Question

A rod of length $l$ moves such that its ends $A$ and $B$ always lie on the lines $3x-y+5=0$ and $y+5=0$ respectively. Then the locus of the point $P$ which divides $AB$ internally in the ratio of $2:1,$ is

• A. $l^2=\frac{1}{4}(3x+3y-5{)}^2+(3y+15{)}^2$
• B. $l^2=\frac{1}{4}(3x-3y+5{)}^2+(3y-5{)}^2$
• C. $l^2=\frac{1}{4}(3x-3y-5{)}^2+(3y-5{)}^2$
• D. $l^2=\frac{1}{4}(3x-3y-5{)}^2+(3y+15{)}^2$

Answer 1

The correct option is D $l^2=\frac{1}{4}(3x-3y-5{)}^2+(3y+15{)}^2$

Equation of first line is
$3x-y+5=0\cdots \left(1\right)$
and equation of other line is
$y=-5\cdots \left(2\right)$

Let the general point on line $\left(1\right)$ be $(\alpha ,3\alpha +5)$ and on line $\left(2\right)$ be $(\beta ,-5).$
Using section formula,
$x=\frac{\alpha +2\beta }{3}$
$\Rightarrow 3x=\alpha +2\beta \cdots \left(3\right)$
$y=\frac{-10+3\alpha +5}{3}$
$\Rightarrow 3y=3\alpha -5\cdots \left(4\right)$

From equations $\left(3\right)$ and $\left(4\right),$ solving $\alpha$ and $\beta$ in terms of $x$ and $y,$
$\alpha =\frac{3y+5}{3},\beta =\frac{9x-3y-5}{6}$
$l^2=A{B}^2=(\alpha -\beta {)}^2+(3\alpha +10{)}^2$
$\Rightarrow l^2=\frac{1}{4}(3x-3y-5{)}^2+(3y+15{)}^2$