Find the locus of point P such that 2PA = 3PB | vedclass

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(D) Let the coordinates of point $P$ be $(x, y)$.Given the condition $2PA = 3PB$,we square both sides to get $4PA^{2} = 9PB^{2}$.Using the distance fo

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$5x^{2} + 5y^{2} - 72x + 54y + 225 = 0$

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(D) Let the coordinates of point $P$ be $(x, y)$.Given the condition $2PA = 3PB$,we square both sides to get $4PA^{2} = 9PB^{2}$.Using the distance formula,$PA^{2} = x^{2} + y^{2}$ and $PB^{2} = (x - 4)^{2} + (y + 3)^{2}$.Substituting these into the equation: $4(x^{2} + y^{2}) = 9[(x - 4)^{2} + (y + 3)^{2}]$.Expanding the terms: $4x^{2} + 4y^{2} = 9[x^{2} - 8x + 16 + y^{2} + 6y + 9]$.$4x^{2} + 4y^{2} = 9x^{2} - 72x + 144 + 9y^{2} + 54y + 81$.Rearranging the terms to one side: $5x^{2} + 5y^{2} - 72x + 54y + 225 = 0$.Thus,the locus of point $P$ is $5x^{2} + 5y^{2} - 72x + 54y + 225 = 0$.

Find the locus of point $P$ such that $2PA = 3PB$,where $A = (0, 0)$ and $B = (4, -3)$.

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