Question

The straight line $\frac{x}{a}+\frac{y}{b}=1$ cuts the axes in $A$ and $B$ and a line perpendicular to $AB$ cuts the axes in $P$ and $Q$. Find the locus of the point of intersection of $AQ$ and $BP$.

Answer 1

Any line $\perp$ to $AB$ is $\frac{x}{b}-\frac{y}{a}=\lambda$ (variable)
$\therefore A(a,0),B(0,b)$
$P(b\lambda ,0),Q(0,-a\lambda )$
By intercepts form the equations of $AQ$ and $BP$ are
$\frac{x}{a}+\frac{y}{-a\lambda }=1$
$\therefore \lambda =\frac{y}{x-a}$
and $\frac{x}{b\lambda }+\frac{y}{b}=1\therefore \lambda =-\frac{x}{y-b}$
In order to find the ocus of the point of intersection of these lines, eliminate the variable $\lambda$
$\therefore \lambda =-\frac{x}{y-b}\therefore \frac{y}{x-a}=-\frac{x}{y-b}=\lambda$
$\therefore x(x-a)+y(y-b)=0$
or $x^2+y^2-ax-by=0$ is the required locus