Two points $A$ and $B$ with coordinates $(1, 1)$ and $(- 2, 3)$ respectively are given. Find the locus of a point $P$ so that the area of $△ P A B$ is $9$ sq. units.

2 x - 3 y - 13 = 0

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2 x + 3 y - 13 = 0

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5 x - 3 y - 13 = 0

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5 x + 3 y - 13 = 0

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Solution

The correct option is B $2 x + 3 y - 13 = 0$
Let the point be $(h, k)$ & A $(1, 1)$ and B $(- 2, 3)$
Area of $△ A B C = \frac{1}{2} ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 - 2 & 3 h & k . 1 & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ = 9 => | 3 + 2 - 2 k - 3 h + h - k | = 18 => 3 k + 2 h = 13$
For locus,replace $k \to y, h \to x$ , $2 x + 3 y - 13 = 0$

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