1.1 Rectangular Coordinates; Graphing Utilities
Objective: Use distance formula
Use midpoint formula
Title: Sep 11­9:58 AM (1 of 9)
1.1 Day 1: The Distance Formula
y
y2
d
y1
x 2­ x1
x1
Title: Sep 11­10:02 AM (2 of 9)
y2 ­ y1
x2
x
The Distance Formula
y
y2
d
y1
y2 ­ y1
x 2­ x1
x1
x2
x
You can use the Pythagorean Theorem to find d.
Title: Sep 11­10:02 AM (3 of 9)
Find the distance d between the points (1, 3) and (5, 6).
2
2
d = (5 ­ 1) + ( 6 ­ 3) 2
2
d = (4) + (3) d = 16 + 9
d = 25
d = 5
Title: Sep 11­10:02 AM (4 of 9)
Consider the three points A = (­2, 1), B = (2, 3), and C = (3, 1).
a) Plot each point and form the triangle ABC.
B
A
Title: Sep 11­10:02 AM (5 of 9)
C
Consider the three points A = (­2, 1), B = (2, 3), and C = (3, 1).
b) Find the length of each side.
d(A, B) =
d(B, C) =
d(A, C) =
Title: Sep 11­10:02 AM (6 of 9)
Consider the three points A = (­2, 1), B = (2, 3), and C = (3, 1).
c) Verify that the triangle is a right triangle.
To show that the triangle is a right triangle, we need to show that the sum of the squares of the lengths of two sides equals the square of the third side.
Check to see whether From the converse of the Pythagorean Theorem, triangle ABC is a right triangle.
Title: Sep 11­10:02 AM (7 of 9)
Consider the three points A = (­2, 1), B = (2, 3), and C = (3, 1).
d) Find the triangle's area.
B
A
Area = C
(Base)(Height)
Title: Sep 11­10:02 AM (8 of 9)
Because the right angle is at vertex B, the sides AB and BC form the base and height.
square units
Homework:
pg. 8
#'s: 15­21, 27­29, 33, 34, 39­44, 46­50
Title: Sep 11­10:02 AM (9 of 9)