7.5: Forms of the Equation of a Line
The previous section explained the equations of vertical and horizontal lines. Now discover three more forms of the equations of a line, namely, the Slope-Intercept Form , the Point-Slope Form , and the Standard Form .
Slope-Intercept Form of the Equation of a Line
The Slope-Intercept Form of the equation of a line is of the form:
\[y = mx + b \nonumber \]
Where \(m\) is the slope of the line and \((0, b)\) is the \(y\)−intercept.
Note that the \(y\)-intercept is the point where the line intersect the \(y\)−axis, that is when \(x = 0\).
Write an equation of the line with the given slopes and \(y\)-intercepts.
- slope = \(5\); \(y\)−intercept \((0, \dfrac{1}{2})\)
- slope = \(−\dfrac{5}{6}\); \(y\)− intercept \((0, −\dfrac{3}{4})\)
Solution
- \(m = 5\) and \(b = \dfrac{1}{2}\)
The equation of a line is of the for \(y = mx + b\). Thus,
\(\begin{array} &&y = mx + b &\text{Slope-intercept form} \\ &= 5x + \dfrac{1}{2} &\text{Substitute \(m = 5\) and \(b = \dfrac{1}{2}\)} \end{array}\)
Therefore, \(y = 5x + \dfrac{1}{2}\) is the equation of the line with the given slope and \(y\)-intercept.
- Given \(m = −\dfrac{5}{6}\) and \(b = −\dfrac{3}{4}\)
Thus,
\(\begin{array} &&y = mx + b &\text{Slope-intercept form} \\ &= −\dfrac{5}{6}x −\dfrac{3}{4} &\text{Substitute values} \end{array}\)
Therefore, \(y = −\dfrac{5}{6}x − \dfrac{3}{4}\) is the equation of the line with the given slope and \(y\)-intercept.
Identify the slope and \(y\)−intercept then, use them to graph each line.
- \(y = −2x + 4\)
- \(5y − 3x = 10\)
Solution
a. Notice that the given linear equation is in the slope-intercept form. So, \(m = −2\) or equivalently, \(m = −\dfrac{2}{1}\) and \(b = 4\)
\(m\) is the slope of the line, then \(m = \dfrac{\text{rise}}{\text{run}} = −\dfrac{2}{1}\). To graph the line, plot at least two points. Start at the \(y\)−intercept \((0, 4)\) and move down \(2\) unit then move to the right \(1\) unit to plot the second point. Now join the two points with a straight line as shown in the figure below.
b. Notice that it is not clear how to identify the slope and \(y\)-intercept in this given linear equation because it is not in the slope-intercept form. Thus, solve for \(y\) to have the equation in the slope-intercept form as follows,
\(\begin{array} &&5y − 3x = −10 &\text{Given} \\ &5y = 3x − 10 &\text{Add \(3x\) to both sides of the equation} \\ &y = \dfrac{3}{5}x − 2 &\text{Divide all terms by \(5\) to isolate \(y\)} \end{array}\)
Now, \(m = \dfrac{3}{5}\) and \(b = −2\). Start by plotting the \(y\)-intercept \((0, −2)\) then move \(3\) units upward and \(5\) units to the right and plot the second point which is \((5, 1)\). Now, join the two points, namely, \((0, −2)\) and \((5, 1)\) to get the graph of the line shown in the figure below.
Write an equation of a line with the given slope and \(y\)-intercept.
- slope: \(2\) \(y\)-intercept: \((0, \dfrac{3}{4})\)
- slope: \(\dfrac{5}{7}\) \(y\)-intercept: \((0, −6)\)
- slope: \(−\dfrac{1}{2}\) \(y\)-intercept: \((0, −\dfrac{7}{11} )\)
Identify the slope and \(y\)-intercept then use them to graph each line.
- \(y = 5x − 3\)
- \(2y = −6x + 1\)
Point-Slope Form of the Equation of a Line
The Point-Slope Form of the equation of a straight line is:
\[y − y_1 = m(x − x_1) \nonumber \]
Where \(m\) is the slope of the line and \((x_1, y_1)\) is any point on the straight line.
Find the equation of each line passing through the given point and given slope.
- Slope \(3\) and point \((−1, 8)\)
- Slope \(−\dfrac{5}{2}\) and point \((\dfrac{4}{3}, \dfrac{1}{3})\)
Solution
- To find the equation of the line through the point \((−1, 8)\) with slope \(m = 3\), use the point-slope form as follows:
\(\begin{array} &&y − y_1 = m(x − x_1) &\text{Point-Slope form} \\ &y − 8 = 3[x − (−1)] &\text{Substitute \(m = 3\), \(x_1 = −1\), and \(y_1 = 8\)} \\ &y − 8 = 3(x + 1) &\text{Simplify} \\ &y − 8 = 3x + 3 &\text{Multiply both terms on the right of the equation by \(3\)} \\ &y = 3x + 11 &\text{Add \(8\) to both sides of the equality to isolate \(y\)} \end{array}\)
Therefore, \(y = 3x + 11\) is the equation of the line with the given slope and point. The line is in the slope-intercept form.
- Similar to part a, use the Point-Slope Form as follows:
\(\begin{array} &&y − y_1 = m(x − x_1) &\text{Point-Slope form} \\ & y−(−\dfrac{1}{3}) = −\dfrac{5}{2} (x −\dfrac{4}{3}) &\text{Substitute \(m = −\dfrac{5}{2},\;\; x_1 = \dfrac{4}{3}\), and \(y_1 = −\dfrac{1}{3}\)} \\ &y + \dfrac{1}{3} = −\dfrac{5}{2}x + \dfrac{20}{6} &\text{Distribute and simplify} \\ &y = −\dfrac{5}{2}x + \dfrac{20}{6} − \dfrac{1}{3} &\text{Subtract \(\dfrac{1}{3}\) from both sides} \\ &y = −\dfrac{5}{2}x + 3 &\text{To combine the two fractions, notice that the LCD \(= 6\).} \\ & &\text{Multiply numerator and denominator of \(\dfrac{1}{3}\) by \(2\) and simplify:} \\ & &\text{\(\dfrac{20}{6} − \dfrac{1(2)}{3(2)} = \dfrac{20}{6} − \dfrac{2}{6} = \dfrac{18}{6} = 3\)} \end{array}\)
Therefore, \(y = −\dfrac{5}{2}x + 3\) is the equation of the line through the give point and the given slope.
Find an equation of the line given points \((2, 4)\) and \((−3, 9)\).
Notice that earlier in this chapter it explained how to find an equation of a line given a slope and \(y\)-intercept. This chapter also explained how to find an equation of a line given any point on the line and a slope. So, in both methods, the slope is given.
Solution
To find an equation of a line given any two points on the line, first find the slope using the slope of the line formula. After, apply the point-slope form with any of the given points. First, use the two points to find the slope of the line. Let \((x_1, y_1) = (2, 4)\) and \((x_2, y_2) = (−3, 9)\). Then,
\(\begin{array} &&m = \dfrac{y_2 − y_1}{x_2 − x_1} &\text{Slope of the line formula} \\ &= \dfrac{9 − 4}{−3 − 2} &\text{Substitute values} \\ &= \dfrac{5}{−5} &\text{Simplify} \\ &= −1 & \end{array}\)
Now the slope has been found so next find the equation of the line using any one of the given points. Thus, \(m = −1\) and consider using point \((2, 4)\).
\(\begin{array} &&y − y_1 = m(x − x_1) &\text{Point-slope form} \\ &y − 4 = −1(x − 2) &\text{Substitute \(m = −1\), \(x_1 = 2\), \(y_1 = 4\)} \\ &y − 4 = −x + 2 &\text{Distribute \(-1\) to both terms on the right} \\ &y = −x + 6 &\text{Add \(4\) to both sides of the equation to isolate \(y\)} \end{array}\)
Therefore, \(y = −x + 6\) is the equation of the line passing through the giving point and has the slope-intercept form.
Find the equation of each line passing through the given point and has the given slope.
- The slope \(−\dfrac{5}{2}\) and the point \((3, 0)\).
- The slope \(\dfrac{1}{2}\) and the point \((−2, −3)\).
Find an equation of the line given the following Points.
- \((−9, −3)\) and \((6, −2)\)
- \((4, 1)\) and \((−2, 2)\)
Standard Form of the Equation of a Line (AKA General Form of a Linear Equation)
The standard form of a non-vertical line is in the form
\[Ax + By = C \nonumber \]
Where \(A\) is a positive integer, \(B\) and \(C\) are integers with \(B \neq 0\).
Graph each line of the following equations:
- \(4x − 3y = 6\)
- \(\dfrac{1}{2} − y + 1 = 0\)
Note that the \(x\)-intercept is the point where the line intersects the \(x\)-axis. That is, when \(y = 0\). Thus, the \(x\)-intercept is a point of the form \((a, 0)\), where \(a\) is any real number.
Solution
- The equation \(4x − 3y = 6\) is in standard form. To graph the line of the given equation it may be possible to use more than one method. For example, solving for \(y\) to get the equation in slope-intercept form, then, graph the line. It’s also possible to find two points, then graph the line. The two easiest points to find quickly are the \(x\) and \(y\) intercepts. So, this method is recommended.
To find the \(x\)-intercept, set \(y = 0\) in the given equation and solve for \(x\) as follows,
\(\begin{array} &&4x − 3y = 6 &\text{Given} \\ &4x − 3(0) = 6 &\text{Substitute \(y = 0\)} \\ &4x = 6 &\text{Simplify} \\ &x = \dfrac{6}{4} &\text{Divide by \(4\) both sides of the equation} \\ &x = \dfrac{3}{2} &\text{Simplify} \end{array}\)
Hence, the \(x\)-intercept is the point \((\dfrac{3}{2}, 0)\)
Now, to find the \(y\)-intercept, set \(x = 0\) as follows,
\(\begin{array} &&4x − 3y = 6 &\text{Given} \\ &4(0) − 3y = 6 &\text{Substitute \(x = 0\)} \\ &−3y = 6 &\text{Simplify} \\ &y = 6 −3 &\text{Divide by \(−3\) both sides of the equation} \\ &y = −2 &\text{Simplify} \end{array}\)
Now, plot the points \((\dfrac{3}{2}, 0)\) and \((0, −2)\) and graph the straight line that passes through them as shown in the figure below.
The equation \(\dfrac{1}{2} x − y + 1 = 0\) is not in standard form. So, subtract \(1\) from both sides of the equation to have \(\dfrac{1}{2}x − y = −1\) which is now in standard form.
Again, similar to part b, find the \(x\) and \(y\)-intercepts. First, find the \(x\)-intercept by setting \(y = 0\) and solve for \(x\) as follows.
\( \begin{array} &&\dfrac{1}{2}x − y = −1 &\text{Standard form of the given equation} \\ &\dfrac{1}{2}x − (0) = −1 &\text{Substitute \(y = 0\)} \\ &\dfrac{1}{2}x = −1 &\text{Simplify} \\ &x = −2 &\text{Multiply by \(2\) both sides of the equation.} \end{array}\)
Thus, the \(x\)-intercept is the point \((−2, 0)\).
Now, set \(x = 0\) to find the \(y\)-intercept, as follows,
\( \begin{array} &&\dfrac{1}{2}x − y = −1 &\text{Standard form of the given equation} \\ &\dfrac{1}{2}(0) − y = −1 &\text{Substitute \(x = 0\)} \\ &−y = −1 &\text{Simplify} \\ &y = 1 &\text{Multiply by \(-1\).} \end{array}\)
Hence, the \(y\)-intercept is \((0, 1)\).
Plot the \(x\) and \(y\)-intercepts, \((−2, 0)\) and \((0, 1)\), then graph the straight line that passes through them as shown in the figure below.
There is no homework for this section.