Question

Find the coordinates of the points of intersection of the straight lines whose equations are
$\frac{x}{a}+\frac{y}{b}=1$ and $\frac{x}{b}+\frac{y}{a}=1$.

Answer 1

$\frac{x}{a}+\frac{y}{b}=1bx+ay=ab......\left(i\right)\frac{x}{b}+\frac{y}{a}=1ax+by=abby=ab-axy=\frac{ab-ax}{b}$

Substituting $y$ in $\left(i\right)$, we get

$bx+a\left(\frac{ab-ax}{b}\right)=ab\frac{b^{2}x+a^{2}b-a^{2}x}{b}=ab(b^2-a^2)x+a^{2}b=a{b}^2(b^2-a^2)x=a{b}^2-a^{2}bx=\frac{ab(b-a)}{(b^2-a^2)}x=\frac{ab(b-a)}{(b+a)(b-a)}\Rightarrow x=\frac{ab}{a+b}y=\frac{ab-ax}{b}by=ab-axby=ab-a\left(\frac{ab}{a+b}\right)by=\frac{ab(a+b)-a^{2}b}{a+b}by=\frac{a^{2}b+a{b}^2-a^{2}b}{a+b}by=\frac{a{b}^2}{a+b}\Rightarrow y=\frac{ab}{a+b}$

So, the point of intersection is $(\frac{ab}{a+b},\frac{ab}{a+b})$