Core Connections Integrated II, 2015
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Core Connections Integrated II, 2015 View details
Chapter Closure

Exercise 123 Page 203

a Let's first draw the segment.


To find the length of this segment we can use the Distance Formula.
d = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)
d = sqrt(( 7 - 3)^2 + ( 2 - ( - 4))^2)
Simplify right-hand side
d = sqrt((7-3)^2 + (2+4)^2)
d = sqrt(4^2 + 6^2)
d = sqrt(16 + 36)
d = sqrt(52)
d = 7.21110...
d ≈ 7.21
b The slope of a segment is the vertical distance divided by the horizontal distance between the segment's endpoints.

Slope: Rise/Run=Δ y/Δ x Let's identify these distances in our diagram and calculate the slope.

c The area of a triangle is the product of its height and base divided by 2. Since a slope triangle is a right triangle, we know that its legs will represent the triangle's height and base. With this information we can calculate the triangle's area.
d First, let's recall the slope-intercept form of the linear function.
y = mx + b In the equation above m is the slope of the line connecting the two points and b is the y-intercept. In Part B we found that the slope of the line segment connecting the two points equals 32, so we can substitute 32 for m in the formula above. y = 3/2x + b To write a complete equation for this line, we also need to determine the y-intercept b. We can do that by substituting one of the given points into the equation and solving for b. Let's choose (3, -4) to do so.
y = 3/2x + b
-4 = 3/2( 3) + b
Solve for b
-4=9/2+b
-8/2=9/2+b
-17/2=b
b= -17/2
Now that we have both the slope and the y-intercept, we can write the final equation. y = 3/2&x + ( -17/2) &⇓ y = 3/2&x - 17/2