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Distance & Midpoint in the Coordinate Plane

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Presentation on theme: "Distance & Midpoint in the Coordinate Plane"— Presentation transcript:

1 Distance & Midpoint in the Coordinate Plane

2 Coordinate Plane Ordered Pair: (x , y) A(2 , -1) Quad. II ( - , +)
y-axis (Dependent) Quad. II ( - , +) Quad. I ( + , +) x-axis (Independent) A Quad. III ( - , -) Quad. IV ( + , -)

3 Graph the following equation.
y = 3x + 1 2 methods x/y chart Slope-Intercept Form y-intercept Slope

4 Graphing and finding the distance
Graph A(2 , -2) & B(2 , 6) Graph C(1 , 2) & D(5 , 6)

5 Distance Formula dAB = √(x2 – x1)2 + (y2 – y1)2
Distance Formula: The distance (d) between two points A(x1 , y1) and B(x2 , y2) is dAB = √(x2 – x1)2 + (y2 – y1)2

6 Ex.1: Find the distance between the following points.
R(-2 , 6) & S(4 , 4) x1 y1 x2 y2 dRS = √(x2 – x1)2 + (y2 – y1)2 = √(4 – (-2))2 + (4 – 6)2 = √(6)2 + (-2)2 = √36 + 4 = √40 = 6.3

7 To the nearest tenth, RS = 11.3.
Find the distance between R(–2, –6) and S(6, –2) to the nearest tenth. Let (x1, y1) be the point R(–2, –6) and (x2, y2) be the point S(6, –2). d = (x2 – x1)2 + (y2 – y1)2 Use the Distance Formula. d = (6 – (–2))2 + (–2 – (–6))2 Substitute. d = (–8)2 Simplify. d = = Use a calculator. To the nearest tenth, RS = 11.3. 1-6

8 Graphing the Midpoint Graph A(2 , -2) & B(2 , 6)
Graph C(1 , 2) & D(5 , 6)

9 ( ) M MidPoint Formula x2 + x1 y2 + y1 , 2 2
Midpoint Formula: The coordinates of the midpoint M of AB with endpoints A(x1 , y1) and B(x2 , y2) are the following: ( ) x2 + x1 y2 + y1 M , 2 2 y-coordinate of a point x-coordinate of a point

10 ( ) ( ) ( ) M(1 , -3) Midpoint Example x2 + x1 y2 + y1 , M 2 2
AB has endpts A(8 , -9) & B(-6 , 3). Find the coordinates of its midpt. ( x2 + x1 y2 + y1 ) , M 2 2 ( (-6) + 8 3 + (-9) ) , M 2 2 ( ) 2 -6 , M 2 2 M(1 , -3)

11 y2 + y1 x2 + x1 2 2 y2 + 4 x2 +1 2 2 Midpoint Example G(-3 , 6) (2)
xm ym G(-3 , 6) The midpt of DG is M(-1 , 5). One endpt is D(1 , 4). Find the coordinates of the other endpts. x1 y1 y2 + y1 x2 + x1 ym = Xm = 2 2 y2 + 4 x2 +1 (2) (2) 5 = (2) -1 = (2) 2 2 10 = y2 + 4 -2 = x2 + 1 6 = y2 -3 = x2

12 ( ) M What have we learned??? dAB = √(x2 – x1)2 + (y2 – y1)2 x2 + x1
Distance Formula Midpoint formula dAB = √(x2 – x1)2 + (y2 – y1)2 ( ) x2 + x1 y2 + y1 M , 2 2 R(-2 , 6) & S(4 , 4) x1 y1 x2 y2


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