McGraw Hill Glencoe Algebra 1, 2012
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McGraw Hill Glencoe Algebra 1, 2012 View details
7. Inverse Linear Functions
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Exercise 44 Page 270

Find the equation of the given line first.

C

Practice makes perfect

Let's find the equation of the given line in slope-intercept form first. Then, we can find the equation of the perpendicular line.

Writing the Given Line's Equation

Equations written in slope-intercept form follow a specific format. y=mx+ b In this form, m is the slope of the line and b is the y-intercept. We need to identify these values. Let's observe the given graph.

We can see that the value of b is 4. Then, to find the slope we can use the Slope Formula. m=y_2-y_1/x_2-x_1 From the graph, we can take two points to find the value of the slope. We will take the ordered pairs (-1,1) and (0,4). Let's substitute this values in the above equation and simplify.
m=y_2-y_1/x_2-x_1
m=4- 1/0-( -1)
m=4-1/0+1
m=3/1
m=3
We found that the value of m is 3. Let's substitute these values. y=3x+ 4

Writing the Perpendicular Line's Equation

Now, we can find the equation of the perpendicular line. Let's find the slope first, and then we can use the given point to complete the equation.

Finding the Slope

Two lines are perpendicular when their slopes are negative reciprocals. This means that the product of a given slope and the slope of a line perpendicular to it will be -1. m_1*m_2=-1 We identified the given line's slope as 3. By substituting this value into our negative reciprocal equation for m_1, we can solve for the slope of the perpendicular line, m_2.
m_1 * m_2 = - 1
3* m_2 = - 1
â–Ľ
Solve for m_2
3 * m_2/3=- 1/3
3/3 * m_2=- 1/3
1 * m_2=- 1/3
m_2=- 1/3
m_2=-1/3
Now, we know that any line perpendicular to the given equation will have a slope of - 13.

Completing the Equation

Using the slope m_2=- 13, we can write a general equation in slope-intercept form for all lines perpendicular to the given equation. y=-1/3x+b By substituting the given point ( 2, 0) into this equation for x and y, we can solve for the y-intercept b of the perpendicular line.
y=-1/3x+b
0=-1/3( 2)+b
â–Ľ
Solve for b
0=-2/3+b
0+ 2/3=-2/3+b+ 2/3
0+2/3=2/3-2/3+b
0/3+2/3=2/3-2/3+b
0+2/3=2-2/3+b
2/3=0/3+b

0/a=0

2/3=0+b
2/3=b
b=2/3
Now that we have the y-intercept, we can complete the equation. y=-1/3x+2/3 The line given by this equation is perpendicular to the given line and passes through the point (2,0). Therefore, our solution corresponds to the option C.