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A curve is traced by a point P(x, y) which moves such that its distance from the point A(-1,1) is three times its distance from the point B(2,-1). Determine the equation of the curve.

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4 votes

Answer:


8x^2+8y^2+43-38x+20y=0

Explanation:

Let
A\left ( x_1,y_1 \right ) and
B\left ( x_2,y_2 \right ) be two points then distance AB is equal to
AB=√(\left ( x_2-x_1 \right )^2+\left ( y_2-y_1 \right )^2)

Here, a curve is traced by a point
P(x,y) which moves such that its distance from the point
A(-1,1) is three times its distance from the point
B(2,-1) i.e
AP=3BP

Using distance formula,


AP=√((-1-x)^2+(1-y)^2)


BP=√((2-x)^2+(-1-y)^2)


AP=3BP\\√((-1-x)^2+(1-y)^2)=3√((2-x)^2+(-1-y)^2)

On squaring both sides, we get


(-1-x)^2+(1-y)^2=9\left [ (2-x)^2+(-1-y)^2 \right ]\\1+x^2+2x+1+y^2-2y=9\left ( 4 +x^2-4x+1+y^2+2y\right )\\1+x^2+2x+1+y^2-2y=36+9x^2-36x+9+9y^2+18y\\8x^2+8y^2+43-38x+20y=0

So, equation of curve is
8x^2+8y^2+43-38x+20y=0

User Iateadonut
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