16 CHAPTER 1 Graphs Using the results in part (b) yields d A B d B C d A C , , 2 5 5 20 5 25 , 2 2 2 2 2 ( ) ( ) ( ) [ ] ( ) [ ] ( ) [ ] + = + = + = = It follows from the converse of the Pythagorean Theorem that triangle ABC is a right triangle. (d) Because the right angle is at vertex B, the sides AB and BC form the base and height of the triangle. Its area is = ⋅ ⋅ = ⋅ ⋅ = Area 1 2 Base Height 1 2 2 5 5 5 square units Now Work PROBLEM 25 2 Use the Midpoint Formula We now derive a formula for the coordinates of the midpoint of a line segment . Let P x y , 1 1 1 ( ) = and ( ) = P x y , 2 2 2 be the endpoints of a line segment, and let ( ) = M x y , be the point on the line segment that is the same distance from P1 as it is from P .2 See Figure 28. The triangles P AM 1 and MBP2 are congruent. [Do you see why? ( ) ( ) = d P M d M P , , 1 2 is given; also, APM BMP 1 2 ∠ = ∠ * and ∠ = ∠ P MA MP B . 1 2 So, we have angle–side–angle.] Because triangles P AM 1 and MBP2 are congruent, corresponding sides are equal in length. That is, − = − = + = + − = − = + = + x x x x x x x x x x y y y y y y y y y y and 2 2 2 2 1 2 1 2 1 2 1 2 1 2 1 2 In Words To find the midpoint of a line segment, average the x -coordinates of the endpoints, and average the y -coordinates of the endpoints. THEOREM Midpoint Formula The midpoint ( ) = M x y , of the line segment from ( ) = P x y , 1 1 1 to ( ) = P x y , 2 2 2 is M x y x x y y , 2 , 2 1 2 1 2 ( ) ( ) = = + + (2) Finding the Midpoint of a Line Segment Find the midpoint of the line segment from P 5, 5 1 ( ) = − to ( ) = P 3, 1 . 2 Plot the points P1 and P2 and their midpoint. Solution EXAMPLE 4 Use the midpoint formula (2) with = − = = x y x 5, 5, 3, 1 1 2 and = y 1. 2 The coordinates ( ) x y , of the midpoint M are = + = − + = − = + = + = x x x y y y 2 5 3 2 1 and 2 5 1 2 3 1 2 1 2 That is, ( ) = − M 1, 3 . See Figure 29. Now Work PROBLEM 31 *A postulate from geometry states that the transversal PP1 2 forms congruent corresponding angles with the parallel line segments PA1 and MB. Figure 28 x1 x2 x y1 x y A = (x, y1) M = (x, y) P1 = (x1, y1) P2 = (x2, y2) B = (x2, y) y y2 x – x1 x2 – x y2 – y y – y1 Figure 29 x y 5 5 –5 P2 5 (3, 1) P1 5 (–5, 5) M 5 (–1, 3)
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