Question

If $A(-3,2)$ and $B(1,-4)$. Find the equation of locus of point P, such that $3PA=2PB$

Answer 1

Given, $A(-3,2)andB(1,-4)$

Let coordinate of P is $(x,y)$

So, According to question

$3PA=2PB$

$3\sqrt{(-3-x{)}^2+(2-y{)}^2}=2\sqrt{(1-x{)}^2+(-4-y{)}^2}$

by squaring both side

$(3\sqrt{(-3-x{)}^2+(2-y{)}^2}{)}^2=(2\sqrt{(1-x{)}^2+(-4-y{)}^2}{)}^2$

$9(x^2+9+6x+y^2+4-4y)=4(1+x^2-2x+y^2+16+8y)$

$9(x^2+6x+y^2+13-4y)=4(x^2-2x+y^2+17+8y)$

$9x^2+54x+9y^2-36y+117=4x^2+4y^2-8x+32y+68$

$5x^2+5y^2+62x-68y+49=0$

So, this is the eq. of locus of P.