Numbers & Geometry Points and Distances

6/3/2013 Numbers and Geometry 2 Distance d between numbers a and b d = Example: | = Points and Distances  20 OR a – b | b – a | = -20 = = 20 | -8 – 12 | 12 – ( -8 ) = 20

6/3/2013 Numbers and Geometry 3 Distance d between points Example: Find the distance between points (-5,-3) and (8,6) x y Points and Distances   (-5,-3) (8,6)  d (8,-3) a b a = | 8 –(-5) b = | 6 – (-3) = 13 a2a2 = 13 2 = 9 b2b2 = 9292 = 81 = a2a2 b2b2 + = 169 = 250

6/3/2013 Numbers and Geometry 4 Distance d between points Example: Find the distance between points (-5,-3) and (8,6) Apply Pythagorean Theorem Right triangle with legs a and b and hypotenuse d x y Points and Distances   (-5,-3) (8,6)  d (8,-3) = 250 a b d2d2 = a2a2 b2b2 + =10  5 = 250  d

6/3/2013 Numbers and Geometry 5 Distance d between points Example: Find the distance between points (-5,-3) and (8,6) x y Points and Distances   (-5,-3) (8,6)  d (8,-3) =10  5 ≈ 5 (3.16) 250  = d a b = 15.8 d

6/3/2013 Numbers and Geometry 6 Distance d between points Distance in general for points (x 1, y 1 ) and (x 2, y 2 ) From Pythagoras Points and Distances x y   (x 1, y 1 ) (x 2, y 2 )  d (x 2, y 1 )    (x 1, y 1 ) (x 2, y 2 ) (x 2, y 1 ) d | x 2 – x 1 | y 2 – y 1 d2d2 | x 2 – x 1 2 | y 2 – y = ( ) y 2 – y 1 2 ( ) x 2 – x = = ( ) x 2 – x 1 2 ( ) y 2 – y √ d = ( ) x 2 – x 1 2 ( ) y 2 – y √ d The Distance Formula

6/3/2013 Numbers and Geometry 7 Line Segment Midpoint Consider segment joining points (x 1, y 1 ), (x 2, y 2 ) The Halfway Point x y (x 1, y 1 ) (x 2, y 2 ) d d       x1x1 x2x2 y1y1 y2y2 m n a a b b a a    b b (m, n) So, midpoint is (m, n) = x 1 + x 2 2 ( y 1 + y 2 2 ), m = x 1 x n = y 1 y 2 2 +

6/3/2013 Numbers and Geometry 8 Example Find the midpoint of the line segment between points (2, -10) and (9, 7) Midpoint coordinates (m, n) : Midpoint is (5.5, -1.5) The Halfway Point x y   (2, -10) (9, 7)  (m, n) m = = 2 11 = 5.5 n = = 2 -3 = -1.5

6/3/2013 Numbers and Geometry 9 Think about it !