SECTION 1.2 The Distance and Midpoint Formulas 15 • A similar argument holds if the line joining P1 and P2 is vertical. See Figure 25(b). ■ Figure 25 x y P2 5 (x2, y1) P1 5 (x1, y1) d(P 1, P2) (a) P2 5 (x1, y2) P1 5 (x1, y1) d(P1, P2) x y (b) ƒ x2 2 x1 ƒ x1 x2 y1 ƒ y2 2 y1 ƒ x1 y1 y2 Solution EXAMPLE 2 Finding the Length of a Line Segment Find the length of the line segment shown in Figure 26. The length of the line segment is the distance between the points ( ) ( ) = = − P x y , 4, 5 1 1 1 and ( ) ( ) = = P x y , 3, 2 . 2 2 2 Using the distance formula (1) with = − = = x y x 4, 5, 3, 1 1 2 and = y 2, 2 the length d is d x x y y 3 4 2 5 7 3 49 9 58 7.62 2 1 2 2 1 2 2 2 2 2 ( ) ( ) ( ) [ ] ( ) ( ) = − + − = − − + − = + − = + = ≈ Exact Approximate Now Work PROBLEMS 11 AND 15 The distance between two points ( ) = P x y , 1 1 1 and ( ) = P x y , 2 2 2 is never a negative number.Also, the distance between two points is 0 only when the points are identical—that is,when = x x 1 2 and = y y . 1 2 And,because ( ) ( ) − = − x x x x 2 1 2 1 2 2 and ( ) ( ) − = − y y y y , 2 1 2 1 2 2 it makes no difference whether the distance is computed from P1 to P2 or from P2 to P;1 that is, d P P d P P , , . 1 2 2 1 ( ) ( ) = Rectangular coordinates enable us to translate geometry problems into algebra problems, and vice versa.The next example shows how algebra (the distance formula) can be used to solve geometry problems. Figure 27 x y –3 3 3 C = (3, 1) B = (2, 3) A = (–2, 1) Using Algebra to Solve a Geometry Problem Consider the three points A B 2, 1 , 2, 3 , ( ) ( ) = − = and ( ) = C 3, 1 . (a) Plot each point and form the triangle ABC. (b) Find the length of each side of the triangle. (c) Show that the triangle is a right triangle. (d) Find the area of the triangle. Solution EXAMPLE 3 (a) Figure 27 shows the points A, B, C and the triangle ABC. (b) To find the length of each side of the triangle, use the distance formula, equation (1). d A B d B C d A C , 2 2 3 1 16 4 20 2 5 , 3 2 1 3 1 4 5 , 3 2 1 1 25 0 5 2 2 2 2 2 2 ( ) ( ) [ ] ( ) ( ) ( ) ( ) ( ) ( ) [ ] ( ) = − − + − = + = = = − + − = + = = − − + − = + = (c) If the sum of the squares of the lengths of two of the sides equals the square of the length of the third side, then the triangle is a right triangle. Looking at Figure 27, it seems reasonable to conjecture that the angle at vertex B might be a right angle. We shall check to see whether d A B d B C d A C , , , 2 2 2 ( ) [ ] ( ) [ ] ( ) [ ] + = (continued) Figure 26 6 0 25 (24, 5) (3, 2) 5
RkJQdWJsaXNoZXIy NjM5ODQ=