3. To _nd the point symmetric about the origin, we replace the x- and y-coordinates with theiropposites to get (2;?3).xyP(?2; 3)(?2;3)(2; 3)(2;3)?3 ?2 ?1 1 2 3?3?2?1123One way to visualize the processes in the previous example is with the concept of a reection. Ifwe start with our point (?2; 3) and pretend that the x-axis is a mirror, then the reection of (?2; 3)across the x-axis would lie at (?2;?3). If we pretend that the y-axis is a mirror, the reectionof (?2; 3) across that axis would be (2; 3). If we reect across the x-axis and then the y-axis, wewould go from (?2; 3) to (?2;?3) then to (2;?3), and so we would end up at the point symmetricto (?2; 3) about the origin. We summarize and generalize this process below.ReectionsTo reect a point (x; y) about the:• x-axis, replace y with ?y.• y-axis, replace x with ?x.• origin, replace x with ?x and y with ?1.1.3 Distance in the PlaneAnother important concept in Geometry is the notion of length. If we are going to unite Algebraand Geometry using the Cartesian Plane, then we need to develop an algebraic understanding ofwhat distance in the plane means. Suppose we have two points, P (x0; y0) and Q(x1; y1) ; in theplane. By the distance d between P and Q, we mean the length of the line segment joining P withQ. (Remember, given any two distinct points in the plane, there is a unique line containing both1.1 Sets of Real Numbers and The Cartesian Coordinate Plane 11points.) Our goal now is to create an algebraic formula to compute the distance between these twopoints. Consider the generic situation below on the left.P (x0; y0)Q(x1; y1)dP (x0; y0)Q(x1; y1)dy.