HOME

David Terr
Ph.D. Math, UC Berkeley

 

Home >> Pre-Calculus >> 1. Graphs, Functions, and Models

<< 1.2. Lines and Slopes

>> 1.4. Circles

 

1.3. Distance and Midpoint Formulas

Given two points P1 = (x1, y1) and P2 = (x2, y2) in the xy-plane, how do we determine the distance between them? It is easy to do so using the Pythagorean Theorem. First we construct a right triangle with vertices at P1, P2, and Q = (x2, y1) as shown.

distance formula derivation

Figure 1.3.1: Derivation of Distance Formula

 

Since the line segment P1Q is horizontal and QP2 is vertical, we see that the angle ∠P1QP2 is a right angle, so the triangle ΔP1QP2 is a right triangle with legs P1Q and QP2 and hypotenuse P1P2. It is easy to see that the legs P1Qand QP2 have lengths a = x2 - x1 and b = y2 - y1 respectively. To find the length c of the hypotenuse P1P2, we now apply the Pythagorean Theorem, which says a2 + b2 = c2, so that c = √a2+b2 = √(x2 - x1)2 + (y2 - y1)2. Thus we have found a formula for the distance between the points P1 and P2, namely

  • (1.3.1) formula (1.3.1)

 

Example 1: Compute the distance between the points P1 = (-3,-2) and P2 = (3,6).

Solution: Using formula (1.3.1), we find d(P1,P2) = √[3 - (-3)]2 + [6 - (-2)]2 = √62 + 82 = √36 + 64 = √100 = 10.

 

How about the midpoint of two given points? By definition, the midpoint of two points is the unique point on the line connecting these points which is equidistant from each point. We claim that the midpoint M of the points P1 = (x1, y1) and P2 = (x2, y2) is given by the formula

  • (1.3.2) M = (xM = (x1+x2)/2, yM = (y1+y2)/2).

How do we show that this is the correct formula? It suffices to show that M lies on the line connecting the points P1 and P2 and that d(M,P1) = d(M,P2). To show that M lies on the line defined by P1 and P2, we must show that yM= mxM + b, where m and b are given by formulas (1.2.1) and (1.2.2) respectively. This is a straighforward but tedious calculation, which we leave as an exercise. To see that d(M,P1) = d(M,P2), note that

d(M,P1)2 = (xM - x1)2 + (yM - y1)2

= [(x1+x2)/2 - x1]2 + [(y1+y2)/2 - y1]2

= [(x2-x1)/2]2 + [(y2-y1)/2]2.

A similar calculation shows that d(M,P2)2 = [(x1-x2)/2]2 + [(y1-y2)/2]2 = d(M,P1)2, so d(M,P2) = d(M,P1).

 

Example 2: Find the midpoint of the points P1 = (-3,-2) and P2 = (3,6).

Solution: Using formula (1.3.2), we find xM = [3 + (-3)] / 2 = 0 and yM = [6 + (-2)] / 2 = 2, so the midpoint is M=(0,2).

 

Home >> Pre-Calculus >> 1. Graphs, Functions, and Models

<< 1.2. Lines and Slopes

>> 1.4. Circles