Question

Find the equation of locus of a point which moves such that its distance from (0,2) is one third distance from (-2,3). ( I WILL MARK BRAINLIEST FOR CORRECT ANSWER )

Answer 1

Answer:

$8(x^2+y^2)-4x-30y+23=0$

Step-by-step explanation:

Distance formula

$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Let P(x, y) = any point on the locus

Let A = (0, 2)    

Let B = (-2, 3)

If a point moves such that its distance from (0, 2) is one third distance from (-2, 3):

$PA=\dfrac{1}{3}PB$

Therefore, using the distance formula:

$\implies \sqrt{(x-0)^2+(y-2)^2}=\dfrac{1}{3}\sqrt{(x-(-2))^2+(y-3)^2}$

Square both sides:

$\implies x^2+(y-2)^2=\dfrac{1}{9}[(x+2)^2+(y-3)^2]$

$\implies x^2+y^2-4y+4=\dfrac{1}{9}(x^2+4x+4+y^2-6y+9)$

Multiply both sides by 9:

$\implies 9x^2+9y^2-36y+36=x^2+4x+4+y^2-6y+9$

$\implies 8x^2+8y^2-4x-30y+23=0$

$\implies 8(x^2+y^2)-4x-30y+23=0$