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
Explanation
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
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