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Question

$$A(1, 2) ,\ B(2, -3),\ C(-2, 3)$$ are three points. A point P moves such that $$PA^2 + PB^2 = 2PC^2$$ Find the locus of P.


Solution

Let point $$P$$ be $$(x,y)$$

Given $$A(1,2),B(2,-3),C(-2,3)$$ and

$$PA^2+PB^2=2PC^2$$

$$\implies (\sqrt{(x-1)^2+(y-2)^2})^2+(\sqrt{(x-2)^2+(y+3)^2})^2=2(\sqrt{(x+2)^2+(y-3)^2})^2$$

$$\implies x^2+1-2x+y^2+4-4y+x^2+4-4x+y^2+9+6y=2(x^2+4+4x+y^2+9-6y)$$

$$\implies 2x^2+2y^2-6x+2y+18=2x^2+2y^2+8x-12y+26$$

$$\implies 14x-14y+8=0$$

$$\implies 7x-7y+4=0$$

Therefore, the locus of point $$P$$ is $$7x-7y+4=0$$

Mathematics

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