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Determinant

Let A = 6,0...

Question

Let $A = (6, 0), B (0, 4)$ and O be the origin. Then find the locus of P such that $A r e a o f △ P O B$ = $2 \times A r e a o f △ P O A$ .

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Solution

Let the point $P$ be $(x, y)$
Area of $△ P O B = 2 \times$ Area of $△ P O A$
$∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} x & y & 1 0 & 0 & 1 0 & 4 & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ = 2 ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} x & y & 1 0 & 0 & 1 6 & 0 & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ⟹ 4 | x | = 12 | y | ⟹ | x | = 3 | y |$
The locus of $P$ is $| x | = 3 | y |$

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