Let A, B and C be three points in a plane. Th | vedclass

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(A) Let the coordinates of the points be $A(x_1, y_1)$,$B(x_2, y_2)$,and $C(x_3, y_3)$. Let $P(x, y)$ be the variable point.Given the condition $PA^2

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straight line

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(A) Let the coordinates of the points be $A(x_1, y_1)$,$B(x_2, y_2)$,and $C(x_3, y_3)$. Let $P(x, y)$ be the variable point.Given the condition $PA^2 + PB^2 = 2PC^2$.Substituting the distance formula: $(x - x_1)^2 + (y - y_1)^2 + (x - x_2)^2 + (y - y_2)^2 = 2[(x - x_3)^2 + (y - y_3)^2]$.Expanding the terms: $(x^2 - 2xx_1 + x_1^2 + y^2 - 2yy_1 + y_1^2) + (x^2 - 2xx_2 + x_2^2 + y^2 - 2yy_2 + y_2^2) = 2(x^2 - 2xx_3 + x_3^2 + y^2 - 2yy_3 + y_3^2)$.Simplifying the equation: $2x^2 + 2y^2 - 2x(x_1 + x_2) - 2y(y_1 + y_2) + x_1^2 + x_2^2 + y_1^2 + y_2^2 = 2x^2 + 2y^2 - 4xx_3 - 4yy_3 + 2x_3^2 + 2y_3^2$.The $x^2$ and $y^2$ terms cancel out,leaving a linear equation in $x$ and $y$ of the form $Ax + By + C = 0$.Therefore,the locus of point $P$ is a straight line.

Let $A, B$ and $C$ be three points in a plane. The locus of a point $P$ moving such that $PA^2 + PB^2 = 2PC^2$ is a

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