Question

The equation of the locus of the foot of perpendicular drawn from $\left(5,6\right)$ on the family of lines $\left(x - 2\right)+\lambda \left(y - 3\right)=0$ (where $\lambda \in R$ ) is

• A. $\left(x - 1\right)\left(x - 3\right)+\left(y - 2\right)\left(y - 6\right)=0$
• B. $\left(x - 5\right)\left(x - 6\right)+\left(y - 2\right)\left(y - 3\right)=0$
• C. $\left(x - 2\right)\left(x - 5\right)+\left(y - 3\right)\left(y - 6\right)=0$
• D. $\left(x + 2\right)\left(x + 5\right)+\left(y + 3\right)\left(y + 6\right)=0$

Answer 1

Answer: (C)

Let $A=\left(5,6\right)$ and the point of concurrency of the family of lines $\left(x - 2\right)+\lambda \left(y - 3\right)=0$ is $\left(2,3\right)=B$ and foot of the perpendicular from $A$ to the family of lines is $P=\left(h , k\right)$
Now, $PA$ is perpendicular to $PB$
$\Rightarrow \left(s l o p e o f P A\right)\times \left(s l o p e o f P B\right)=-1$
$\Rightarrow \frac{k - 6}{h - 5}\times \frac{k - 3}{h - 2}=-1\Rightarrow \left(k - 6\right)\left(k - 3\right)=-\left(h - 5\right)\left(h - 2\right)$
$\Rightarrow $ locus is $\left(x - 2\right)\left(x - 5\right)+\left(y - 3\right)\left(y - 6\right)=0$