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SectionFinding Linear Functions

SubsectionFinding a Linear Equation from a Graph

We can also use the slope-intercept form to find the equation of a line from its graph. First, we note the value of the \(y\)-intercept from the graph, and then we calculate the slope using two convenient points.

Example90

Find an equation for the line shown in Figure91.

graph of line
Figure91
Solution

The line crosses the \(y\)-axis at the point \((0, 3200)\text{,}\) so \(b=3200\text{.}\) To calculate the slope of the line, we locate another point, say \((20, 6000)\text{,}\) and compute:

\begin{align*}m \amp = \frac{\Delta y}{\Delta x} = \frac{6000 - 3200}{20 - 0}\\ \amp =\frac{2800}{20}= 140 \end{align*} The slope-intercept form of the equation, with \(m = 140\) and \(b = 3200\text{,}\) is

\begin{equation*} y = 3200 + 140x\text{.} \end{equation*}
graph of line
Figure92
Example93

Find an equation for the line shown in Figure94.

graph of line
Figure94
Solution

The line crosses the \(y\)-axis at the point \((0, 80)\text{,}\) so \(b=80\text{.}\) To calculate the slope of the line, we locate another point, say \((20,30)\text{,}\) and compute:

\begin{equation*} \begin{aligned}m \amp = \frac{\Delta y}{\Delta x} = \frac{30 - 80}{20 - 0}\\ \amp =\frac{-50}{20}= -\frac{5}{2} \end{aligned} \end{equation*}

The slope-intercept form of the equation is

\begin{equation*} y = 80 - \frac{5}{2}x\text{.} \end{equation*}

SubsectionPoint-Slope Form

We can find an equation for a line if we know its slope and \(y\)-intercept. What if we do not know the \(y\)-intercept, but instead know some other point on the line? There is only one line that passes through a given point and has a given slope, so we should be able to find an equation for this line.

For example, we can graph the line of slope \(\dfrac{-3}{4}\) that passes through the point \((1, -4)\text{.}\) We first plot the given point, \((1, -4)\text{,}\) as shown in Figure95.

Then we use the slope to find another point on the line. The slope is

\begin{equation*} m =\frac{-3}{4}= \frac{\Delta y}{\Delta x} \end{equation*}

so we move down \(3\) units and then \(4\) units to the right, starting from \((1, -4)\text{.}\) This brings us to the point \((5, -7)\text{.}\) We can then draw the line through these two points.

point slope
Figure95

We can also find an equation for the line, as shown in Example96.

Example96

Find an equation for the line that passes through \((1, -4)\) and has slope \(\dfrac{-3}{4}\text{.}\)

Solution

We will use the formula for slope,

\begin{equation*} m = \frac{y_2 - y_1}{x_2 - x_1}. \end{equation*}

We substitute \(\dfrac{-3}{4}\) for the slope, \(m\text{,}\) and \((1, -4)\) for \((x_1, y_1)\text{.}\) For the second point, \((x_2, y_2)\text{,}\) we use the variable point \((x, y)\text{.}\) Substituting these values into the slope formula gives us

\begin{equation*} \frac{-3}{4}= \frac{y - (-4)}{x - 1}=\frac{y + 4}{x - 1}. \end{equation*}

To solve for \(y\) we first multiply both sides by \(x - 1\text{.}\) \begin{align*} \alert{(x-1)}\frac{-3}{4}\amp =\frac{y +4}{x - 1}\alert{(x-1)} \amp\amp\\ \frac{-3}{4}(x-1)\amp=y+4 \amp\amp\text{Apply the distributive law.}\\ \frac{-3}{4}x+\frac{3}{4}\amp=y+4 \amp\amp \text{Subtract 4 from both sides.}\\ \frac{-3}{4}x-\frac{13}{4}\amp=y \amp\amp \frac{3}{4}-4=\frac{3}{4}-\frac{16}{4}=\frac{-13}{4} \end{align*} An equation of the line is \(y=\dfrac{-13}{4}-\dfrac{3}{4}x.\)

When we use the slope formula in this way to find an equation of a line, we substitute a variable point \((x, y)\) for the second point. This version of the formula,

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

is called the point-slope form for a linear equation. It is sometimes stated in another form obtained by clearing the fraction to get \begin{align*} \alert{(x-x_1)}m\amp =\frac{y -y_1}{x - x_1}\alert{(x-x_1)} \amp\amp \text{Multiply both sides by }(x - x_1)\\ (x - x_1) m\amp=y-y_1 \amp\amp\text{Clear fractions and solve for }y\text{.}\\ y\amp=y_1+m(x-x_1). \end{align*}

Point-Slope Form

An equation of the line that passes through the point \((x_1, y_1)\) and has slope \(m\) is

\begin{equation*} y= y_1 + m(x- x_1). \end{equation*}
Example97

Use the point-slope form to find an equation of the line that passes through the point \((-3, 5)\) and has slope \(-1.4\text{.}\)

Solution

Subsituting \(m=-1.4\) and \((-3,5)\) for \((x_1,y_1)\) into the point-slope form of a line, we have

\begin{equation*} \begin{aligned} y\amp=5-1.4(x-(-3)) \\ y\amp=0.8-1.4x \end{aligned} \end{equation*}

The point-slope form is useful for modeling linear functions when we do not know the initial value but do know some other point on the line.

Example98

Under a proposed graduated income tax system, single taxpayers would owe $1500 plus 20% of the amount of their income over $13,000. (For example, if your income is $18,000, you would pay $1500 plus 20% of $5000.)

  1. Complete the table of values for the tax, \(T\text{,}\) on various incomes, \(I\text{.}\)

    \(I\) \(15,000\) \(20,000\) \(22,000\)
    \(T\)
  2. Write a linear equation in point-slope form for the tax, \(T\text{,}\) on an income \(I\text{.}\)
  3. Write the equation in slope-intercept form.
Solution
  1. On an income of $15,000, the amount of income over $13,000 is $15,000 - $13,000 = $2000, so you would pay $1500 plus 20% of $2000, or

    \begin{equation*} T = 1500 + 0.20(2000) = 1900. \end{equation*}

    You can compute the other function values in the same way.

    \(I\) \(15,000\) \(20,000\) \(22,000\)
    \(T\) \(1900\) \(2900\) \(3300\)
  2. On an income of \(I\text{,}\) the amount of income over $13,000 is \(I - \$13,000\text{,}\) so you would pay $1500 plus 20% of \(I - 13,000\text{,}\) or
    \begin{equation*} T = 1500 + 0.20 (I - 13,000). \end{equation*}
  3. Simplify the right side of the equation to get \begin{align*} T \amp = 1500 + 0.20I - 2600\\ T \amp = -1100 + 0.20I. \end{align*}
Example99

A healthy weight for a young woman of average height, 64 inches, is 120 pounds. To calculate a healthy weight for a woman taller than 64 inches, add 5 pounds for each inch of height over 64.

  1. Write a linear equation in point-slope form for the healthy weight, \(W\text{,}\) for a woman of height, \(H\text{,}\) in inches.

  2. Write the equation in slope-intercept form.

Solution
  1. \(W = 120 + 5(H - 64)\)

  2. \(W = -200 + 5H\)

If you would like more information on linear equations, refer to the following two sections from Intermediate Algebra:

  1. The Slope of a Line
  2. Modeling Linear Functions

SubsectionExercises