Question

The locus of $P$ such that the area of $\Delta PAB$ is $12$ square units where $A=\left( 2,2 \right)$ and $B\left( -4,5 \right)$ is
$\left( A \right)\text{ }{{x}^{2}}+4xy+4{{y}^{2}}-12x-24y-28=0$
$\left( B \right)\text{ }{{x}^{2}}-6xy+9{{y}^{2}}+22x+66y+23=0$
$\left( C \right)\text{ }{{x}^{2}}+6xy+9{{y}^{2}}-22x-66y-23=0$
$\left( D \right)\text{ }{{x}^{2}}-6xy+9{{y}^{2}}-22x-66y-23=0$

Answer 1

In this question we have been given two points of a triangle which are $A\left( 2,2 \right)$ and $B\left( -4,5 \right)$. We have also been given the area of the triangle as $12$ square units. We have to find the locus of $P$ from the given data. We know that the area of triangle with coordinates $A\left( {{x}_{1}},{{y}_{1}} \right)$, $B\left( {{x}_{2}},{{y}_{2}} \right)$ and $C\left( {{x}_{3}},{{y}_{3}} \right)$ is given by the formula $A=\left| \dfrac{1}{2}\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \\\ {{x}_{2}} & {{y}_{2}} & 1 \\\ {{x}_{3}} & {{y}_{3}} & 1 \\\ \end{matrix} \right| \right|$. Since we have with us two coordinates out of the three and the area provided, we will consider the third coordinate to be $\left( x,y \right)$ and get the required equation of the locus.

Complete step by step solution:
We know that the area of $\Delta PAB$ is $12$ square units and the coordinates of the two sides of the triangle are $A\left( 2,2 \right)$ and $B\left( -4,5 \right)$.
Consider the third point to be $\left( x,y \right)$.
On substituting the values in the formula, we get:
$\Rightarrow 12=\left| \dfrac{1}{2}\left| \begin{matrix} 2 & 2 & 1 \\\ -4 & 5 & 1 \\\ x & y & 1 \\\ \end{matrix} \right| \right|$
On expanding the determinant, we get:
$\Rightarrow 12=\left| \dfrac{1}{2}\times 2\left( 5-y \right)-2\left( -4-x \right)+1\left( -4y-5x \right) \right|$
On simplifying, we get:
$\Rightarrow 12=\left| \dfrac{1}{2}\times \left( 10-2y+8+2x-4y-5x \right) \right|$
On simplifying the terms, we get:
$\Rightarrow 12=\left| \dfrac{1}{2}\times \left( 18-6y-3x \right) \right|$
On taking $3$ common, we get:
$\Rightarrow 12=\left| \dfrac{3}{2}\times \left( 6-2y-x \right) \right|$
On rearranging the terms and multiplying, we get:
$\Rightarrow 8=\left| 6-2y-x \right|$
Now to remove the modulus sign, we will square both the sides, we get:
$\Rightarrow {{8}^{2}}={{\left( 6-2y-x \right)}^{2}}$
on expanding, we get:
$\Rightarrow 64=36+4{{y}^{2}}+{{x}^{2}}-24y+4xy-12x$
On rearranging the terms, we get:
$\Rightarrow {{x}^{2}}+4{{y}^{2}}+4xy+36-12x-24y=64$
On transferring the term $64$ to the left-hand side, we get:
$\Rightarrow {{x}^{2}}+4{{y}^{2}}+4xy-12x-24y-64+36=0$
On simplifying, we get:
$\Rightarrow {{x}^{2}}+4xy+4{{y}^{2}}-12x-24y-28=0$, which is the required solution

So, the correct answer is “Option A”.

Note: It is to be remembered that locus is a set of all points satisfying some condition. This type of question can also be solved by using the distance formula. It is to be noted that we take a modulus of the value since area cannot be negative. On squaring a term the modulus sign gets removed.