The locus of $P$ such that area of $△ P A B$ is $12$ square units where $A = (2, 2)$ and $B (- 4, 5)$ is

x^{2} + 4 x y + 4 y^{2} - 12 x - 24 y - 28 = 0

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x^{2} - 6 x y + 9 y^{2} + 22 x + 66 y + 23 = 0

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x^{2} + 6 x y + 9 y^{2} - 22 x - 66 y - 23 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x^{2} - 6 x y + 9 y^{2} - 22 x - 66 y - 23 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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Solution

The correct option is A $x^{2} + 4 x y + 4 y^{2} - 12 x - 24 y - 28 = 0$
$A B = \sqrt{36 + 9} = \sqrt{45} = 3 \sqrt{5}$
$\Rightarrow \frac{1}{2} A B \times h = 12 s q u n i t s$
$\Rightarrow h = \frac{8}{\sqrt{5}}$
$∴ P$ lies on line paralle to $A B$ at distance of $\frac{8}{\sqrt{5}} .$
Equation of $A B = (y - 5) = \frac{3}{- 6} (x + 4)$
$\Rightarrow - 6 y + 30 = 3 x + 12$
$3 x + 6 y - 18 = 0$
$x + 2 y - 6 = 0$
Let $x + 2 y + d = 0$ be equation of required line -
$∴ \frac{| d + 6 |}{\sqrt{1 + 4}} = \frac{8}{\sqrt{5}}$
$| d + 6 | = 8$
$\Rightarrow d = 2, d = - 14$
$∴$ locus of $P = (x + 2 y + 2) (x + 2 y - 14) = 0$
$\Rightarrow x^{2} + 4 x y + 4 y^{2} - 12 x - 24 y - 28 = 0$
Hence, the answer is $x^{2} + 4 x y + 4 y^{2} - 12 x - 24 y - 28 = 0.$

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