Question

The line $\frac{x}{a}+\frac{y}{b}=1$ cuts the axis at $A$ and $B$,another line perpendicular to $AB$ cuts the axes at$P,Q$respectively.Locus of points of intersection of $AQ$ and $BP$ is

• A. $x^2+y^2+ax+by=0$
• B. $x^2+y^2-ax-by=0$
• C. $x^2+y^2-ax+by=0$
• D. $x^2+y^2+ax-by=0$

Answer 1

The correct option is B $x^2+y^2-ax-by=0$

Let equation of line perpendicular to $AB$ be
$PQ\equiv \frac{x}{b}-\frac{y}{a}=c$
Where $c$ is some constant.$\therefore P=(cb,0),Q=(0,-ac)$
Let the locus of intersection point be $T(h,k)$
$\therefore$ Using intercept form line $BP\equiv \frac{x}{cb}+\frac{y}{b}=1$ and $AQ\equiv \frac{x}{a}-\frac{y}{ac}=1$
$\because$ both the lines passes through $T\equiv (h,k)$
$\therefore \frac{h}{cb}+\frac{k}{b}=1\Rightarrow c=\frac{h}{b-k}$
and $\frac{h}{a}-\frac{k}{ac}=1\Rightarrow c=\frac{k}{h-a}$
$\therefore$ locus will be $\frac{x}{b-y}=\frac{y}{x-a}$
$\Rightarrow x^2+y^2-ax-by=0$