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1.3. Distance and Midpoint Formulas

Given two points P_{1} = (x_{1}, y_{1}) and P_{2} = (x_{2}, y_{2}) 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 P_{1}, P_{2}, and Q = (x_{2}, y_{1}) as shown.

Figure 1.3.1: Derivation of Distance Formula

Since the line segment P_{1}Q is horizontal and QP_{2} is vertical, we see that the angle ∠P_{1}QP_{2} is a right angle, so the triangle ΔP_{1}QP_{2} is a right triangle with legs P_{1}Q and QP_{2} and hypotenuse P_{1}P_{2}. It is easy to see that the legs P_{1}Qand QP_{2} have lengths a = x_{2} - x_{1} and b = y_{2} - y_{1} respectively. To find the length c of the hypotenuse P_{1}P_{2}, we now apply the Pythagorean Theorem, which says a^{2} + b^{2} = c^{2}, so that c = √a^{2}+b^{2} = √(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}. Thus we have found a formula for the distance between the points P_{1} and P_{2}, namely

**(1.3.1)**

**Example 1:** Compute the distance between the points P_{1} = (-3,-2) and P_{2} = (3,6).

**Solution: **Using formula (1.3.1), we find d(P_{1},P_{2}) = √[3_{} - (-3)_{}]^{2} + [6_{} - (-2)]^{2} = √6^{2} + 8^{2} = √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 P

_{1}= (x

_{1}, y

_{1}) and P

_{2}= (x

_{2}, y

_{2}) is given by the formula

**(1.3.2)**M = (x_{M}= (x_{1}+x_{2})/2, y_{M}= (y_{1}+y_{2})/2).

How do we show that this is the correct formula? It suffices to show that M lies on the line connecting the points P_{1} and P_{2} and that d(M,P_{1}) = d(M,P_{2}). To show that M lies on the line defined by P_{1} and P_{2}, we must show that y_{M}= mx_{M} + 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,P_{1}) = d(M,P_{2}), note that

d(M,P_{1})^{2} = (x_{M} - x_{1})^{2} + (y_{M} - y_{1})^{2}

= [(x_{1}+x_{2})/2 - x_{1}]^{2} + [(y_{1}+y_{2})/2 - y_{1}]^{2}

= [(x_{2}-x_{1})/2]^{2} + [(y_{2}-y_{1})/2]^{2}.

A similar calculation shows that d(M,P_{2})^{2} = [(x_{1}-x_{2})/2]^{2} + [(y_{1}-y_{2})/2]^{2 }= d(M,P_{1})^{2}, so d(M,P_{2}) = d(M,P_{1}).

**Example 2: **Find the midpoint of the points P_{1 }= (-3,-2) and P_{2} = (3,6).

**Solution: **Using formula (1.3.2), we find x_{M} = [3 + (-3)_{}] / 2 = 0 and y_{M} = [6 + (-2)_{}] / 2 = 2, so the midpoint is M=(0,2).

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