Core Connections Integrated II, 2015
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Core Connections Integrated II, 2015 View details
2. Section 8.2
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Exercise 48 Page 448

Practice makes perfect
a Let's plot the given points in a coordinate plane.
To find the distance between points A and B, we need to use the Distance Formula.
d = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)
d = sqrt(( 10 - 4)^2 + ( 9 - 1)^2)
Evaluate right-hand side
d = sqrt(6^2 + 8^2)
d = sqrt(36 + 64)
d = sqrt(100)
d = 10
b Let's take a look at A and B on a coordinate plane again. To get to point B from A, we need to take 6 steps to the right and 8 steps up.

We want to find a point P(x, y) on the line segment AB that is 34 of the way from point A to point B. This means that to get from A to P, we want to make 34 the steps we made to get from A to B.

As we can see, to get from A to P we need to take 4.5 steps to the right and 6 steps up. Therefore, to find the coordinates of P, we can add 4.5 to the x -coordinate of A and 6 to the y -coordinate of A. x &= 4 + 4.5 = 8.5 y &= 1 + 6 = 7 Therefore, point P(8.5, 7) is the point on AB that is 34 of the way from A to B.

c In the same coordinate plane from Part A, we will plot point C. Let's also label ∠ CAB as θ.

Since points B and C have the same x-coordinate and AC is horizontal, we can create a right triangle ABC by connecting B and C. Additionally, since AC is horizontal and BC is vertical, their lengths are the difference between the x-coordinates and y-coordinates, respectively.

Since we know the leg's length, we can find the measure of θ by using the tangent ratio.
tan θ = Opposite/Hypotenuse
tan θ = 8/6
Solve for θ
tan θ = 4/3

tan^(-1)(LHS) = tan^(-1)(RHS)

θ = sin^(-1) 4/3
θ = 53.130102... ^(∘)
θ ≈ 53.13 ^(∘)