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bagirrra123 [75]
2 years ago
15

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 )
Mathematics
1 answer:
scoray [572]2 years ago
3 0

Answer:

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

Step-by-step explanation:

<u />

<u>Distance formula</u>

\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

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