Geometry 13 5 The Midpoint Formula The Midpoint

Geometry 13. 5 The Midpoint Formula

The Midpoint Formula The midpoint of the segment that joins points (x 1, y 1) and (x 2, y 2) is the point • (1, 5) • (-4, 2) • (6, 8)

How does it work? Step 1: Coordinate of the Midpoint of AB A ● B ● 4 + 12 2 4 8 12 =8

How does it work? Step 2: Coordinate of the Midpoint of AC 7 A● 1+7 4 1 2 C● 4 8 12 =4

How does it work? Step 2: Coordinate of the Midpoint of BC 7 A ● 4 1 B ● ● 4 + 12 1+7 2 2 , C● 4 8 12 (8, 4)

Exercises 1. A (3, 5) B (7, -5) midpoint: 3+7 5+(-5) 2 , 2 (5, 0) 2. A (0, 4) B (4, 3) midpoint: 0+4 4+3 2 , 2 (2, 7 2 )

Exercises 3. M (3, 5) A (0, 1) B (x, y) This is the midpoint To find the coordinates of B: x-coordinate: y-coordinate: 0+x 3= 2 6=0+x x=6 1+y 5= 2 10 = 1 + y y=9 (6, 9)

Exercises 4. A (p, g) B (p + 4, g) midpoint: 2 p+4 2 g 2 , 2 (p + 2, g) Do #5 and #6 on your own and check answers below: 5. k -m 2, 2 6. Point B: (6, 1)

7. Given points A (5, 2) and B(3, -6) show that P (0, -1) is on the perpendicular bisector of AB. (5, 2) ●A P ● (0, -1) M ●(4, -2) ●B Midpoint of AB: 5+3 (3, -6) 2+6 Slope of AB: m = 5 -3 = -2+1 Slope of PM: m = 4 -0 = In order to be on the perpendicular bisector, point P would have to be on the segment that is both perpendicular to AB and goes through it’s midpoint. 4 -¼ 2 , 2+(-6) 2 Perpendicular (4, -2)

Homework pg. 545 WE #1 -19 pg. 538 #15, 17