Question

Let P and Q be any two points on the lines represented by $2x-3y=0$ and $2x+3y=0$ respectively. If the area of triangle OPQ (where O is origin) is 5, then the equation of the locus of mid – point of PQ can be equal to
(a) $4{{x}^{2}}-9{{y}^{2}}+20=0$
(b) $9{{x}^{2}}-4{{y}^{2}}+20=0$
(c) $9{{x}^{2}}-4{{y}^{2}}-20=0$
(d) $4{{x}^{2}}-9{{y}^{2}}-30=0$

Answer 1

For solving this type of question you should know about the general properties of line and the area of a triangle. In this problem first we will find the points P and Q and then put these in the equations of lines and satisfy them. And then find the equation of the locus of mid – point of line PQ.

Complete step by step answer:
According to our question it is asked to find the locus of mid – point of line PQ if the points P and Q are on the lines represented by $2x-3y=0$ and $2x+3y=0$ respectively.
Let, ${{L}_{1}}=2x-3y=0P\left( \dfrac{k}{2},\dfrac{k}{3} \right)$
${{L}_{2}}=2x+3y=0Q\left( \dfrac{h}{2},\dfrac{-h}{3} \right)$
Put $P\left( \dfrac{k}{2},\dfrac{k}{3} \right)$ in ${{L}_{1}}\Rightarrow$ satisfied
And $Q\left( \dfrac{h}{2},\dfrac{-h}{3} \right)$ in ${{L}_{2}}\Rightarrow$ satisfied
Now, ${{\Delta }_{OPQ}}=\dfrac{1}{2}\left| \begin{matrix} 0 & 0 & 1 \\\ \dfrac{k}{2} & \dfrac{k}{3} & 1 \\\ \dfrac{h}{2} & \dfrac{-h}{3} & 1 \\\ \end{matrix} \right|=5$
If we solve this, then we find
$\Rightarrow \left| \dfrac{-kh}{6}-\dfrac{kh}{6} \right|=10$
On further simplification, we can write the above equation as following, that is
$\Rightarrow kh\left( \dfrac{2}{6} \right)=10$
$\Rightarrow kh=10\times 3\Rightarrow kh=30$
From our knowledge of Mid - point, we can say that the coordinates of mid - point of PQ can be written as
$=\left( \dfrac{\dfrac{k+h}{2}}{2},\dfrac{\dfrac{k-h}{3}}{2} \right)=\left( \dfrac{k+h}{4},\dfrac{k-h}{6} \right)$
So, $\dfrac{k+h}{4}=x$ and $\dfrac{k-h}{6}=y$
$k+h=4x$ and $k-h=6y$
Now, we know that above mentioned two separate equations can combinedly expressed as the following, therefore, we get

& {{\left( 4x \right)}^{2}}-{{\left( 6y \right)}^{2}}={{k}^{2}}+{{h}^{2}}+2kh-\left( {{k}^{2}}+{{h}^{2}}-2kh \right) \\\ & \Rightarrow {{\left( 4x \right)}^{2}}-{{\left( 6y \right)}^{2}}=4kh=120 \\\ & \Rightarrow 16{{x}^{2}}-36{{y}^{2}}=120 \\\ & \Rightarrow 16{{x}^{2}}-36{{y}^{2}}-120=0 \\\ & \Rightarrow 4{{x}^{2}}-9{{y}^{2}}-30=0 \\\ \end{aligned}$$ **So, the correct answer is “Option d”.** **Note:** While solving these types of questions you have to mind if any equation of any line or any plane is given. And it is also given a point of middle or any place on that plane or line then it must satisfy the given equations always. And if this is not satisfying then it is not on that line or in that plane.