Given points, $A(6,0), B(0,4)$ and $O$ as the origin, find the locus of a point $P$ such that area of…
Given points, $A(6,0), B(0,4)$ and $O$ as the origin, find the locus of a point $P$ such that area of $\triangle P O B$ is 2 times the area of $\triangle P O A$.
$x^2-3 y^2=0$
$x^2+3 y^2=0$
$x^2-9 y^2=0$
$x^2-4 y^2=0$
Solution
Given points,
Let $P$ be $(x, y)$
$\because \quad \operatorname{ar}(\triangle P O B)=2 \cdot \operatorname{ar}(\triangle P O A)$ ...(i)
Now, $\operatorname{ar}(\triangle P O B)$
$\begin{aligned} & =\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right| \\ & =\frac{1}{2}|x(0-4)+0+0| \\ & =2 x\end{aligned}$
$\operatorname{ar}(\triangle P O A)=\frac{1}{2}|x(0-0)+0+6(y-0)|$
$= \pm 3 y$
From Eq. (i), we get
$2 x= \pm 2 \cdot 3 y$
$\Rightarrow \quad x= \pm 3 y$
Hence, the required locus of both the parts is
$(x-3 y)(x+3 y)=0$
$\therefore \quad x^2-9 y^2=0$