Graphs Rectangular Coordinates Use the distance formula Use

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 n 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 n n

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

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

Distance Formula n (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 2 42 + 3 2 = c 2 16 + 9 = c 2 25 = c 2 5=c

Distance Formula n D = sqrt ((x 2 – x 1)2 + (y 2 – y 1)2)) n Remember: This is an application of the Pythagorean Theorem n n 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. n 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 n To find the midpoint of a line segment, average the x-coordinates and average the ycoordinates of the endpoints. n M(x, y) = ( (x 1 + x 2)/2 , (y 1 + y 2)/2)) n (-5, 5) to (3, 1) = (-5 + 3)/2 , (5 + 1)/2 n (-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 + (-3 - -3)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. n Does (sqrt 25)2 + (sqrt 16)2 = (sqrt 41)2? n Yes to it is a right triangle. n Area = ½ l w so Area =. 5 (5) (4) = 10 n n n

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