MATH 13100 Lecture Notes - Lecture 2: Pythagorean Theorem, Cartesian Coordinate System, Hypotenuse
20 Jun 2018
School
Department
Course
Professor
MATH 131
Week 2
The Rectangular Coordinate System (0.3)
Graphs of equations (0.4)
Domain, Range (0.5)
Operations on Functions (0.6)
I. Cartesian coordinates
A. Must be in the form of an ordered pair (a, b), where a=x and b=y
II. Distance formula
A. The distance formula can be derived from the pythagorean theorem; see:
b ca2 + 2= 2
Can be transformed into….
(P, ) d Q = √(x1 2) y2 1)− x2+ ( − y2
How? Here’s the proof:
●b ca2 + 2= 2
Where and x2 1) a= ( − x y2 1) b= ( − y
(See diagram)
Your goal is to solve for the hypotenuse (the dashed line). We
can substitute the a and b in the original pythagorean theorem
equation to get…
x2 1) y2 1)( − x2+ ( − y2=c2
Now, simplify so you are solving for d.
d= √(x1 2) y2 1)− x2+ ( − y2
End of proof.
III. Equation of a circle
A. Definition of a circle: a set of points that lie at a fixed distance (the radius) from a
fixed point (the center).
B. The standard equation for a circle x)y)( − h2+ ( − k2=r2
How did we get that equation for a circle?
So, using the distance formula, we see that d is actually equal to the radius of the circle.
→ (P, ) d Q = √(x1 2) y2 1)− x2+ ( − y2 r= √(x− ) y)a2+ ( − b2
The standard equation of a circle, therefore, can be described by:
x)y)( − h2+ ( − k2=r2
Where h and k are the coordinates of the center of the circle.
For example, for a circle with a center (1, -5) and a radius of 5, the equation would be:
x)y) 5( − 1 2+ ( + 5 2= 2
Let’s assume there is an x-coordinate of 2. Substituting 2 into the equation will
get…
2 ) y) 5( − 1 2+ ( + 5 2= 2
...and can be simplified to…
y= -5 +/- 2 .
√6
Your next step is to plug all the values into this equation:
Document Summary
Cartesian coordinates: must be in the form of an ordered pair (a, b), where a=x and b=y. Distance formula: the distance formula can be derived from the pythagorean theorem; see: a2 + 2 = 2 b c. 1) a2 + 2 = 2 c b. Your goal is to solve for the hypotenuse (the dashed line). We can substitute the a and b in the original pythagorean theorem equation to get x2. Now, simplify so you are solving for d. d = (x1. So, using the distance formula, we see that d is actually equal to the radius of the circle. d (p , The standard equation of a circle, therefore, can be described by: r = (x ) Q = (x1 a 2 + ( b 2. 1) y2 y ( h 2 + ( k 2 = r2 x y. Where h and k are the coordinates of the center of the circle.
