Using coordinates, write a detailed step-by-step proof that the set of points equidistant from two fixed points, A and B, is the perpendicular bisector of the segment AB.[br][br]Let AB be a line segment with endpoints A and B. Let C be the midpoint of AB such that C is centered at the origin. Thus, we know that C = (0, 0). We will denote A and B as follows: A = (x, 0) and B = (-x, 0).[br][br]Now, let D be another point which is equidistant from line segment AB. Denote D as [math](x_1,y_1)[/math].[br]Using the distance formula, we see that:[br][br]1) [math]dist\left(A,D\right)=\sqrt[]{\left(x-x_1\right)^2+\left(0-y_1\right)^2}[/math][br]2)[math]dist\left(B,D\right)=\sqrt[]{\left(-x-x_1\right)^2+\left(0-y_1\right)^2}[/math][br][br][br]Since we know that D is equidistance from both A and B, we can say that dist(A,D) = dist(B,D).[br][br]Consider the following:[br][br][math]\sqrt[]{\left(x-x_1\right)^2+\left(0-y_1\right)^2}=\sqrt[]{\left(-x-x_1\right)^2+\left(0-y_1\right)^2}[/math][br][math]\Longrightarrow\left(x-x_1\right)^2+\left(0-y_1\right)^2=\left(-x-x_1\right)^2+\left(0-y_1\right)^2[/math][br][math]\Longrightarrow\left(x-x_1\right)^2=\left(-x-x_1\right)^2[/math][br][math]\Longrightarrow\left(x-x_1\right)^{ }=\left(-x-x_1\right)^{ }[/math][br][math]\Longrightarrow2x=x_1-x_1[/math][br][math]\Longrightarrow x=0[/math].[br][br]Thus, for any point D that we choose equidistant from A and B, the x coordinate must always be zero. We know that C is the midpoint of AB, and the x coordinate of C is zero, thus lying on the y-axis. Note that by definition, the y-axis is perpendicular to the x-axis. Thus, any point D equidistant from A and B forms the line CD which is perpendicular to AB since point points lie on the y-axis. Therefore, the set of all points equidistant from AB must lie on line CD, the perpendicular bisector of AB.[br][br][br][br][br]