The coordinates of the points O,A,and B are ( | vedclass

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(A) Let the coordinates of the moving point $P$ be $(x, y)$.The coordinates of the given points are $O(0, 0)$,$A(0, 4)$,and $B(6, 0)$.The area of a tr

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$(x - 3y)(x + 3y) = 0$

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(A) Let the coordinates of the moving point $P$ be $(x, y)$.The coordinates of the given points are $O(0, 0)$,$A(0, 4)$,and $B(6, 0)$.The area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.Area of $\Delta POA = \frac{1}{2} |0(4 - y) + 0(y - 0) + x(0 - 4)| = \frac{1}{2} |-4x| = 2|x|$.Area of $\Delta POB = \frac{1}{2} |0(0 - y) + 6(y - 0) + x(0 - 0)| = \frac{1}{2} |6y| = 3|y|$.According to the problem,Area of $\Delta POA = 2 	imes$ Area of $\Delta POB$.Therefore,$2|x| = 2 	imes 3|y|$.$|x| = 3|y|$.Squaring both sides,we get $x^2 = 9y^2$,which implies $x^2 - 9y^2 = 0$.This can be factored as $(x - 3y)(x + 3y) = 0$.

The coordinates of the points $O$,$A$,and $B$ are $(0,0)$,$(0,4)$,and $(6,0)$ respectively. If a point $P$ moves such that the area of $\Delta POA$ is always twice the area of $\Delta POB$,then the equation to both parts of the locus of $P$ is

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