JAMB - Mathematics (2019 - No. 52)

Find the equation of the locus of a point A(x, y) which is equidistant from B(0, 2) and C(2, 1)

4x + 2y = 3

4x - 3y = 1

4x - 2y = 1

4x + 2y = -1

Since A(x, y) is the point of equidistance between B and C, then

AB = AC

(AB)\(^2\) = (AC)\(^2\)

Using the distance formula,

(x - 0)\(^2\) + (y - 2)\(^2\) = (x - 2)\(^2\) + (y - 1)\(^2\)

x\(^2\) + y\(^2\) - 4y + 4 = x\(^2\) - 4x + 4 + y\(^2\) - 2y + 1

x\(^2\) - x\(^2\) + y\(^2\) - y\(^2\) + 4x - 4y + 2y = 5 - 4

4x - 2y = 1