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

Answer:

Step-by-step explanation:

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