14 CHAPTER 1 Graphs The distance formula provides a straightforward method for computing the distance between two points. THEOREM Distance Formula The distance between two points P x y , 1 1 1 ( ) = and P x y , , 2 2 2 ( ) = denoted by d P P , , 1 2 ( ) is d P P x x y y , 1 2 2 1 2 2 1 2 ( ) ( ) ( ) = − + − (1) Proof of the Distance Formula Let x y , 1 1 ( ) denote the coordinates of point P1 and let x y , 2 2 ( ) denote the coordinates of point P .2 • Assume that the line joining P1 and P2 is neither horizontal nor vertical. Refer to Figure 24(a).The coordinates of P3 are x y , . 2 1 ( ) The horizontal distance from P1 to P3 equals the absolute value of the difference of the x -coordinates, − x x . 2 1 The vertical distance from P3 to P2 equals the absolute value of the difference of the y -coordinates, y y . 2 1 − See Figure 24(b). The distance d P P , 1 2 ( ) is the length of the hypotenuse of the right triangle, so, by the Pythagorean Theorem, it follows that d P P x x y y x x y y d P P x x y y , , 1 2 2 2 1 2 2 1 2 2 1 2 2 1 2 1 2 2 1 2 2 1 2 ( ) [ ] ( ) ( ) ( ) ( ) ( ) = − + − = − + − = − + − Figure 24 P1 = (x1, y1) (a) P3 = (x2, y1) y y x x P2 = (x2, y2) y2 y1 (b) P2 = (x2, y2) P1 = (x1, y1) d(P1, P2) P3 = (x2, y1) x1 x2 x1 x2 y2 y1 ƒ x2 - x1 ƒ ƒ y2 - y1 ƒ • If the line joining P1 and P2 is horizontal, then the y -coordinate of P1 equals the y -coordinate of P ;2 that is, = y y . 1 2 Refer to Figure 25(a) on the next page. In this case, the distance formula (1) still works, because for = y y , 1 2 it reduces to ( ) ( ) ( ) = − + = − = − d P P x x x x x x , 0 1 2 2 1 2 2 2 1 2 2 1 of length 4 (since − = 5 1 4), and the other is of length 3 (since − = 6 3 3). By the Pythagorean Theorem, the square of the distance d that we seek is = + = + = = = d d 4 3 16 9 25 25 5 2 2 2 Figure 23 x y 3 (1, 3) (5, 6) 3 4 d 6 (b) 6 3 (5, 3) x y 3 (1, 3) (5, 6) d 6 (a) 6 3 Need to Review? The Pythagorean Theorem and its converse are discussed in Sect ion A .2, pp. A14–A15. In Words To compute the distance between two points, find the difference of the x -coordinates, square it, and add this to the square of the difference of the y -coordinates. The square root of this sum is the distance.
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