# 2.2 Linear equations in one variable  (Page 5/15)

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Given $\text{\hspace{0.17em}}m=4,$ find the equation of the line in slope-intercept form passing through the point $\text{\hspace{0.17em}}\left(2,5\right).$

$y=4x-3$

## Finding the equation of a line passing through two given points

Find the equation of the line passing through the points $\text{\hspace{0.17em}}\left(3,4\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(0,-3\right).\text{\hspace{0.17em}}$ Write the final equation in slope-intercept form.

First, we calculate the slope using the slope formula and two points.

$\begin{array}{ccc}\hfill m& =& \frac{-3-4}{0-3}\hfill \\ & =& \frac{-7}{-3}\hfill \\ & =& \frac{7}{3}\hfill \end{array}$

Next, we use the point-slope formula with the slope of $\text{\hspace{0.17em}}\frac{7}{3},$ and either point. Let’s pick the point $\text{\hspace{0.17em}}\left(3,4\right)\text{\hspace{0.17em}}$ for $\text{\hspace{0.17em}}\left({x}_{1},{y}_{1}\right).$

In slope-intercept form, the equation is written as $\text{\hspace{0.17em}}y=\frac{7}{3}x-3.$

## Standard form of a line

Another way that we can represent the equation of a line is in standard form . Standard form is given as

$Ax+By=C$

where $\text{\hspace{0.17em}}A,$ $B,$ and $\text{\hspace{0.17em}}C$ are integers. The x- and y- terms are on one side of the equal sign and the constant term is on the other side.

## Finding the equation of a line and writing it in standard form

Find the equation of the line with $\text{\hspace{0.17em}}m=-6\text{\hspace{0.17em}}$ and passing through the point $\text{\hspace{0.17em}}\left(\frac{1}{4},-2\right).\text{\hspace{0.17em}}$ Write the equation in standard form.

We begin using the point-slope formula.

$\begin{array}{ccc}\hfill y-\left(-2\right)& =& -6\left(x-\frac{1}{4}\right)\hfill \\ \hfill y+2& =& -6x+\frac{3}{2}\hfill \end{array}$

From here, we multiply through by 2, as no fractions are permitted in standard form, and then move both variables to the left aside of the equal sign and move the constants to the right.

$\begin{array}{ccc}\hfill 2\left(y+2\right)& =& \left(-6x+\frac{3}{2}\right)2\hfill \\ \hfill 2y+4& =& -12x+3\hfill \\ \hfill 12x+2y& =& -1\hfill \end{array}$

This equation is now written in standard form.

Find the equation of the line in standard form with slope $\text{\hspace{0.17em}}m=-\frac{1}{3}\text{\hspace{0.17em}}$ and passing through the point $\text{\hspace{0.17em}}\left(1,\frac{1}{3}\right).$

$x+3y=2$

## Vertical and horizontal lines

The equations of vertical and horizontal lines do not require any of the preceding formulas, although we can use the formulas to prove that the equations are correct. The equation of a vertical line is given as

$x=c$

where c is a constant. The slope of a vertical line is undefined, and regardless of the y- value of any point on the line, the x- coordinate of the point will be c .

Suppose that we want to find the equation of a line containing the following points: $\text{\hspace{0.17em}}\left(-3,-5\right),\left(-3,1\right),\left(-3,3\right),$ and $\text{\hspace{0.17em}}\left(-3,5\right).\text{\hspace{0.17em}}$ First, we will find the slope.

$m=\frac{5-3}{-3-\left(-3\right)}=\frac{2}{0}$

Zero in the denominator means that the slope is undefined and, therefore, we cannot use the point-slope formula. However, we can plot the points. Notice that all of the x- coordinates are the same and we find a vertical line through $\text{\hspace{0.17em}}x=-3.\text{\hspace{0.17em}}$ See [link] .

The equation of a horizontal line is given as

$y=c$

where c is a constant. The slope of a horizontal line is zero, and for any x- value of a point on the line, the y- coordinate will be c .

Suppose we want to find the equation of a line that contains the following set of points: $\text{\hspace{0.17em}}\left(-2,-2\right),\left(0,-2\right),\left(3,-2\right),$ and $\text{\hspace{0.17em}}\left(5,-2\right).$ We can use the point-slope formula. First, we find the slope using any two points on the line.

$\begin{array}{ccc}\hfill m& =& \frac{-2-\left(-2\right)}{0-\left(-2\right)}\hfill \\ & =& \frac{0}{2}\hfill \\ & =& 0\hfill \end{array}$

Use any point for $\text{\hspace{0.17em}}\left({x}_{1},{y}_{1}\right)\text{\hspace{0.17em}}$ in the formula, or use the y -intercept.

$\begin{array}{ccc}\hfill y-\left(-2\right)& =& 0\left(x-3\right)\hfill \\ \hfill y+2& =& 0\hfill \\ \hfill y& =& -2\hfill \end{array}$

The graph is a horizontal line through $\text{\hspace{0.17em}}y=-2.\text{\hspace{0.17em}}$ Notice that all of the y- coordinates are the same. See [link] . The line x = −3 is a vertical line. The line y = −2 is a horizontal line.

## Finding the equation of a line passing through the given points

Find the equation of the line passing through the given points: $\text{\hspace{0.17em}}\left(1,-3\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(1,4\right).$

The x- coordinate of both points is 1. Therefore, we have a vertical line, $\text{\hspace{0.17em}}x=1.$

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