+b}\right)$. By two point formula the equation of the line joining this point and $(a,b)$ is
$a^{2}y-b^{2}x=ab(a-b)$
Sum of intercepts
$=ab(a-b)\left[\frac{1}{a^2}\frac{1}{b^2}\right]=\frac{a-b}{ab}(b^2-a^2)$ etc.