Algebra Period 3 (Tuesday)

CHAPTER 11-7: DISTANCE FORMULA
(based on the Pythagorean Theorem): see p. 513 in book
Reviewed from last week:

The distance between any two points on the coordinate plane (x y plane)
The distance is the hypotenuse of a right triangle that you can draw using any two points on the coordinate plane
The formula is:

distance = SQRT[( difference of the two x's)² + (difference of the two y's)²]
The difference between the 2x’s is the length of the leg parallel to the y axis and the difference between the 2y’s is the length of the leg parallel to the x axis

EXAMPLE: What is the distance between (3, -10) and (-7, -2)?
d = SQRT[(3 - -7)² + (-10 - -2)²]
d = SQRT[10² +( -8²)]
d = SQRT(164)
Simplifying: 2rad41

CHAPTER 11-8: USING THE PYTHAGOREAN THEOREM - WORD PROBLEMS
There are many real life examples where you can use the Pythagorean Theorem to find a length.
EXAMPLE: HOW HIGH A 10 FOOT LADDER REACHES ON A HOUSE
A 10 ft ladder is placed on a house 5 ft away from the base of the house.
Find how high up the house the ladder reaches.
The ladder makes a right triangle with the ground being one leg, the house the other, and the ladder is the hypotenuse ( see drawing in #1 on p. 515) You need to find the distance on the house, so you're finding one leg.

You're flying your kite for the kite project and you want to know how long the kite string must be so that it can reach a height of 13 ft in the air if you're standing 9 feet away from where the kite is in the air. The string represents the hypotenuse. You know one leg is the height in the air (13 ft) and the other leg is how far on the ground you are standing away from where the kite is flying (9 ft)
You need to find the hypotenuse.