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College Algebra/Graphing Lines/Forms of Linear Equations

Graphing Lines

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Forms of Linear Equations

Slope-Intercept Form

The equation of a nonvertical line with slope mm and yy-intercept bb can be written in the slope-intercept form y=mx+by = mx + b.
The slope-intercept form of a line is a linear equation, where mm is the slope and bb is the yy-intercept of the line, expressed as:
y=mx+by=mx+b
As the name suggests, the slope-intercept form gives two important pieces of information: the slope and the yy-intercept. This information can be used to graph the line without making a table and plotting points.
Step-By-Step Example
Graphing Positive Slope in Slope-Intercept Form
Graph the line:
y=32x+5y=\frac{3}{2}x+5
Step 1
The equation is in slope-intercept form, where mm is the slope and bb is the yy-intercept:
y=mx+by=mx+b
Identify the slope and yy-intercept of the equation.
Step 2

The yy-intercept is 5, so plot one point at (0,5)(0, 5).

Then use the rise and run of the slope, mm, to locate another point.
m=RiseRun=32\begin{aligned}m&=\frac{\text{Rise}}{\text{Run}}\\&=\frac{3}{2}\end{aligned}
The rise is 3, so move up 3 units from the first point (0,5)(0, 5). The run is 2, so move right 2 units. This means that another point on the line is (2,8)(2, 8).
Solution
Connect the two points to graph the line.
Step-By-Step Example
Graphing with Negative Slope in Slope-Intercept Form
Graph the line:
y=−2x−3y=-2x-3
Step 1
The equation is in slope-intercept form, where mm is the slope and bb is the yy-intercept. Identify the slope and yy-intercept of the equation:
y=mx+by=−2x−3\begin{aligned}y&=mx+b\\y&=-2x-3\end{aligned}
The slope is –2. The yy-intercept is –3.
Step 2

The yy-intercept is –3, so plot one point at (0,−3)(0,-3).

Then use the rise and run of the slope, mm, to locate another point. Write the slope as a fraction with a denominator of 1.
m=RiseRun=−21\begin{aligned}m&=\frac{\text{Rise}}{\text{Run}}\\&=\frac{-2}{1}\end{aligned}
The rise is –2, so move 2 units down. The run is 1, so move 1 unit to the right. So, another point on the line is (1,−5)(1,-5).
Solution
Connect the two points to graph the line.

Point-Slope Form

The equation of a nonvertical line with slope mm that passes through the point (x1,y1)(x_1, y_1) can be written in the point-slope form y−y1=m(x−x1)y-y_1=m(x-x_1).
The point-slope form of a line is a linear equation, where mm is the slope and (x1,y1)(x_1, y_1) is a point on the line, expressed as:
y−y1=m(x−x1)y-y_1=m(x-x_1)
As the name suggests, the point-slope form is convenient to use in writing the equation of a line when the slope and the coordinates of a point on the line are given.
Step-By-Step Example
Writing an Equation with a Given Slope and Point
Write the equation of a line with a slope of 3 that passes through the point (−4,1)(-4,1).
Step 1
Substitute 3 for mm, –4 for x1x_1, and 1 for y1y_1 in point-slope form.
y−y1=m(x−x1)y−1=3(x−(−4))y−1=3(x+4)\begin{aligned}y-y_1&=m(x-x_1)\\y-1&=3(x-(-4))\\y-1&=3(x+4)\end{aligned}
Step 2
Although the point-slope form of a line is a correct form, it is common to rewrite the equation of a line in slope-intercept form. First, distribute the 3 on the right side.
y−1=3(x+4)y−1=3x+12\begin{aligned}y-1&=3(x+4)\\y-1&=3x+12\end{aligned}
Solution
Add 1 to both sides.
y−1=3x+12y+ 1‾=3x + 1‾y=3x+13\begin{aligned}y-1&=3x+12\\\underline{\phantom{y}+\ 1}&\phantom{=\:}\underline{\phantom{3x\ }+\ \: 1}\\y&=3x+13\end{aligned}
Step-By-Step Example
Graphing a Line in Point-Slope Form
Graph the line of the equation:
y−4=−3(x+2)y-4=-3(x+2)
Step 1

Identify the slope and a point on the line.

The equation uses the point-slope form, where mm is the slope and x1x_1 and y1y_1 are coordinates of a point on the line, or (x1,y1)(x_1, y_1):
y−y1=m(x−x1)y-y_1=m(x-x_1)
When x1x_1 or y1y_1 are being subtracted in the equation, the coordinates are positive.
When x1x_1 or y1y_1 are being added in the equation, the coordinates are negative.
Step 2
Plot the point at (−2,4)(-2, 4). Then use the rise and run of the slope, mm, to locate another point. Write the slope as:
m=RiseRun=−31\begin{aligned}m&=\frac{\text{Rise}}{\text{Run}}\\&=\frac{-3}{1}\end{aligned}
The rise is –3, so move down 3 units. The run is 1, so move right 1 unit.
Solution
Connect the two points to graph the line.
Step-By-Step Example
Determining the Equation of a Line through Two Points
Identify the equation of a line that passes through point (3,1)(3, 1) and point (5,9)(5, 9).
Step 1
Use the coordinates of the two points to identify the slope.
m=y2−y1x2−x1=9−15−3=82=4\begin{aligned}m &=\frac{y_2-y_1}{x_2-x_1}\\&=\frac{9-1}{5-3}\\&=\frac{8}{2}\\&=4\end{aligned}
Step 2
Choose one of the points on the line to write an equation in point-slope form. Note that using either point will result in correct equations.
(x1,y1)=(3,1)(x_1,y_1)=(3,1)
Solution
Use the slope and one of the points on the line to write the point-slope form:
y−1=4(x−3)y-1=4(x-3)
It is common to rewrite the equation in slope-intercept form. First, distribute the 4. Then add 1 to both sides.
y−1=4(x−3)y−1=4x−12y−1+1=4x−12+1y=4x−11\begin{aligned}y-1&=4(x-3)\\y-1&=4x-12\\y-1+1&=4x-12+1\\y&=4x-11\end{aligned}

Standard Form of a Line

The standard form of a line is Ax+By=CAx + By = C. It is typically used for determining the xx- and yy-intercepts of a line.
The standard form of a line is a linear equation, where AA, BB, and CC are integers, AA is nonnegative, and AA and BB are both not zero, expressed as:
Ax+By=CAx+By=C
Although the standard form of a line is not the most commonly used for graphing, it is called standard because it is used in other applications, such as solving systems of equations. It is also a useful form for finding the intercepts of a line. The same process is used to find the intercepts of a line in any form.
  • To calculate the xx-intercept, set y=0y=0 and solve for xx.
  • To calculate the yy-intercept, set x=0x=0 and solve for yy.
Step-By-Step Example
Identifying the Intercepts of a Line in Standard Form
Identify the xx-intercept and the yy-intercept of the equation:
3x+4y=123x + 4y = 12
Step 1
Determine the xx-intercept. Substitute zero for yy, and solve for xx.
3x+4y=123x+4(0)=123x=12x=4\begin{aligned}3x + 4y &= 12\\3x+4(0)&=12\\3x&=12\\x&=4\end{aligned}
Step 2
Determine the yy-intercept. Substitute zero for xx, and solve for yy.
3x+4y=123(0)+4y=124y=12y=3\begin{aligned}3x + 4y &= 12\\3(0)+4y&=12\\4y&=12\\y&=3\end{aligned}
Solution
The xx-intercept is (4,0)(4,0). The yy-intercept is (0,3)(0,3).

Although it is possible to graph a line in standard form, it is sometimes convenient to rewrite the equation in slope-intercept form. One reason is to determine the slope. Another reason is that it is easier to compare two different lines when they are both written in slope-intercept form.

For a line that is given in standard form Ax+By=CAx+By=C, to rewrite the equation in slope-intercept form, first subtract −Ax-Ax from both sides. Then divide both sides by BB, and simplify.

Step-By-Step Example
Rewriting the Standard Form of a Line in Slope-Intercept Form
Write the linear equation in slope-intercept form:
3x+4y=123x + 4y = 12
Step 1
The equation is in standard form:
Ax+By=C3x+4y=12\begin{aligned}Ax+By&=C\\3x+4y&=12\end{aligned}
Rewrite the equation to the slope-intercept form:
y=mx+by=mx+b
Subtract −3x-3x from both sides to isolate the yy-term. Since the slope-intercept form of a line has the xx-term before the constant, write the xx-term first on the right side of the equation.
3x+4y=123x+4y−3x=12−3x4y=−3x+12\begin{aligned}3x+4y&=12\\3x+4y-3x&=12-3x\\4y&=-3x+12\end{aligned}
Step 2
Divide both sides by 4.
4y=−3x+124y4=−3x+124\begin{aligned}4y&={-3x+12}\\{\frac{4y}{4}}&=\frac{-3x+12}{4}\end{aligned}
Solution
Simplify the equation.
4y4=−3x+124y=−34x+3\begin{aligned}\frac{4y}{4}&=\frac{-3x+12}{4}\\y&=-\frac{3}{4}x+3\end{aligned}
The slope is −34-\frac{3}{4}. The yy-intercept is 3.
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