Question

if A(-3,2) and B(1,-4), find the equation of the locus of point P such that 3(PA)=2( PB)

Answer 1

use distance formula, if two points A(x₁,y₁) and B(x₂,y₂) are given, then
AB = $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

now, come to the question,
Let locus of point P(α,β) ,
Given, 3(PA) = 2(PB)
taking square both sides,
9PA² = 4PB²

now, PA= $\sqrt{(\alpha+3)^2+(\beta-2)^2}$
so, PA² = (α + 3)² + (β -2 )²

similarly, PB = $\sqrt{(\alpha-1)^2+(\beta+4)^2}$
so, PB² = (α - 1)² + (β + 4)²

Then, 9[(α + 3)² + (β -2 )² ] = 4[(α - 1)² + (β + 4)²
⇒9(α + 3)² + 9(β - 2)² = 4(α -1)² + 4(β + 4)²
⇒ {3(α+3)}² - {2(α-1)}² = {2(β+4)}²-{3(β-2)}²
⇒(3α + 9 - 2α + 2)(3α + 9 + 2α - 2) = (2β + 8 - 3β + 6)(2β + 8 + 3β -6)
⇒ (α + 11)(5α + 7) = (-β + 14)(5β + 2)
⇒5α² + 62α + 77 = -5β² +68β + 28
⇒5α² + 5β² + 62α - 68β + 49 = 0

Now, put α = x and β = y
Then, 5x² + 5y² + 62x - 68y + 49 = 0 is the locus of point P e.g., circle