Question

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

Answer 1

Let P $=(x,y)$

Distanec between P and A $=\sqrt{(x-0{)}^2+(y-0{)}^2}=\sqrt{x^2+Y^2}$

Distanec between P and B $=\sqrt{(x-4{)}^2+(y-(-3){)}^2}$

given, 2PA=3PB

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

So, Locus of P is $5x^2+5y^2-72x+54y+225=0$