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Precalculus for Team-Based Inquiry Learning PREVIEW Edition

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Section 3.2 Equations of Lines (LF2)

Subsection 3.2.1 Activities

Activity 3.2.1.

Consider the graph of two lines.
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(e)
What is the same about the two lines?
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Answer.
Lines A and B have the same slope.
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(f)
What is different about the two lines?
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Answer.
Lines A and B have different \(y\)-intercepts.
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Remark 3.2.2.

Notice that in Activity 3.2.1 the lines have the same slope but different \(y\)-intercepts. It is not enough to just know one piece of information to determine a line, you need both a slope and a point.
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Definition 3.2.3.

Linear functions can be written in slope-intercept form
\begin{equation*} f(x)=mx+b \end{equation*}
where \(b\) is the \(y\)-intercept (or starting value) and \(m\) is the slope (or constant rate of change).
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Activity 3.2.4.

Write the equation of each line in slope-intercept form.
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(a)
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  1. \(\displaystyle y= -3x+1\)
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  2. \(\displaystyle y= -x+3\)
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  3. \(\displaystyle y= -\dfrac{1}{3}x+1\)
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  4. \(\displaystyle y= -\dfrac{1}{3}x+3\)
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Answer.
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(c)
Two points on the line are \((0,1)\) and \((2,4)\text{.}\)
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  1. \(\displaystyle y= 2x+1\)
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  2. \(\displaystyle y= -\dfrac{3}{2}x+4\)
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  3. \(\displaystyle y= \dfrac{3}{2}x+1\)
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  4. \(\displaystyle y= \dfrac{3}{2}x+4\)
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Answer.
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(d)
\(x\) \(f(x)\)
\(-2\) \(-8\)
\(0\) \(-2\)
\(1\) \(1\)
\(4\) \(10\)
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  1. \(\displaystyle f(x)= -3x-2\)
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  2. \(\displaystyle f(x)= -\dfrac{1}{3}x-2\)
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  3. \(\displaystyle f(x)= 3x+1\)
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  4. \(\displaystyle f(x)= 3x-2\)
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Answer.
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Activity 3.2.5.

Let’s try to write the equation of a line given two points that don’t include the \(y\)-intercept.
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(a)
Plot the points \((2,1)\) and \((-3,4)\text{.}\)
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Answer.
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(c)
When you draw a line connecting the two points, it’s often hard to draw an accurate enough graph to determine the \(y\)-intercept of the line exactly. Let’s use the slope-intercept form and one of the given points to solve for the \(y\)-intercept. Try using the slope and one of the points on the line to solve the equation \(y=mx+b\) for \(b\text{.}\)
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  1. \(\displaystyle 2\)
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  2. \(\displaystyle \dfrac{11}{5}\)
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  3. \(\displaystyle \dfrac{5}{2}\)
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  4. \(\displaystyle 3\)
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Answer.
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(d)
Write the equation of the line in slope-intercept form.
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Answer.
\(y=-\dfrac{3}{5}x+\dfrac{11}{5}\)
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Remark 3.2.6.

It would be nice if there was another form of the equation of a line that works for any points and does not require the \(y\)-intercept.
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Definition 3.2.7.

Linear functions can be written in point-slope form
\begin{equation*} y-y_0=m(x-x_0) \end{equation*}
where \((x_0, y_0)\) is any point on the line and \(m\) is the slope.
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Activity 3.2.8.

Write an equation of each line in point-slope form.
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(a)
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  1. \(\displaystyle y= \dfrac{1}{3}x+\dfrac{2}{3}\)
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  2. \(\displaystyle y-1= 3(x-1)\)
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  3. \(\displaystyle y-1= \dfrac{1}{3}(x-1)\)
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  4. \(\displaystyle y+2= \dfrac{1}{3}(x+2)\)
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  5. \(\displaystyle y= \dfrac{1}{3}(x+2)\)
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Answer.
C or E
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(d)
\(x\) \(f(x)\)
\(-2\) \(-8\)
\(1\) \(1\)
\(4\) \(10\)
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  1. \(\displaystyle y+8= 3(x-2)\)
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  2. \(\displaystyle y-1= -\dfrac{1}{3}(x-1)\)
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  3. \(\displaystyle y+8= -\dfrac{1}{3}(x+2)\)
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  4. \(\displaystyle y-10= 3(x-4)\)
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Answer.
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Activity 3.2.9.

Consider again the two points from Activity 3.2.5, \((2,1)\) and \((-3,4)\text{.}\)
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(a)
Use point-slope form to find an equation of the line.
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  1. \(\displaystyle y=-\dfrac{3}{5}x+\dfrac{11}{5}\)
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  2. \(\displaystyle y-1=-\dfrac{3}{5}(x-2)\)
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  3. \(\displaystyle y-4 =-\dfrac{3}{5}(x+3)\)
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  4. \(\displaystyle y-2=-\dfrac{3}{5}(x-1) \)
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Answer.
B or C
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(b)
Solve the point-slope form of the equation for \(y\) to rewrite the equation in slope-intercept form. Identify the slope and intercept of the line.
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Answer.
The slope-intercept form is: \(y=-\dfrac{3}{5}x+\dfrac{11}{5}\text{.}\) The slope is \(-\dfrac{3}{5}\) and the \(y\)-intercept is \(\dfrac{11}{5}\text{.}\)
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Remark 3.2.10.

Notice that it was possible to use either point to find an equation of the line in point-slope form. But, when rewritten in slope-intercept form the equation is unique.
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Activity 3.2.11.

For each of the following lines, determine which form (point-slope or slope-intercept) would be "easier" and why. Then, write the equation of each line.
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(a)
The line whose graph is given below.
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Answer.
Slope-intercept: \(y=\dfrac{3}{4}x+2\)
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(b)
The line whose slope is \(-\dfrac{1}{2}\) and passes through the point \((1,-3)\text{.}\)
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Answer.
Point-slope: \(y+3=-\dfrac{1}{2}(x-1)\)
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(c)
The line that passes through the points \((0,3)\) and \((2,0)\text{.}\)
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Answer.
Slope-intercept: \(y=-\dfrac{3}{2}x+3\)
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Remark 3.2.12.

It is always possible to use both forms to write the equation of a line and they are both valid. Although, sometimes the given information lends itself to make one form easier.
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Activity 3.2.13.

Write the equation of each line.
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(c)
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  1. \(\displaystyle x=-2 \)
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  2. \(\displaystyle y-2=x\)
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  3. \(\displaystyle y=-2x-2\)
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  4. \(\displaystyle y=-2\)
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Answer.
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Definition 3.2.14.

A horizontal line has a slope of zero and has the form \(y=k\) where \(k\) is a constant. A vertical line has an undefined slope and has the form \(x=h\) where \(h\) is a constant.
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Definition 3.2.15.

The equation of a line can also be written in standard form. Standard form looks like \(Ax+By=C\text{.}\)
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Remark 3.2.16.

It is possible to rearrange a line written in standard form to slope-intercept form by solving for \(y\text{.}\)
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Activity 3.2.17.

Given a line in standard form
\begin{equation*} 5x+4y=2. \end{equation*}
Find the slope and \(y\)-intercept of the line.
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Answer.
Slope: \(-\dfrac{5}{4}\)
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\(y\)-intercept: \(\dfrac{2}{5}\)
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Subsection 3.2.2 Exercises

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