Find the equation of the locus of a point A(x, y) which is equidistant from B(0, 2) and C(2, 1)
-
A.
4x + 2y = 3 -
B.
4x – 3y = 1 -
C.
4x – 2y = 1 -
D.
4x + 2y = -1
Correct Answer: Option C
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
There is an explanation video available below.