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Notice that both forms rely on knowing the slope. If we are given two points on the line we may still find the equation of the line passing through them by first finding the slope of the line, then using the point-slope form.
It is customary to use either the slope-intercept form or the general form for the final form of the line. We will use the slope-intercept form as the final form.
Find the equation of the line using the given information.
$m=\mathrm{6\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}},\text{}\text{}\text{}\text{}y\text{-intercept\hspace{0.17em}}\text{}(0,4)$
Since we’re given the slope and the $y\text{-intercept,}$ we’ll use the slope-intercept form. $m=6,b=4.$
$\begin{array}{l}y=mx+b\hfill \\ y=6x+4\hfill \end{array}$
$m=-\frac{3}{4},\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\text{}\text{}\text{}\text{}y\text{-intercept\hspace{0.17em}}\text{}\left(0,\frac{1}{8}\right)$
Since we’re given the slope and the $y\text{-intercept,}$ we’ll use the slope-intercept form. $m=\frac{-3}{4},$
$\begin{array}{ccc}b& =& \frac{1}{8}.\\ y& =& mx+b\\ y& =& -\frac{3}{4}x+\frac{1}{8}\end{array}$
$\begin{array}{cc}m=2,& \text{the\hspace{0.17em}}\text{}\text{point}(4,3).\end{array}$
Write the equation in slope-intercept form.
Since we’re given the slope and some point, we’ll use the point-slope form.
$\begin{array}{llll}y-{y}_{1}\hfill & =\hfill & m(x-{x}_{1})\hfill & \text{Let\hspace{0.17em}}\text{}({x}_{1},{y}_{1})\hspace{0.17em}\text{}\text{be\hspace{0.17em}}\text{}\text{(4,3)}\text{.}\hfill \\ y-3\hfill & =\hfill & 2(x-4)\hfill & \text{Put\hspace{0.17em}}\text{}\text{this\hspace{0.17em}}\text{}\text{equation\hspace{0.17em}}\text{}\text{in\hspace{0.17em}}\text{}\text{slope-intercept\hspace{0.17em}}\text{}\text{form\hspace{0.17em}}\text{}\text{by\hspace{0.17em}}\text{}\text{solving\hspace{0.17em}}\text{}\text{for\hspace{0.17em}}\text{}y.\hfill \\ y-3\hfill & =\hfill & 2x-8\hfill & \hfill \\ \hfill y& =\hfill & 2x-5\hfill & \hfill \end{array}$
$\begin{array}{cc}m=-5,& \text{the\hspace{0.17em}}\text{}\text{point}(-3,0).\end{array}$
Write the equation in slope-intercept form.
Since we’re given the slope and some point, we’ll use the point-slope form.
$\begin{array}{llll}y-{y}_{1}\hfill & =\hfill & m(x-{x}_{1})\hfill & \text{Let\hspace{0.17em}}\text{}({x}_{1},{y}_{1})\hspace{0.17em}\text{}\text{be\hspace{0.17em}}\text{}\text{(-3,0)}\text{.}\hfill \\ y-0\hfill & =\hfill & -5\left[x-(-3)\right]\hfill & \hfill \\ \hfill y& =\hfill & -5(x+3)\hfill & \text{Solve\hspace{0.17em}}\text{}\text{for\hspace{0.17em}}\text{}y.\hfill \\ \hfill y& =\hfill & -5x-15\hfill & \hfill \end{array}$
$\begin{array}{cc}m=-1,& \text{the\hspace{0.17em}}\text{}\text{point}(0,7).\end{array}$
Write the equation in slope-intercept form.
We’re given the slope and a point, but careful observation reveals that this point is actually the $y\text{-intercept}\text{.}$ Thus, we’ll use the slope-intercept form. If we had not seen that this point was the $y\text{-intercept}$ we would have proceeded with the point-slope form. This would create slightly more work, but still give the same result.
$\begin{array}{ll}\underset{\xaf}{\text{Slope-intercept\hspace{0.17em}}\text{}\text{form}}\hfill & \underset{\xaf}{\text{Point-slope\hspace{0.17em}}\text{}\text{form}}\hfill \\ \begin{array}{lll}y\hfill & =\hfill & mx+b\hfill \\ y\hfill & =\hfill & -1x+7\hfill \\ y\hfill & =\hfill & -x+7\hfill \\ \hfill & \hfill & \hfill \end{array}\hfill & \begin{array}{lll}y-{y}_{1}\hfill & =\hfill & m(x-{x}_{1})\hfill \\ y-7\hfill & =\hfill & -1(x-0)\hfill \\ y-7\hfill & =\hfill & -x\hfill \\ y\hfill & =\hfill & -x+7\hfill \end{array}\hfill \end{array}$
The two points
$(4,1)$ and
$(3,5).$
Write the equation in slope-intercept form.
Since we’re given two points, we’ll find the slope first.
$m=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}=\frac{5-1}{3-4}=\frac{4}{-1}=-4$
Now, we have the slope and two points. We can use either point and the point-slope form.
Using $(4,\hspace{0.17em}\text{}1)$ | Using $(3,\hspace{0.17em}\text{}5)$ |
$\begin{array}{lll}y-{y}_{1}\hfill & =\hfill & m(x-{x}_{1})\hfill \\ y-1\hfill & =\hfill & -4(x-4)\hfill \\ y-1\hfill & =\hfill & -4x+16\hfill \\ y\hfill & =\hfill & -4x+17\hfill \end{array}$ | $\begin{array}{lll}y-{y}_{1}\hfill & =\hfill & m(x-{x}_{1})\hfill \\ y-5\hfill & =\hfill & -4(x-3)\hfill \\ y-5\hfill & =\hfill & -4x+12\hfill \\ y\hfill & =\hfill & -4x+17\hfill \end{array}$ |
We can see that the use of either point gives the same result.
Find the equation of each line given the following information. Use the slope-intercept form as the final form of the equation.
$m=5,\hspace{0.17em}\text{}y\text{-intercept\hspace{0.17em}}\text{}(0,8).$
$y=5x+8$
$m=-8,\hspace{0.17em}\text{}y\text{-intercept\hspace{0.17em}}\text{}(0,3).$
$y=-8x+3$
$m=2,\hspace{0.17em}\text{}y\text{-intercept\hspace{0.17em}}\text{}(0,-7).$
$y=2x-7$
$m=1,\hspace{0.17em}\text{}y\text{-intercept\hspace{0.17em}}\text{}(0,-1).$
$y=x-1$
$m=-1,\hspace{0.17em}\text{}y\text{-intercept\hspace{0.17em}}\text{}(0,-10).$
$y=-x-10$
Find the equation of the line passing through the point $(4,-7)$ having slope 0.
We’re given the slope and some point, so we’ll use the point-slope form. With $m=0$ and $({x}_{1},{y}_{1})$ as $(4,-7),$ we have
$\begin{array}{lll}y-{y}_{1}\hfill & =\hfill & m(x-{x}_{1})\hfill \\ y-(-7)\hfill & =\hfill & 0(x-4)\hfill \\ y+7\hfill & =\hfill & 0\hfill \\ y\hfill & =\hfill & -7\hfill \end{array}$
This is a horizontal line.
Find the equation of the line passing through the point $(1,3)$ given that the line is vertical.
Since the line is vertical, the slope does not exist. Thus, we cannot use either the slope-intercept form or the point-slope form. We must recall what we know about vertical lines. The equation of this line is simply $x=1.$
Find the equation of the line passing through the point $(-2,9)$ having slope 0.
$y=9$
Find the equation of the line passing through the point $(-1,6)$ given that the line is vertical.
$x=-1$
Reading only from the graph, determine the equation of the line.
The slope of the line is $\frac{2}{3},$ and the line crosses the $y\text{-axis}$ at the point $(0,-3).$ Using the slope-intercept form we get
$y=\frac{2}{3}x-3$
Reading only from the graph, determine the equation of the line.
$y=\frac{-2}{3}x+4$
For the following problems, write the equation of the line using the given information in slope-intercept form.
$m=3,\hspace{0.17em}\text{}y\text{-intercept\hspace{0.17em}}\text{}(0,4)$
$y=3x+4$
$m=2,\hspace{0.17em}\text{}y\text{-intercept\hspace{0.17em}}\text{}(0,5)$
$m=8,\hspace{0.17em}\text{}y\text{-intercept\hspace{0.17em}}\text{}(0,1)$
$y=8x+1$
$m=5,\hspace{0.17em}\text{}y\text{-intercept\hspace{0.17em}}\text{}(0,-3)$
$m=-6,\hspace{0.17em}\text{}y\text{-intercept\hspace{0.17em}}\text{}(0,-1)$
$y=-6x-1$
$m=-4,\hspace{0.17em}\text{}y\text{-intercept\hspace{0.17em}}\text{}(0,0)$
$m=-\frac{3}{2},\hspace{0.17em}\text{}y\text{-intercept\hspace{0.17em}}\text{}(0,0)$
$y=-\frac{3}{2}x$
$m=3,\hspace{0.17em}\text{}(1,4)$
$m=2,\hspace{0.17em}\text{}(1,4)$
$m=-3,\hspace{0.17em}\text{}(3,0)$
$m=-6,\hspace{0.17em}\text{}(0,0)$
$(0,0),\hspace{0.17em}\text{}(3,2)$
$(4,1),\hspace{0.17em}\text{}(6,3)$
$(5,-3),\hspace{0.17em}\text{}(6,2)$
$(2,3),\hspace{0.17em}\text{}(5,3)$
$y=\mathrm{3\hspace{0.17em}}\text{}\left(\text{horizontal\hspace{0.17em}}\text{}\text{line}\right)$
$(-1,5),\hspace{0.17em}\text{}(4,5)$
$(4,1),\hspace{0.17em}\text{}(4,2)$
$x=\mathrm{4\hspace{0.17em}}\text{}\left(\text{vertical\hspace{0.17em}}\text{}\text{line}\right)$
$(2,7),\hspace{0.17em}\text{}(2,8)$
$(0,0),\hspace{0.17em}\text{}(1,1)$
$(-2,4),\hspace{0.17em}\text{}(3,-5)$
$y=-\frac{9}{5}x+\frac{2}{5}$
$(1,6),\hspace{0.17em}\text{}(-1,-6)$
$(0,-4),\hspace{0.17em}\text{}(5,0)$
For the following problems, read only from the graph and determine the equation of the lines.
(
[link] ) Graph the equation
$x-3=0.$
(
[link] ) Supply the missing word. The point at which a line crosses the
$y\text{-axis}$ is called the
$y\text{-intercept}$
(
[link] ) Supply the missing word. The
( [link] ) Find the slope of the line that passes through the points $(4,0)$ and $(-2,-6).$
$m=1$
(
[link] ) Graph the equation
$3y=2x+3.$
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