Graphs Rectangular Coordinates Use the distance formula. Use the midpoint formula.

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Presentation transcript:

Graphs Rectangular Coordinates Use the distance formula. Use the midpoint formula.

Graphing ordered pairs. x axis: horizontal line y axis: vertical line Origin: point of intersection of the two axes Rectangular (Cartesian) Coordinate Plane: plane formed by the x-axis and the y-axis Divided into four sections called quadrants. 1 st is upper right, 2 nd is upper left, 3 rd is lower left, 4 th is lower right

Ordered Pair: (x, y) First value is the x coordinate (abscissa): tells right and left movement Second value is the y coordinate (ordinate): tells up and down movement

Graph the following ordered pairs and state what quadrant they are in (-2, 5) Left 2, up 5: 2 nd quadrant (-7, -2.5) Left 7, down 2.5 (approximate the.5): 3 rd quadrant (0,8) No motion right and left, up 8: not in a quadrant (3, -6) Right 3, down 6: 4 th quadrant

Distance Formula (1, 3) and (5, 6) Find horizontal distance by subtracting x’s. 5 – 1 = 4 Find vertical distance by subtracting y’s. 6 – 3 = 3 Pythagorean Theorem a 2 + b 2 = c = c = c 2 25 = c 2 5 = c

Distance Formula D = sqrt ((x 2 – x 1 ) 2 + (y 2 – y 1 ) 2 )) Remember: This is an application of the Pythagorean Theorem Find the distance between (-4,5) and (3,2) Sqrt ((-4 – 3) 2 + (5 – 2) 2 ) = d Sqrt (49 + 9) = d Sqrt 58 = d

Determine if the triangle formed by the coordinates (-2, 1), (2, 3), and (3,1) is an isosceles triangle. Find the length of each side by using the distance formula. Sqrt((-2 – 2) 2 + (1 – 3) 2 )) = Sqrt 20 Sqrt((2 – 3) 2 + (3 – 1) 2 ) = Sqrt 5 Sqrt((3 - -2) 2 + (1 – 1) 2 = Sqrt 25 = 5 Not isosceles since no two sides are equal. If you test pythagorean theorem you will find that it is a right triangle. (sqrt 20) 2 + (sqrt 5) 2 = (5) 2

Midpoint Formula To find the midpoint of a line segment, average the x-coordinates and average the y- coordinates of the endpoints. M(x, y) = ( (x 1 + x 2 )/2, (y 1 + y 2 )/2)) (-5, 5) to (3, 1) = (-5 + 3)/2, (5 + 1)/2 (-2/2, 6/2) => (-1, 3)

Verify that the following is a right triangle, then find the area. (4, -3), (0, -3), (4, 2) Sqrt ((4 - 0) 2 + ( ) 2 ) = sqrt 16 = 4 Sqrt ((0 – 4) 2 + (2 - -3) 2 ) = sqrt (41) Sqrt ((4 – 4) 2 + (2 - -3) 2 ) = sqrt (25) = 5 Right triangle if a 2 + b 2 = c 2. When testing remember that ‘a’ and ‘b’ are the legs of the triangle, the shorter sides. Does (sqrt 25) 2 + (sqrt 16) 2 = (sqrt 41) 2 ? Yes to it is a right triangle. Area = ½ l w so Area =.5 (5) (4) = 10

Assignment Page 163 #11, 13, 17, 21, 27, 33, 37, 43, 45, 53, 61, 65