For two points A(2,1) and B(1,2),P is a point | vedclass

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(B) Let the coordinates of point $P$ be $(x, y)$.Given $PA:PB = 2:1$,we have $PA^2 = 4PB^2$.Using the distance formula,$PA^2 = (x-2)^2 + (y-1)^2$ and

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$3x^2+3y^2-4x-14y+15=0$

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(B) Let the coordinates of point $P$ be $(x, y)$.Given $PA:PB = 2:1$,we have $PA^2 = 4PB^2$.Using the distance formula,$PA^2 = (x-2)^2 + (y-1)^2$ and $PB^2 = (x-1)^2 + (y-2)^2$.Substituting these into the equation:$(x-2)^2 + (y-1)^2 = 4[(x-1)^2 + (y-2)^2]$$x^2 - 4x + 4 + y^2 - 2y + 1 = 4[x^2 - 2x + 1 + y^2 - 4y + 4]$$x^2 + y^2 - 4x - 2y + 5 = 4[x^2 + y^2 - 2x - 4y + 5]$$x^2 + y^2 - 4x - 2y + 5 = 4x^2 + 4y^2 - 8x - 16y + 20$Rearranging the terms to one side:$3x^2 + 3y^2 - 4x - 14y + 15 = 0$.

For two points $A(2,1)$ and $B(1,2)$,$P$ is a point such that $PA:PB = 2:1$. Then the locus of $P$ is:

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