Find the locus of a particle which moves in the first quadrant so that it is equidistant from the lines x = 0 and y = 0 (where k is a constant)

Answer Details

The locus of a particle refers to the path that the particle follows in the coordinate plane. In this case, we want to find the path that a particle follows if it always remains at the same distance from the lines x = 0 and y = 0. Since the particle is equidistant from the x-axis and y-axis, it must lie on the perpendicular bisector of the line segment joining the origin to any point on the x-axis or y-axis. Let's consider the point (a,0) on the x-axis. The distance between this point and the particle is given by the distance formula: √[ (a-x)^2 + (0-y)^2 ] Similarly, for the point (0,b) on the y-axis, the distance between this point and the particle is given by: √[ (0-x)^2 + (b-y)^2 ] Since the particle is equidistant from both lines, we can set these two distances equal to each other: √[ (a-x)^2 + (0-y)^2 ] = √[ (0-x)^2 + (b-y)^2 ] Squaring both sides of the equation, we get: (a-x)^2 + y^2 = x^2 + (b-y)^2 Simplifying the equation yields: 2xy - 2bx - 2ay + a^2 + b^2 = 0 This is the equation of a hyperbola with the x-axis and y-axis as its asymptotes. However, we need to take into account that the particle is in the first quadrant, so we only consider the part of the hyperbola that lies in the first quadrant. Thus, the correct answer is x - y = 0.