Given points A(6,0),B(0,4),and O as the origi | vedclass

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(C) Let the coordinates of point $P$ be $(x, y)$.Given that $\operatorname{ar}(	riangle POB) = 2 \cdot \operatorname{ar}(	riangle POA)$.The area of

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$x^2-9y^2=0$

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(C) Let the coordinates of point $P$ be $(x, y)$.Given that $\operatorname{ar}(	riangle POB) = 2 \cdot \operatorname{ar}(	riangle POA)$.The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is $\frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$.For $	riangle POB$ with vertices $P(x, y), O(0, 0), B(0, 4)$:$\operatorname{ar}(	riangle POB) = \frac{1}{2} |x(0-4) + 0(4-y) + 0(y-0)| = \frac{1}{2} |-4x| = 2|x|$.For $	riangle POA$ with vertices $P(x, y), O(0, 0), A(6, 0)$:$\operatorname{ar}(	riangle POA) = \frac{1}{2} |x(0-0) + 0(0-y) + 6(y-0)| = \frac{1}{2} |6y| = 3|y|$.Substituting these into the given condition:$2|x| = 2 \cdot 3|y| \Rightarrow |x| = 3|y|$.Squaring both sides,we get $x^2 = 9y^2$.Thus,the locus is $x^2 - 9y^2 = 0$.

Given points $A(6,0)$,$B(0,4)$,and $O$ as the origin,find the locus of a point $P(x, y)$ such that the area of $\triangle POB$ is $2$ times the area of $\triangle POA$.

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