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

To find the locus of a point A that is equidistant from two given points B and C, we need to find the perpendicular bisector of the line segment joining B and C. The locus of point A will be the line that is perpendicular to the line segment BC and passes through the midpoint of BC. First, we need to find the midpoint of the line segment BC: Midpoint = ((0+2)/2, (2+1)/2) = (1, 1.5) The slope of the line segment BC is (1-2)/(2-0) = -1/2. The slope of a line perpendicular to this line is the negative reciprocal of the slope, which is 2. Using the point-slope form of a line with the midpoint (1, 1.5) and slope 2, we get: y - 1.5 = 2(x - 1) Simplifying the equation, we get: y = 2x - 2 + 1.5 y = 2x - 0.5 Therefore, the equation of the locus of point A that is equidistant from B(0, 2) and C(2, 1) is 4x - 2y = 1, which is option (C) in the given choices.