LESSON 5 THE CARTESIAN PLANE (RECTANGULAR COORDINATE SYSTEM)
You should learn to:
1. Plot points in the Cartesian plane.
3. Use the Midpoint Formula.
2. Use the Distance Formula.
4. Find and use the equation of a circle.
Terms to know: x-axis, y-axis, origin, quadrants, ordered pair, x-coordinate, y-coordinate, distance formula,
midpoint formula, standard form of the equation for a circle
Example 1: Plot the points A: (2,3) , B: (0, 2) , C: (3, 0) , D: (3, 1) , and E: (0, 0) in the coordinate plane.
A
C
E
D
B
The distance formula (for finding the distance between 2 points ( x1 , y1 ) and ( x2 , y2 ) in a coordinate plane)
can be developed from the Pythagorean Theorem.
y
( x2 , y2 )
x
( x1 , y1 )
Distance Formula: d  ( x2  x1 )2  ( y2  y1 )2
Example 2: Find the distance between the points (2,5) and (1, 3) to 3 decimal place accuracy.
d  (1  (2)) 2  (3  5) 2
d  32  (8) 2
d  9  64
d  73
d  8.544 units
The midpoint of a line segment connecting the 2 points ( x1 , y1 ) and ( x2 , y2 ) in the coordinate plane can be
found by finding the average x-value and the average y-value for the points.
 x1  x2 y1  y2 
,

2 
 2
Midpoint Formula: Midpoint = 
Example 3: Find the midpoint of the segment connecting the points (3, 5) and (7, 2) .
 3  (7) 5  2 
midpoint: 
,

2
2 

 10 3 
midpoint: 
, 
 2 2 
3 

midpoint:  5, 
2 

Example 4: Paul’s Sporting Goods had annual sales of $5.326 million in 2000 and $8.490 million in
2006. Estimate the annual sales for 2003.
(2000,5.326) (2006,8.490)
 2000  2006 5.326  8.490 
,


2
2


(2003, 6.908)
So, in 2003 there was $6.908 million in sales.
r
Let (x, y) represent any point on a circle with center at the origin and radius r.
By the Pythagorean Theorem x 2  y 2  r 2 . This is the equation of any circle
with radius r and center at the origin.
Example 5: Graph the circle x 2  y 2  9
center:
radius :
 0, 0 
9 3
y
x
You can write the equation for any circle if you know its radius and coordinates of its center. In the equation the
center of the circle is (h, k) and the radius is r.
Standard form of the equation for a circle :
( x  h)2  ( y  k )2  r 2
Example 6: Identify the center and radius in each equation. Then graph the circle.
a. ( x  2)  ( y  1)  4
2
b. x  ( y  2)  9
2
2
center: (2, 1)
radius:
2
center: (0, -2)
4 2
radius:
9 3
Example 7: The point (1, 2) lies on a circle whose center is at (2, 4) . Write the standard form equation of the
circle.
radius (distance) = (2  ( 1)) 2  ( 4  2) 2
radius (distance)= 32  ( 6) 2
radius (distance) = 9  36
radius (distance) = 45
center: (2, -4)
( x  2) 2  ( y  4) 2  45
ASSIGNMENT 5
A stands for Appendix (located in the back of the book in Section B1)
Pages A32-A35: (Vocabulary Check 1-5, 2, 4, 7-11, 13, 15,19, 23 (you don’t need to
verify), 50 (3 decimal place accuracy), 55, 61, 63, 73, 76, 96,).
Page: A59 (14, 15, 26)