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

 Page 2 / 2

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.

## Sample set a

Find the equation of the line using the given information.

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

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

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

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

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

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

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

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

$\begin{array}{llll}y-{y}_{1}\hfill & =\hfill & m\left(x-{x}_{1}\right)\hfill & \text{Let\hspace{0.17em}}\text{}\left({x}_{1},{y}_{1}\right) \text{}\text{be\hspace{0.17em}}\text{}\text{(4,3)}\text{.}\hfill \\ y-3\hfill & =\hfill & 2\left(x-4\right)\hfill & \text{Put\hspace{0.17em}}\text{}\text{this\hspace{0.17em}}\text{}\text{equation\hspace{0.17em}}\text{}\text{in\hspace{0.17em}}\text{}\text{slope-intercept\hspace{0.17em}}\text{}\text{form\hspace{0.17em}}\text{}\text{by\hspace{0.17em}}\text{}\text{solving\hspace{0.17em}}\text{}\text{for\hspace{0.17em}}\text{}y.\hfill \\ y-3\hfill & =\hfill & 2x-8\hfill & \hfill \\ \hfill y& =\hfill & 2x-5\hfill & \hfill \end{array}$

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

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

$\begin{array}{llll}y-{y}_{1}\hfill & =\hfill & m\left(x-{x}_{1}\right)\hfill & \text{Let\hspace{0.17em}}\text{}\left({x}_{1},{y}_{1}\right) \text{}\text{be\hspace{0.17em}}\text{}\text{(-3,0)}\text{.}\hfill \\ y-0\hfill & =\hfill & -5\left[x-\left(-3\right)\right]\hfill & \hfill \\ \hfill y& =\hfill & -5\left(x+3\right)\hfill & \text{Solve\hspace{0.17em}}\text{}\text{for\hspace{0.17em}}\text{}y.\hfill \\ \hfill y& =\hfill & -5x-15\hfill & \hfill \end{array}$

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

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

The two points $\left(4,1\right)$ and $\left(3,5\right).$
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 $\left(4, \text{}1\right)$ Using $\left(3, \text{}5\right)$ $\begin{array}{lll}y-{y}_{1}\hfill & =\hfill & m\left(x-{x}_{1}\right)\hfill \\ y-1\hfill & =\hfill & -4\left(x-4\right)\hfill \\ y-1\hfill & =\hfill & -4x+16\hfill \\ y\hfill & =\hfill & -4x+17\hfill \end{array}$ $\begin{array}{lll}y-{y}_{1}\hfill & =\hfill & m\left(x-{x}_{1}\right)\hfill \\ y-5\hfill & =\hfill & -4\left(x-3\right)\hfill \\ y-5\hfill & =\hfill & -4x+12\hfill \\ y\hfill & =\hfill & -4x+17\hfill \end{array}$

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

## Practice set a

Find the equation of each line given the following information. Use the slope-intercept form as the final form of the equation.

$m=5, \text{}y\text{-intercept\hspace{0.17em}}\text{}\left(0,8\right).$

$y=5x+8$

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

$y=-8x+3$

$m=2, \text{}y\text{-intercept\hspace{0.17em}}\text{}\left(0,-7\right).$

$y=2x-7$

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

$y=x-1$

$m=-1, \text{}y\text{-intercept\hspace{0.17em}}\text{}\left(0,-10\right).$

$y=-x-10$

$m=4,$ the point $\left(5,2\right).$

$y=4x-18$

$m=-6,$ the point $\left(-1,0\right).$

$y=-6x-6$

$m=-1,$ the point $\left(-5,-5\right).$

$y=-x-10$

The two points $\left(4,1\right)$ and $\left(6,5\right).$

$y=2x-7$

The two points $\left(-7,-1\right)$ and $\left(-4,8\right).$

$y=3x+20$

## Sample set b

Find the equation of the line passing through the point $\left(4,-7\right)$ having slope 0.

We’re given the slope and some point, so we’ll use the point-slope form. With $m=0$ and $\left({x}_{1},{y}_{1}\right)$ as $\left(4,-7\right),$ we have

$\begin{array}{lll}y-{y}_{1}\hfill & =\hfill & m\left(x-{x}_{1}\right)\hfill \\ y-\left(-7\right)\hfill & =\hfill & 0\left(x-4\right)\hfill \\ y+7\hfill & =\hfill & 0\hfill \\ y\hfill & =\hfill & -7\hfill \end{array}$

This is a horizontal line.

Find the equation of the line passing through the point $\left(1,3\right)$ 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 $x=1.$

## Practice set b

Find the equation of the line passing through the point $\left(-2,9\right)$ having slope 0.

$y=9$

Find the equation of the line passing through the point $\left(-1,6\right)$ given that the line is vertical.

$x=-1$

## Sample set c

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 $\left(0,-3\right).$ Using the slope-intercept form we get

$y=\frac{2}{3}x-3$ ## Practice set c

Reading only from the graph, determine the equation of the line. $y=\frac{-2}{3}x+4$

## Exercises

For the following problems, write the equation of the line using the given information in slope-intercept form.

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

$y=3x+4$

$m=2, \text{}y\text{-intercept\hspace{0.17em}}\text{}\left(0,5\right)$

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

$y=8x+1$

$m=5, \text{}y\text{-intercept\hspace{0.17em}}\text{}\left(0,-3\right)$

$m=-6, \text{}y\text{-intercept\hspace{0.17em}}\text{}\left(0,-1\right)$

$y=-6x-1$

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

$m=-\frac{3}{2}, \text{}y\text{-intercept\hspace{0.17em}}\text{}\left(0,0\right)$

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

$m=3, \text{}\left(1,4\right)$

$m=1, \text{}\left(3,8\right)$

$y=x+5$

$m=2, \text{}\left(1,4\right)$

$m=8, \text{}\left(4,0\right)$

$y=8x-32$

$m=-3, \text{}\left(3,0\right)$

$m=-1, \text{}\left(6,0\right)$

$y=-x+6$

$m=-6, \text{}\left(0,0\right)$

$m=-2, \text{}\left(0,1\right)$

$y=-2x+1$

$\left(0,0\right), \text{}\left(3,2\right)$

$\left(0,0\right), \text{}\left(5,8\right)$

$y=\frac{8}{5}x$

$\left(4,1\right), \text{}\left(6,3\right)$

$\left(2,5\right), \text{}\left(1,4\right)$

$y=x+3$

$\left(5,-3\right), \text{}\left(6,2\right)$

$\left(2,3\right), \text{}\left(5,3\right)$

$y=3 \text{}\left(\text{horizontal\hspace{0.17em}}\text{}\text{line}\right)$

$\left(-1,5\right), \text{}\left(4,5\right)$

$\left(4,1\right), \text{}\left(4,2\right)$

$x=4 \text{}\left(\text{vertical\hspace{0.17em}}\text{}\text{line}\right)$

$\left(2,7\right), \text{}\left(2,8\right)$

$\left(3,3\right), \text{}\left(5,5\right)$

$y=x$

$\left(0,0\right), \text{}\left(1,1\right)$

$\left(-2,4\right), \text{}\left(3,-5\right)$

$y=-\frac{9}{5}x+\frac{2}{5}$

$\left(1,6\right), \text{}\left(-1,-6\right)$

$\left(14,12\right), \text{}\left(-9,-11\right)$

$y=x-2$

$\left(0,-4\right), \text{}\left(5,0\right)$

For the following problems, read only from the graph and determine the equation of the lines. $y=\frac{2}{5}x+1$  $y=\frac{1}{4}x+1$  $x=-4$  $y=-3x-1$

## Exercises for review

( [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 of a line is a measure of the steepness of the line.

( [link] ) Find the slope of the line that passes through the points $\left(4,0\right)$ and $\left(-2,-6\right).$

$m=1$

( [link] ) Graph the equation $3y=2x+3.$ Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
Difference between extinct and extici spicies
Researchers demonstrated that the hippocampus functions in memory processing by creating lesions in the hippocampi of rats, which resulted in ________.
The formulation of new memories is sometimes called ________, and the process of bringing up old memories is called ________.
Please keep in mind that it's not allowed to promote any social groups (whatsapp, facebook, etc...), exchange phone numbers, email addresses or ask for personal information on QuizOver's platform. By By By By By    By Rhodes   By