# 7.5: Forms of the Equation of a Line

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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

##### Definition: Slope-Intercept Form

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 $$x = 0$$.

##### Example 7.5.1

Write an equation of the line with the given slopes and $$y$$-intercepts.

1. slope = $$5$$; $$y$$−intercept $$(0, \dfrac{1}{2})$$
2. slope = $$−\dfrac{5}{6}$$; $$y$$− intercept $$(0, −\dfrac{3}{4})$$

Solution

1. $$m = 5$$ 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.

1. 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.

##### Example 7.5.2

Identify the slope and $$y$$−intercept then, use them to graph each line.

1. $$y = −2x + 4$$
2. $$5y − 3x = 10$$

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.

##### Exercise 7.5.1

Write an equation of a line with the given slope and $$y$$-intercept.

1. slope: $$2$$ $$y$$-intercept: $$(0, \dfrac{3}{4})$$
2. slope: $$\dfrac{5}{7}$$ $$y$$-intercept: $$(0, −6)$$
3. slope: $$−\dfrac{1}{2}$$ $$y$$-intercept: $$(0, −\dfrac{7}{11} )$$
##### Exercise 7.5.2

Identify the slope and $$y$$-intercept then use them to graph each line.

1. $$y = 5x − 3$$
2. $$2y = −6x + 1$$

## Point-Slope Form of the Equation of a Line

##### Definition: Point-Slope Form

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.

##### Example 7.5.3

Find the equation of each line passing through the given point and given slope.

1. Slope $$3$$ and point $$(−1, 8)$$
2. Slope $$−\dfrac{5}{2}$$ and point $$(\dfrac{4}{3}, \dfrac{1}{3})$$

Solution

1. 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.

1. 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.

##### Example 7.5.4

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.

##### Exercise 7.5.3

Find the equation of each line passing through the given point and has the given slope.

1. The slope $$−\dfrac{5}{2}$$ and the point $$(3, 0)$$.
2. The slope $$\dfrac{1}{2}$$ and the point $$(−2, −3)$$.
##### Exercise 7.5.4

Find an equation of the line given the following Points.

1. $$(−9, −3)$$ and $$(6, −2)$$
2. $$(4, 1)$$ and $$(−2, 2)$$

## Standard Form of the Equation of a Line (AKA General Form of a Linear Equation)

##### Definition: Standard Form

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 $$B \neq 0$$.

##### Example 7.5.5

Graph each line of the following equations:

1. $$4x − 3y = 6$$
2. $$\dfrac{1}{2} − y + 1 = 0$$

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

1. 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.