# 1.3: Determining a Linear Equation

- Page ID
- 147254

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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)In this section, you will learn to:

- Find an equation of a line if a point and the slope are given.
- Find an equation of a line if two points are given.

Before you get started, take this prerequisite quiz.

1. Simplify each expression:

a. \(3(2x-5)\)

b. \(-7(4x-2)\)

**Click here to check your answer**-
a. \(6x-15\)

b. \(-28x+14\)

If you missed any part of this problem,

__review here__. (Note that this will open a different textbook in a new window.)

2. Find the slope of the line containing each given pair of points.

a. (1, 4) and (5, -2)

b. (-2, 5) and (10, -4).

**Click here to check your answer**-
a. \(-\dfrac{3}{2}\)

b. \(-\dfrac{3}{4}\)

If you missed any part of this problem,

__review Section 1.2__. (Note that this will open in a new window.)

3. Solve: \(\frac { 1 } { 3 } x + \frac { 1 } { 5 } = \frac { 1 } { 5 } x - 1\).

**Click here to check your answer**-
\(x=-9\)

If you missed this problem,

__review Section 1.1__. (Note that this will open in a new window.)

So far, we were given an equation of a line and were asked to give information about it. For example, we were asked to find points on the line, find its slope and even find intercepts. Now we are going to reverse the process. That is, we will be given either two points, or a point and the slope of a line, and we will be asked to find its equation.

An equation of a line can be written in three forms, the **slope-intercept form**, the **point-slope form,** or the **standard form**. We will discuss each of them in this section.

A line is completely determined by two points, or by a point and slope. The information we are given about a particular line will influence which form of the equation is most convenient to use. Once we know any form of the equation of a line, it is easy to re-express the equation in the other forms if needed.

### THE SLOPE-INTERCEPT FORM OF A LINE: \(y = mx + b\)

In the last section we learned that the equation of a line whose slope = \(m\) and \(y\)-intercept = \(b\) is \[y=mx+b.\] This is called the** slope-intercept form** of the line and is the most commonly used form.

Find an equation of a line whose slope is 5, and \(y\)-intercept is 3.

**Solution**

Since the slope is \(m = 5\), and the \(y\)- intercept is \(b = 3\), the equation is \(y = 5x + 3\).

Find the equation of the line that passes through the point (\(2, 7\)) and has slope \(3\).

**Solution**

Since \(m = 3\), the partial equation is \(y = 3x + b\).

Now \(b\) can be determined by substituting the point (\(2, 7\)) in the equation \(y = 3x + b\).

\begin{aligned}

&7=3(2)+b \nonumber \\

&b=1 \nonumber

\end{aligned}

Therefore, the equation is \(y = 3x + 1\).

Find an equation of the line that passes through the points (-1, 2), and (1, 8).

**Solution**

We need to find the slope of the line containing these points first. Remember that if (\(x_1\), \(y_1\)) and (\(x_2\), \(y_2\)) are two different points on a line, the **slope** of the line is

\[\text{slope}=m=\frac{y_2-y_1}{x_2-x_1} \label{slope}\nonumber\]

\(m=\frac{8-2}{1-(-1)}=\frac{6}{2}=3\). So the partial equation is \(y = 3x + b\).

We can use either of the two points (-1, 2) or (1, 8), to find \(b\). Substituting (-1, 2) gives

\begin{aligned}

&2=3(-1)+b \nonumber \\

&5=b \nonumber

\end{aligned}

So the equation is \(y = 3x +5\).

Find an equation of the line that has \(x\)-intercept 3, and \(y\)-intercept 4.

**Solution**

\(x\)-intercept = 3, and \(y\)-intercept = 4 correspond to the points (3, 0), and (0, 4), respectively.

\[ m=\frac{4-0}{0-3} = -\frac{4}{3} \nonumber \]

We are told the \(y\)-intercept is 4; thus \(b\) = 4

Therefore, the equation is \(y = -\frac{4}{3} x + 4\).

Use slope-intercept form to graph \(y=2x-3\).

**Solution**

Since this equation is already in \(y=mx+b\) form, we can identify the slope as 2 (the coefficient to the x-term) and the y-intercept as -3, or the point \((0,-3)\).

To graph this line, we'll begin with the y-intercept as that is a known point on the graph.

We'll then use the slope to find additional points. The slope is 2, which is more helpful to write in its fractional form of \(2/1\). This tells us that the rise (vertical change between points) is 2 and the run (horizontal change between points) is 1. We can find additional points by going up 2 and right 1, as well as down 2 and left 1.

The graph is

### THE POINT-SLOPE FORM OF A LINE: \(y - y_1 = m(x - x_1)\)

The **point-slope** form is useful when we know two points on the line and want to find the equation of the line.

The definition of a slope leads us to the point-slope formula. Using two points (\(x_1\), \(y_1\)) and any other (\(x,y\)), the slope is \( \frac{y-y_1}{x-x_1}= m\).

Multiplying both sides by (\(x-x_1\)) gives the point-slope form: \(y - y_1 = m(x - x_1)\).

A line with slope * which contains a specific point (\(x_1, y_1)\) can be expressed in the form \[y - y_1 = m(x - x_1).\]*

Find the point-slope form of the equation of the line given in Example \(\PageIndex{2}\). (Find the equation of the line that passes through the point (\(2, 7\)) and has slope \(3\).) Show that the two forms of the equations are equivalent.

**Solution**

Substituting the point \((x_1,y_1) = (2,7)\) and \(m= 3\) in the point-slope formula, we get

\[\mathbf{y - y_1 = m(x - x_1)} \nonumber \]

\[y - 7 = 3(x - 2) \nonumber \]

We can show that the forms are equivalent by solving this equation for \(y\).

\[y - 7 = 3(x - 2) \nonumber \]

\[y - 7 = 3x - 6 \nonumber \]

\[y = 3x + 1 \nonumber \]

This is the slope-intercept form of the equation that we found in Example \(\PageIndex{2}\).

Find the point-slope form of the equation of a line that has slope \(\frac{1}{2}\) and passes through the point (12,4). Then write the equation in slope-intercept form.

**Solution**

Substituting the point \((x_1,y_1) = (12,4)\) and \(m= \frac{1}{2}\) in the point-slope formula, we get

\[\mathbf{y - y_1 = m(x - x_1)} \nonumber \]

\[y - 4 = \frac{1}{2}(x - 12) \nonumber \]

\[y - 4 = \frac{1}{2}x - 6 \nonumber \]

\[y = \frac{1}{2}x - 2 \nonumber \]

### THE STANDARD FORM OF A LINE: \(Ax + By = C\)

Another useful form of the equation of a line is the standard form.

If we know the equation of a line in point-slope form, \(y - y_1 = m(x - x_1)\), or if we know the equation of the line in slope-intercept form \(y = mx + b\), we can simplify the formula to have all terms for the \(x\) and \(y\) variables on one side of the equation, and the constant on the other side of the equation.

The result is referred to as the **standard form** of the line: \(Ax + By = C\) where \(A, B,\) and \(C\) are integers and \(A>0\).

We should always be able to convert from one form of an equation to another. For example, if we are given a line in the slope-intercept form, we should be able to express it in the standard form, and vice versa.

Write the equation \(y = -\frac{2}{3}x + 3\) in the standard form.** **

**Solution**

Multiplying both sides of the equation by 3, we get

\begin{aligned}

&3y = -2x + 9 \\

&2x + 3y = 9 \quad \text { Standard Form }

\end{aligned}

Write the equation \(3x - 4y = 10\) in the slope-intercept form.

**Solution**

Solving for \(y\), we get

\begin{aligned}

&-4y = -3x + 10 \\

&y = \frac{3}{4}x - \frac{5}{2} \quad \text { Standard Form }

\end{aligned}

Using the point-slope formula, find the standard form of an equation of the line that passes through the point (2, 3) and has slope \(-3/5\).** **

**Solution:*** *Substituting the point (2, 3) and \(m= - \frac{3}{5}\) in the point-slope formula, we get

\[y - 3 = - \frac{3}{5}(x - 2) \nonumber \]

Multiplying both sides by 5 gives us

\begin{aligned}

&5(y-3)=-3(x-2)\\

&5 y-15=-3 x+6\\

&3 x+5 y=21 \quad \text { Standard Form }

\end{aligned}

Find the standard form of the line that passes through the points (0, -2), and (4, 0).

**Solution**

First we find the slope: \(m = \frac{0-(-2)}{4-0} = \frac{2}{4}= \frac{1}{2}\)

Then, the point-slope form is:

\begin{aligned}

y - (-2) &= \frac{1}{2}(x -0) \\

y+2 &= \frac{1}{2}x \\

\end{aligned}

Multiplying both sides by 2 gives us

\begin{aligned} 2(y+2) &= 2(\frac{1}{2}x) \\

2y+4 &=1x\\

-1x+2y+4& =0\\

-1x+2y &= -4 \quad \text{ Standard Form } \end{aligned}

Standard form is useful to find both intercepts for a line.

Use standard form to graph \(2x-4y=12\).

**Solution**

To find the x-intercept, we let \(y=0\).

\begin{aligned} 2x-4(0) &= 12 \\

2x &= 12\\

x &= 6\\

\end{aligned}

The x-intercept is \((6,0)\).

To find the y-intercept, we let \(x=0\).

\begin{aligned} 2(0)-4y &= 12 \\

-4y &= 12\\

y &= -3\\

\end{aligned}

The y-intercept is \((0,-3)\).

We graph the line by graphing both intercepts and connecting the points.

We summarize the forms for equations of a line below:

**Slope Intercept form: **\(\mathbf{y = mx + b}\),

where \(m\) = slope, \(b\) = \(y\)-intercept

**Point Slope form:** \(\mathbf{y - y_1 = m(x - x_1)}\),

where \(m\) = slope, \((x_1,y_1)\) is a point on the line

**Standard form: **\(\mathbf{Ax + By = C}\)

**Horizontal Line:** \(\mathbf{y = b}\)

where \(b\) = \(y\)-intercept

**Vertical Line:** \(\mathbf{x = a}\)

where \(a\) = \(x\)-intercept