Question

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

Answer 1

$\textbf{Given:}$

$\textsf{A(2,1) and B(1,2) and PA:PB=2:1}$

$\textbf{To find:}$

$\textsf{The locus of P}$

$\textbf{Solution:}$

$\textsf{Let the coordinates of the moving point P be (h,k)}$

$\mathsf{Consider,}$

$\mathsf{\dfrac{PA}{PB}=\dfrac{2}{1}}$

$\implies\mathsf{PA=2\,PB}$

$\implies\mathsf{\sqrt{(h-2)^2+(k-1)^2}=2\,\sqrt{(h-1)^2+(k-2)^2}}$

$\textsf{Squaring on bothsides, we get}$

$\mathsf{(h-2)^2+(k-1)^2=4[(h-1)^2+(k-2)^2]}$

$\mathsf{h^2+4-4h+k^2+1-2k=4[h^2+1-2h+k^2+4-4k]}$

$\mathsf{h^2+4-4h+k^2+1-2k=4h^2+4-8h+4k^2+16-16k}$

$\mathsf{h^2-4h+k^2-2k+5=4h^2-8h+4k^2-16k+20}$

$\mathsf{3h^2+3k^2-4h-14k+15=0}$

$\therefore\textsf{The locus of P is}$

$\boxed{\mathsf{3x^2+3x^2-4x-14y+15=0}}$

$\textbf{Find more:}$

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

https://brainly.in/question/3201572

A(c,0) and B(-c,0) are two points. If P is a point such that PA + PB = 2a where 0 < c < a, then find the locus of P.​

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