0.10: Review - Find the Equation of a Line

Find an Equation of a Line Given a Graph

Sometimes, the slope and intercept need to be determined from the graph.

Example \(\PageIndex{1}\) Equation of a line given a graph

Find the equation of the line shown in the graph.

a.

b.

c.

a. Solution

We need to find the slope and y-intercept of the line from the graph so we can substitute the needed values into the slope-intercept form, \(y=mx+b\).

To find the slope, we choose two points on the graph.

The y-intercept is \((0,−4)\) and the graph passes through \((3,−2)\).

Step 1. Find the slope, by counting the rise and run.	\( m=\dfrac{\text{rise}}{\text{run}}= {\color{red}{\dfrac{2}{3}}} \)
Step 2. Find the y-intercept.	y-intercept is \( (0, {\color{Cerulean}{-4}} ) \)
Step 3. Substitute the values into \(y=mx+b\).	\( y = {\color{red}{m}}x + \color{Cerulean}{b} \) \( y = {\color{red}{\dfrac{2}{3}}}x  \color{Cerulean}{-4} \)

b. Solution

The graph is a horizontal line, we know the form of the equation for the line is \(y=b\). The \(y\) intercept is \((0,-4)\) so the equation is \(y=-4\).

The slope of this line is zero because the \(y\) coordinates of all the points on the line are the same value of \(-4\).  Thus the rise, which is the numerator of the slope equation, is zero. The \(y\)-intercept is \((0,-4)\) so substitution in the equation  \(y=mx+b\) produces \(y = 0x+(-4)\) or \(y=-4\). 

c. Solution

The graph is a vertical line, so we know the form of the equation for the line is \(x=a\). The \(x\) intercept is \((3,0)\) so the equation is \(x=3\). 

The slope of this line is undefined because the \(x\) coordinates of all the points on the line are the same value of \(3\). Thus the run, which is the denominator of the slope equation, is zero, and division by zero is undefined, and thus the slope-intercept form of an equation for a line cannot be used.

Try It \(\PageIndex{2}\)

Find the equation of the following lines. 

a.

b.

c.

d.

Answer

a.    \(y=\frac{3}{5}x+1\)        b.    \(y=\frac{4}{3}x−5\)        c.    \(x=-1\)        d.    \(y=3\)

Find an Equation of the Line Given the Slope and y-Intercept

Sometimes we are told what the slope and y-intercept are.

Find an Equation of the Line Given the Slope and a Point

Finding an equation of a line using the slope-intercept form of the equation works well when you are given the slope and y-intercept or when you read them off a graph. When given a point on the line instead of the y-intercept a slightly different approach must be used to obtain the equation of the line.

One approach directly uses the familiar slope-intercept form of an equation for a line \( y=mx+b \). First substitute the values of the given slope \(m\), and coordinates \(x\) and \(y\) of the given point into the formula \( y = mx + b \). Solve for the value for \(b\). Then rewrite the slope-intercept formula for a line with the now known values for \(m\) and \(b\).

Alternatively, a different form of an equation for a line, called the point-slope form, can be used. The derivation of this equation begins by supposing we have a line that has slope \(m\) and that contains some specific point \((x_1,y_1)\) and some other point, which we will just call \((x,y)\). The slope of this line is \(m= \dfrac{y-y_1}{x-x_1}\). Multiplying both sides of this equation by \((x−x_1)\) produces \( m(x-x_1)=y-y_1 \), which is the point-slope form of an equation for a line.

Find an Equation of the Line Given Two Points

When we are not given the slope nor a point but rather just two points on the line, the value of the slope must be calculated.

Find an Equation of a Line Parallel to a Given Line

 Suppose we need to find an equation of a line that passes through a specific point and is parallel to a given line. We can use the fact that parallel lines have the same slope. So we will have a point and the slope—just what we need to use the point-slope equation.

First, let’s look at this graphically.

Given the graph for \(y=2x−3,\) we want to graph a line parallel to this line and passing through the point \((−2,1)\).

We know that parallel lines have the same slope. We’ll use the notation \(m_∥\) to represent the slope of a line parallel to a line with slope \(m\). (Notice that the subscript || looks like two parallel lines.)

So the second line will have the same slope as \(y=2x−3\). That slope is \(m_∥=2\).

The second (parallel) line will pass through \((−2,1)\) and have slope \(m=2\).

To graph the line, we start at \((−2,1)\) and count out the rise and run.

With \(m=2\) (or \(m=\frac{2}{1}\)), we count out the rise 2, the run 1, and draw the line, as shown in the graph.

Now let’s see how to do this algebraically.

Find an Equation of a Line Perpendicular to a Given Line

Now, let’s consider perpendicular lines. Suppose we need to find a line passing through a specific point and which is perpendicular to a given line. We can use the fact that perpendicular lines have slopes that are negative reciprocals. We will again use the point-slope equation, like we did with parallel lines.

First, let’s look at this graphically.

Given the graph of \(y=2x−3\), we want to graph a line perpendicular to this line and passing through \((−2,1)\).

We know that perpendicular lines have slopes that are negative reciprocals. We’ll use the notation \(m_⊥\) to represent the slope of a line perpendicular to a line with slope m. (Notice that the subscript \(⊥\) looks like the right angles made by two perpendicular lines.)

So the second line will have a slope that is the negative reciprocal of the given line \(y=2x−3 \). The slope of the given line is \(m=2\) so the slope of the wanted line will be \(m_⊥=−\dfrac{1}{2} \).

We now know the second (perpendicular) line will pass through \((−2,1)\) and have slope \(m_⊥=−\dfrac{1}{2}\).

To graph the line, we start at \((−2,1)\) and count out the rise \(−1\), the run \(2\) and draw the line.

Now, let’s see how to do this algebraically.

Key Concepts

• How to determine the slope of a line if it is not given
• If given a graph of a line, choose two points on the graph and count the rise (up is positive) and the run (right is positive to get from one point to the other point. The slope is the ratio, \(m = \dfrac{\text{Rise}}{\text{Run}}\).
• If two points \((x_1,y_1)\) and \((x_2,y_2)\) on the line are given rather than the slope, the slope is \(m=\dfrac{y_2−y_1}{x_2−x_1}\)
• If given a parallel line, identify the slope of the given line. The slope of the wanted line has the same slope.
• If given a perpendicular line, identify the slope of the given line. The slope of the wanted line is the negative reciprocal of the given line.
• If the slope of the given line is \(0\), the given line is horizontal and its equation is \(y=b\).
• If the slope of the given line is undefined, the given line is vertical and its equation is \(x=c\).

• Vertical lines (\(x = c)\) and horizontal lines \((y = b)\) are perpendicular to each other.

• How to find an equation of a line using the point-slope equation.

1. Identify the slope, \(m\).
2. Identify a point \((x_1,y_1)\) on the line. If given two points, either point may be chosen. 
3. Substitute the values of \( y_1\), \( x_1\), and \(m\) into the point-slope equation,\( y−y_1=m(x−x_1)\). 
4. Solve for \(y\)  and write the equation in slope-intercept form, \(y=mx+b\).

• How to find an equation of a line using the slope-intercept equation.

1. Identify the slope, \(m\).
2. Identify a point \((x,y)\) on the line. If given two points, either point may be chosen.
3. Substitute the values of \(x\), \(y\), and \(m\) into the slope-intercept equation,\( y=mx+b\).  
4. Solve for \(b\).  Substitute the values of \(m\) and \(b\) in the slope-intercept form, \(y=mx+b\).

Glossary

point-slope form

The point-slope form of an equation of a line with slope m and containing the point \((x_1,y_1)\) is \(y−y_1=m(x−x_1)\).

slope-intercept form

The slope-intercept form of an equation of a line with slope \(m\) and \(y\)-intercept \( (0,b) \)  is \( y=mx+b \). The \(y\)-intercept is a point on the line where it crosses the \(y\)-axis.