7.6 Finding the equation of a line  (Page 2/2)

$m=\mathrm{6\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}},y\text{-intercept\hspace{0.17em}}(0,4)$

Since we’re given the slope and the $y\text{-intercept,}$ we’ll use the slope-intercept form. $m=6,b=4.$

$\begin{array}{l}y=mx+b\hfill \\ y=6x+4\hfill \end{array}$

$m=-\frac{3}{4},\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}y\text{-intercept\hspace{0.17em}}\left(0,\frac{1}{8}\right)$

Since we’re given the slope and the $y\text{-intercept,}$ we’ll use the slope-intercept form. $m=\frac{-3}{4},$

$\begin{array}{ccc}b& =& \frac{1}{8}.\\ y& =& mx+b\\ y& =& -\frac{3}{4}x+\frac{1}{8}\end{array}$

$\begin{array}{cc}m=2,& \text{the\hspace{0.17em}}\text{point}(4,3).\end{array}$
Write the equation in slope-intercept form.

Since we’re given the slope and some point, we’ll use the point-slope form.

$\begin{array}{llll}y-y_1\hfill & =\hfill & m(x-x_1)\hfill & \text{Let\hspace{0.17em}}(x_1,y_1)\hspace{0.17em}\text{be\hspace{0.17em}}\text{(4,3)}\text{.}\hfill \\ y-3\hfill & =\hfill & 2(x-4)\hfill & \text{Put\hspace{0.17em}}\text{this\hspace{0.17em}}\text{equation\hspace{0.17em}}\text{in\hspace{0.17em}}\text{slope-intercept\hspace{0.17em}}\text{form\hspace{0.17em}}\text{by\hspace{0.17em}}\text{solving\hspace{0.17em}}\text{for\hspace{0.17em}}y.\hfill \\ y-3\hfill & =\hfill & 2x-8\hfill & \hfill \\ \hfill y& =\hfill & 2x-5\hfill & \hfill \end{array}$

$\begin{array}{cc}m=-5,& \text{the\hspace{0.17em}}\text{point}(-3,0).\end{array}$
Write the equation in slope-intercept form.

Since we’re given the slope and some point, we’ll use the point-slope form.

$\begin{array}{llll}y-y_1\hfill & =\hfill & m(x-x_1)\hfill & \text{Let\hspace{0.17em}}(x_1,y_1)\hspace{0.17em}\text{be\hspace{0.17em}}\text{(-3,0)}\text{.}\hfill \\ y-0\hfill & =\hfill & -5\left[x-(-3)\right]\hfill & \hfill \\ \hfill y& =\hfill & -5(x+3)\hfill & \text{Solve\hspace{0.17em}}\text{for\hspace{0.17em}}y.\hfill \\ \hfill y& =\hfill & -5x-15\hfill & \hfill \end{array}$

$\begin{array}{cc}m=-1,& \text{the\hspace{0.17em}}\text{point}(0,7).\end{array}$
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.

$\begin{array}{ll}\underset{¯}{\text{Slope-intercept\hspace{0.17em}}\text{form}}\hfill & \underset{¯}{\text{Point-slope\hspace{0.17em}}\text{form}}\hfill \\ \begin{array}{lll}y\hfill & =\hfill & mx+b\hfill \\ y\hfill & =\hfill & -1x+7\hfill \\ y\hfill & =\hfill & -x+7\hfill \\ \hfill & \hfill & \hfill \end{array}\hfill & \begin{array}{lll}y-y_1\hfill & =\hfill & m(x-x_1)\hfill \\ y-7\hfill & =\hfill & -1(x-0)\hfill \\ y-7\hfill & =\hfill & -x\hfill \\ y\hfill & =\hfill & -x+7\hfill \end{array}\hfill \end{array}$

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.

We can see that the use of either point gives the same result.

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 $x=1.$

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

$y=\frac{2}{3}x-3$