Question

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

• A. $x(x-7)+y(y+5)=0$
• B. $x(x-7)-y(y+5)=0$
• C. $x(x+7)+y(y-5)=0$
• D. $x(x-7)+y(y-5)=0$

Answer 1

The correct option is A $x(x-7)+y(y+5)=0$

The given lines cuts the $x-axis$ at $A(7,0)$ and the $y-axis$ is $B(0,-5)$.

Hence, the equation of the line $AB$ is $\frac{x}{7}+\frac{y}{-5}=\mathrm{1............}\left(1\right)$

Hence, the equation becomes $5x-7y=35$

We know that,

the equation of the line perpendicular to $AB$ is $7x+5y=\mu .$

Hence, it wil meet $x-axis$ at $P\left(\frac{\mu }{7},0\right)andy-axisatQ\left(0,\frac{\mu }{5}\right)$

The equations of lines $AQ$ and $BP$ are $\frac{x}{7}+\frac{5y}{\mu }=1and\frac{7x}{\mu }-\frac{y}{5}=1$

Let $R(h,k)$ be the point of intersection of lines $AQ$ and $BP$.

$\begin{array}{c}Then,\frac{h}{7}+\frac{5k}{\mu }=1\\ \frac{7h}{\mu }-\frac{k}{5}=1\\ \frac{1}{5k}\left(1-\frac{h}{7}\right)=\frac{1}{7h}\left(1+\frac{k}{5}\right)\\ so,h\left(7-h\right)=k\left(5+k\right)\\ h^2+k^2-7h+5k=0\end{array}$

Hence, the locus of $R$ is $x^2+y^2-7x+5y=0$