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

Hint: We will assume the point P to be $(x,y)$ and will apply the distance formula between the points P and A to get PA and then to point P and point B to get PB. Now we know PA and PB, we will use the given relation $2PA = 3PB$ to get the locus of point P.

Formula Used:
If there are two points A having coordinates $({x_1},{y_1})$and B be $({x_2},{y_2})$ then distance between these points (AB) is given by
$AB = \sqrt {{{({x_1} - {x_2})}^2} + {{({y_2} - {y_1})}^2}} $

Complete step by step solution:
Given, A is $(0,0)$and B is $(4, - 3)$.Let the point P have coordinates $(x,y)$.
We know that the distance between two points $({x_1},{y_1})$ and $({x_2},{y_2})$ can be calculated using the distance formula as
Distance =$\sqrt {{{({x_1} - {x_2})}^2} + {{({y_2} - {y_1})}^2}} $
Hence the distance between points A$(0,0)$ and P$(x,y)$ can be calculated as –
$PA = \sqrt {{{\left( {x - 0} \right)}^2} + {{\left( {y - 0} \right)}^2}} $
$ \Rightarrow PA = \sqrt {{x^2} + {y^2}} $
Again, using the distance formula between points P$(x,y)$and B$(4, - 3)$ to calculate PB
$PB = \sqrt {{{\left( {x - 4} \right)}^2} + {{\left( {y - \left( { - 3} \right)} \right)}^2}} $
$ \Rightarrow PB = \sqrt {{{\left( {x - 4} \right)}^2} + {{\left( {y + 3} \right)}^2}} $
Given that $2PA = 3PB$
Hence, substituting $PA = \sqrt {{x^2} + {y^2}} $ and $PB = \sqrt {{{\left( {x - 4} \right)}^2} + {{\left( {y + 3} \right)}^2}} $in the above given relation
$2\sqrt {{x^2} + {y^2}} = 3\sqrt {{{\left( {x - 4} \right)}^2} + {{\left( {y + 3} \right)}^2}} $
Squaring on the both sides of the relation
${\left( {2\sqrt {{x^2} + {y^2}} } \right)^2} = 9{\left( {\sqrt {{{\left( {x - 4} \right)}^2} + {{\left( {y + 3} \right)}^2}} } \right)^2}$
$ \Rightarrow 4\left( {{x^2} + {y^2}} \right) = 9{\left( {x - 4} \right)^2} + {\left( {y + 3} \right)^2}$
Simplifying the expression
$4{x^2} + 4{y^2} = 9\left( {\left( {{x^2} + 16 - 8x} \right) + \left( {{y^2} + 9 + 6y} \right)} \right)$
Rearranging the terms
$9{x^2} + 144 - 72x + 9{y^2} + 81 + 54y = 4{x^2} + 4{y^2}$
$ \Rightarrow 5{x^2} + 5{y^2} - 72x + 54y + 225 = 0$
Hence the locus of point P is $5{x^2} + 5{y^2} - 72x + 54y + 225 = 0$.

Note: In the problems of finding locus, students may get confused as to where to start. In such types of questions, it is advised to start by assuming the point as a general point $(x,y)$ and then apply all the conditions given in the question to solve for locus.