Question

If a point P(x,y) is the equidistant from the points A(2,3) and B(6,1), find the equation of locus of moving point P.​



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Answer 1

Answer:

$y=2x-6$

Step-by-step explanation:

A locus can be defined as a curve or figure formed by all the points satisfying a particular equation of the relation between coordinates.

The condition stated in the question is such that a generic (x,y) point of the curve is equidistant from the points A(2,3) and B(6,1).

The distance d1 from (x,y) to (2,3) is:

$d_1=\sqrt{(x-2)^2+(y-3)^2}$

The distance d2 from (x,y) to (6,1) is:

$d_2=\sqrt{(x-6)^2+(y-1)^2}$

Since d1=d2:

$\sqrt{(x-2)^2+(y-3)^2}=\sqrt{(x-6)^2+(y-1)^2}$

Squaring both sides:

$(x-2)^2+(y-3)^2=(x-6)^2+(y-1)^2$

Operating:

$x^2-4x+4+y^2-6y+9=x^2-12x+36+y^2-2y+1$

Simplifying all the squares:

$-4x+4x+4-6y+9=-12x+36-2y+1$

Moving the variables to the left side and the numbers to the right side:

$-4x+12x-6y+2y=36+1-4-9$

Simplifying:

$8x-4y=24$

Dividing by 4:

$2x-y=6$

Or, equivalently:

$\boxed{y=2x-6}$