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Section 1.5 Equations of Lines

Subsection 1. Calculate slope using a formula

Recall that the subscripts on the coordinates in \(~P_1(x_1,y_1)~\) and \(~P_2(x_2,y_2)~\) just mean "first point" and "second point".

Two-Point Formula for Slope.

The slope of the line joining points \(~P_1(x_1,y_1)~\) and \(~P_2(x_2,y_2)~\) is

\begin{equation*} m = \dfrac{\Delta y}{\Delta x} = \dfrac{y_2-y_1}{x_2-x_1}~~~~~~\text{if}~~~~x_2 \not= x_1 \end{equation*}

Subsubsection Example

Example 1.70.

Compute the slope of the line joining \((-6,2)\) and \((3,-1)\text{.}\)

Solution

It doesn't matter which point is \(P_1\) and which is \(P_2\text{,}\) so we choose \(P_1\) to be \((-6,2)\text{.}\) Then \((x_1,y_1)=(-6,2)\) and \((x_2,y_2)=(3,-1)\text{.}\) Thus,

\begin{align*} m \amp = \dfrac{y_2-y_1}{x_2-x_1}\\ \amp = \dfrac{-1-2}{3-(-6)} = \dfrac{-3}{9} = \dfrac{-1}{3} \end{align*}
Caution 1.71.

Make sure that you subtract both the \(x\) and \(y\) coordinates in the same order! That is, do NOT calculate

\begin{equation*} \dfrac{y_{\alert{2}}-y_{\alert{1}}}{x_{\alert{1}}-x_{\alert{2}}}~~~~~~~~~~\blert{\text{Incorrect!}} \end{equation*}

or your slope will have the wrong sign.

Subsubsection Exercises

Compute the slope of the line joining the points \((5,2)\) and \((8,7)\text{.}\)

Answer
\(\dfrac{5}{3}\)

Compute the slope of the line joining the points \((-3,-4)\) and \((-7,1)\text{.}\)

Answer
\(\dfrac{-5}{4}\)

Subsection 2. Slope-Intercept Form

Because the \(y\)-intercept \((0,b)\) is the "starting value" of a linear model, and its rate of change is measured by its slope,\(m\text{,}\) the equation for a linear model

\begin{equation*} y = \text{starting value} + \text{rate} \times x \end{equation*}

can be expressed in symbols as

\begin{equation*} y = b + mx \end{equation*}
Slope-Intercept Form.

If we write the equation of a linear function in the form,

\begin{equation*} f (x) = b + mx \end{equation*}

then \(m\) is the slope of the line, and \(b\) is the \(y\)-intercept.

Subsubsection Examples

Example 1.74.

The temperature inside a pottery drying oven starts at 70 degrees and is rising at a rate of 0.5 degrees per minute. Write a function for the temperature, \(H\text{,}\) inside the oven after \(t\) minutes.

Solution

At \(t=0\text{,}\) the temperature is 70 degrees, so \(b=70\text{.}\)

The slope is given by the rate of increase of \(H\text{,}\) so \(m=0.5\text{.}\)

Thus, the function is

\begin{equation*} H=70+0.5t \end{equation*}
Example 1.75.

A perfect score on a driving test is 120 points, and you lose 4 points for each wrong answer. Write a function for your score, \(S\text{,}\) if you give \(n\) wrong answers.

Solution

If \(n=0\text{,}\) your score is 120, so \(b=120\text{.}\)

Your score decreases by 4 points per wrong answer, so \(m=\dfrac{\Delta S}{\Delta n} = 4\text{.}\)

The function is

\begin{equation*} S=120-4n \end{equation*}

Subsubsection Exercises

Monica has saved $7800 to live on while she attends college. She spends $600 a month. Write a function for the amount, \(S\text{,}\) in Monica's savings account after \(t\) months.

Answer
\(b=7800~~\) and \(~~m=-400~~\text{,}\) so \(~~S=7800-600t\)

Jesse opened a new doughnut shop in an old store-front. He invested $2400 in remodeling and set-up, and he makes about $400 per week from the business. Write a function giving the shop's financial standing, \(F\text{,}\) after \(w\) weeks.

Answer
\(b=-2400~~\) and \(~~m=400~~\text{,}\) so \(~~F=-2400+400w\)

Subsection 3. Point-Slope Form

If we don't know the \(y\)-intercept of a line but we do know one other point and the slope, we can still find an equation for the line.

Point-Slope Formula.

To find an equation for the line of slope \(m\) passing through the point \((x_1,y_1)\text{,}\) use the point-slope formula

\begin{equation*} \dfrac{y-y_1}{x-x_1} = m \end{equation*}

or

\begin{equation*} y-y_1=m(x-x_1) \end{equation*}

Subsubsection Example

Example 1.78.

Find an equation for the line that passes through \((1,3)\) and has slope \(-2\text{.}\)

Solution

We substitute \(x_1=\alert{1}\text{,}\) \(y_1=\alert{3}\text{,}\) and \(m=\alert{-2}\) into the point-slope formula.

\begin{align*} y-\alert{3} \amp = \alert{-2} (x-\alert{1}) \amp \amp \blert{\text{Apply the distributive law.}}\\ y-3 \amp = -2x + 2 \amp \amp \blert{\text{Add 3 to both sides.}}\\ y \amp = -2x + 5 \end{align*}
Example 1.79.

Find an equation for the line of slope \(\dfrac{-1}{2}\) that passes through \((-3,-2)\text{.}\)

Solution

We substitute \(x_1=\alert{-3}\text{,}\) \(y_1=\alert{-2}\text{,}\) and \(m=\alert{\dfrac{-1}{2}}\) into the point-slope formula.

\begin{align*} \dfrac{y-(\alert{-2})}{x-(\alert{-3})} \amp = \alert{\dfrac{-1}{2}} \amp \amp \blert{\text{Simplify the left side.}}\\ \dfrac{y+2}{x+3} \amp = \dfrac{-1}{2} \amp \amp \blert{\text{Cross-multiply.}}\\ 2(y+2) \amp = -1(x+3) \amp \amp \blert{\text{Apply the distributive law.}}\\ 2y+4 \amp = -x-3 \amp \amp \blert{\text{Subtract 4 from both sides.}}\\ 2y \amp = -x-7 \amp \amp \blert{\text{Divide both sides by 2.}}\\ y \amp = \dfrac{-1}{2}x-\dfrac{7}{2} \end{align*}

Subsubsection Exercises

Find an equation for the line of slope \(-4\) that passes through \((2,-5)\text{.}\)

Answer
\(y=-4x+3\)

Find an equation for the line of slope \(\dfrac{2}{3}\) that passes through \((-6,1)\text{.}\)

Answer
\(y=\dfrac{2}{3}x+5\)

Subsection 4. Graphing a linear equation by the point-slope method

If we know one point on a line and its slope, we can sketch its graph without having to make a table of values.

Subsubsection Examples

Example 1.82.

Graph the line \(~y=\dfrac{3}{4}x-2\)

Solution

Step1: Begin by plotting the \(y\)-intercept, \((0,-2)\text{.}\)

Step 2: We use the slope, \(\dfrac{\Delta y}{\Delta x} = \dfrac{3}{4}\text{,}\) to find another point on the line, as follows. Start at the point \((0,-2)\) and move 3 units up and 4 units to the right. Plot a second point here, at \((4,1)\text{.}\)

Step 3: Find a third point by writing the slope as \(\dfrac{\Delta y}{\Delta x} = \dfrac{-3}{-4}\text{:}\) from \((0,-2)\text{,}\) move down 3 units and 4 units to the left. Plot a third point here, at \((-4,-5)\text{.}\)

line

Finally, draw a line through the three points.

Example 1.83.

Graph the line of slope \(\dfrac{-1}{2}\) that passes through \((-3,-2)\text{.}\)

Solution

Step 1: Begin by plotting the point \((-3,-2)\text{.}\)

Step 2: Use the slope, \(\dfrac{\Delta y}{\Delta x} = \dfrac{-1}{2}\text{,}\) to find another point on the line, as follows. Start at the point \((-3,-2)\) and move 1 unit down and 2 units to the right. Plot a second point here, at \((-1,-3)\text{.}\)

Step 3: Find a third point by writing the slope as \(\dfrac{\Delta y}{\Delta x} = \dfrac{1}{-2}\text{:}\) from \((-3,-2)\text{,}\) move 1 unit up and 2 units to the left. Plot a third point here, at \((-5,-1)\text{.}\)

line

Finally, draw a line through the three points.

Subsubsection Exercises

Graph the line \(~y=\dfrac{-1}{3}x-3\)

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Graph the line with slope \(m=\dfrac{3}{2}\) passing through \((-1,-2)\text{.}\)

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