Core Connections Algebra 1, 2013
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Core Connections Algebra 1, 2013 View details
2. Section 6.2
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Exercise 57 Page 272

Practice makes perfect
a Let's begin by graphing the points (- 3, 2), and (5,- 4).
These two points determine the line. We can calculate the slope of this line by substituting the points into the Slope Formula. m = y_2-y_1/x_2-x_1 Note that it does not matter which point we choose to use for ( x_1, y_1) and ( x_2, y_2), since the result of using the Slope Formula will be the same.
m=y_2-y_1/x_2-x_1
m=- 4- 2/5-( - 3)
â–Ľ
Add and subtract terms
m=- 4 -2/5+3
m=- 6/8
â–Ľ
Simplify
m=- 6/8
m=- 3/4
The slope of the line that passes through the given points is - 34.
b To find the distance between the points, we will use the Pythagorean Theorem.
In the formula a and b are the legs and c is the hypotenuse of a right triangle. Now, let's look at the graph from the Part A. We will draw the lines that form a right-angled triangle using the given points as two of the corners.
Now we can find the lengths of the horizontal and vertical sides of the right triangle. We just need to calculate the absolute value of the difference between the x- and y-coordinates. a=| 5 - (- 3)| = 8 b=| 2 - (- 4)| = 6 We obtained a triangle with a=8 and b=6. Let's substitute these values into the formula.
a^2+b^2=c^2
8^2+ 6^2=c^2
â–Ľ
Solve for c
64+36=c^2
100=c^2
± sqrt(100)=c

c > 0

sqrt(100)=c
c=sqrt(100)
c=10
The distance between the given points is 10 units.