Core Connections Geometry, 2013
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Core Connections Geometry, 2013 View details
2. Section 6.2
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Exercise 59 Page 369

Practice makes perfect
a The line is written in slope-intercept form which means it follows a certain format.

y= mx+ b In this form, m is the slope and b is the y-intercept. To graph the given equation, we will start at the y-intercept, (0,1), and then use the line's slope to plot a second point.

Finally, to find the slope angle, θ, we will use the tangent ratio on the slope triangle.

Let's solve this equation for θ.

tan θ = 3/1
â–¼
Solve for θ
tan θ = 3

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

θ = tan^(- 1) 3
θ = 71.56505... ^(∘)
θ ≈ 71.57 ^(∘)

The slope angle is about 71.57 ^(∘).

b To write the equation, we have to find it's slope, m, and y-intercept, b. When we know this, we can write the equation in slope-intercept form.

y= mx+ b In addition to the y-intercept, we know the angle with which the line increases. Let's demonstrate this in a graph. Note that a line's slope is defined as the vertical change when you move 1 unit in the positive horizontal direction. Therefore, from the y-intercept, we will also draw a slope triangle with a horizontal side of 1 and an angle of 45^(∘).

If we can find m, we can get the slope of the line. Since we know the adjacent leg to the reference angle, we can use the tangent ratio to find the value of m.

tan θ = opposite/adjacent
tan 45^(∘) = m/1
â–¼
Solve for m
tan 45^(∘) = m
m=tan 45^(∘)
m =1

The slope is 1. With this, we can complete the equation. y=x+3

c If we combine the equations, we get a system of equations.

y=3x+1 & (I) y=x+3 & (II)Since both equations are solved for y, we can solve the system by using the Substitution Method.

y=3x+1 & (I) y=x+3 & (II)
y=3x+1 3x+1=x+3
â–¼
(II): Solve for x
y=3x+1 2x+1=3
y=3x+1 2x=2
y=3x+1 x=1

Having solved for x, we can substitute this into the first equation and solve for y.

y=3x+1 x=1
y=3( 1)+1 x=1
y=3+1 x=1
y=4 x=1

The solution is (1,4).