Graphs
Rectangular Coordinates
Use the distance formula.
Use the midpoint formula.
Graphing ordered pairs.
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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.
1st is upper right, 2nd is upper left, 3rd is lower
left, 4th is lower right
 Ordered Pair: (x, y)
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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: 2nd quadrant
 (-7, -2.5)
Left 7, down 2.5 (approximate the .5): 3rd
quadrant
 (0,8)
No motion right and left, up 8: not in a quadrant
 (3, -6)
Right 3, down 6: 4th 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
a2 + b 2 = c 2 42 + 3 2 = c 2
16 + 9 = c2 25 = c2
5=c
Distance Formula
 D = sqrt ((x2 – x1)2 + (y2 – y1)2))
 Remember: This is an application of the
Pythagorean Theorem
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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 ycoordinates of the endpoints.
 M(x, y) = ( (x1 + x2)/2 , (y1 + y2)/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.
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(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 a2 + b2 = c2. 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