Question

A line cuts the x - axis at A(7, 0) and the y - axis at B(0, -5). A variable line PQ is drawn perpendicular to AB. Cutting the x - axis at P and the y - axis at Q.
If AQ and BP intersect at R, the locus of R is

• A. $x^2+y^2+7x-5y=0$
• B. $x^2+y^2-7x+5y=0$
• C. 5x - 7y = 35
• D. none of these

Answer 1

The correct option is B

$x^2+y^2-7x+5y=0$

Let P(a, 0) and Q(0, b)
Slope of PQ $=\frac{-b}{a}$
$-\frac{b}{a}\times \frac{5}{7}=-1\Rightarrow \frac{a}{b}=\frac{5}{7}$
Equation of AQ is $\frac{x}{7}+\frac{y}{b}=1\Rightarrow b=\frac{7y}{7-x}$
Equation of BP is $\frac{x}{a}-\frac{y}{5}=1\Rightarrow a=\frac{5x}{5+y}$
so that $\frac{5x}{5+y}\times \frac{7-x}{7y}=\frac{a}{b}=\frac{5}{7}\Rightarrow x(7-x)=y(5+y)$
$\Rightarrow x^2+y^2-7x+5y=0$
which is the locus of R(x, y).