Find the equation of the locus of a point P(x, y) that moves such that it is equidistant f… — JAMB UTME

Find the equation of the locus of a point $P(x, y)$ that moves such that it is equidistant from the fixed points $A(-1, 3)$ and $B(5, -1)$ .

Answer: 3x - 2y - 4 = 0 / 3x-2y-4=0 / 3x - 2y = 4 / 3x-2y=4 / 12x - 8y - 16 = 0 / y = (3/2)x - 2 / y = 1.5x - 2

The equation of the locus is

3x - 2y - 4 = 0

(or

3x - 2y = 4

The locus of a point equidistant from two fixed points

A(-1, 3)

and

B(5, -1)

is the perpendicular bisector of the line segment joining them. Equating the squared distances

(x+1)^2 + (y-3)^2 = (x-5)^2 + (y+1)^2

and simplifying yields the linear equation

3x - 2y - 4 = 0

Set up the distance equality condition using the distance formula.

\\sqrt{(x - (-1))^2 + (y - 3)^2} = \\sqrt{(x - 5)^2 + (y - (-1))^2}

Since point

P(x, y)

is equidistant from

A

and

B

PA = PB

Square both sides to remove the radical signs.

(x + 1)^2 + (y - 3)^2 = (x - 5)^2 + (y + 1)^2

Squaring both sides eliminates square roots and simplifies polynomial expansion.

Expand all squared terms on both sides.

x^2 + 2x + 1 + y^2 - 6y + 9 = x^2 - 10x + 25 + y^2 + 2y + 1

Expanding allows gathering like terms.

Cancel

x^2

and

y^2

from both sides and collect all terms on one side.

(2x + 10x) + (-6y - 2y) + (10 - 26) = 0 \\Rightarrow 12x - 8y - 16 = 0

Combining like terms simplifies the locus equation into standard linear form.

Divide the entire equation by the common factor of 4.

3x - 2y - 4 = 0

Expressing the linear equation in its simplest form gives the perpendicular bisector of line segment

AB

Locus equidistant from two fixed points (Perpendicular Bisector)

Estimated Time:2m 0s

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