# 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] .

## 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, $\text{\hspace{0.17em}}x=1.$

f(x)=x/x+2 given g(x)=1+2x/1-x show that gf(x)=1+2x/3
proof
AUSTINE
sebd me some questions about anything ill solve for yall
how to solve x²=2x+8 factorization?
x=2x+8 x-2x=2x+8-2x x-2x=8 -x=8 -x/-1=8/-1 x=-8 prove: if x=-8 -8=2(-8)+8 -8=-16+8 -8=-8 (PROVEN)
Manifoldee
x=2x+8
Manifoldee
×=2x-8 minus both sides by 2x
Manifoldee
so, x-2x=2x+8-2x
Manifoldee
then cancel out 2x and -2x, cuz 2x-2x is obviously zero
Manifoldee
so it would be like this: x-2x=8
Manifoldee
then we all know that beside the variable is a number (1): (1)x-2x=8
Manifoldee
so we will going to minus that 1-2=-1
Manifoldee
so it would be -x=8
Manifoldee
so next step is to cancel out negative number beside x so we get positive x
Manifoldee
so by doing it you need to divide both side by -1 so it would be like this: (-1x/-1)=(8/-1)
Manifoldee
so -1/-1=1
Manifoldee
so x=-8
Manifoldee
Manifoldee
so we should prove it
Manifoldee
x=2x+8 x-2x=8 -x=8 x=-8 by mantu from India
mantu
lol i just saw its x²
Manifoldee
x²=2x-8 x²-2x=8 -x²=8 x²=-8 square root(x²)=square root(-8) x=sq. root(-8)
Manifoldee
I mean x²=2x+8 by factorization method
Kristof
I think x=-2 or x=4
Kristof
x= 2x+8 ×=8-2x - 2x + x = 8 - x = 8 both sides divided - 1 -×/-1 = 8/-1 × = - 8 //// from somalia
Mohamed
hii
Amit
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Dorbor
well
Biswajit
can u tell me concepts
Gaurav
Find the possible value of 8.5 using moivre's theorem
which of these functions is not uniformly cintinuous on (0, 1)? sinx
which of these functions is not uniformly continuous on 0,1
solve this equation by completing the square 3x-4x-7=0
X=7
Muustapha
=7
mantu
x=7
mantu
3x-4x-7=0 -x=7 x=-7
Kr
x=-7
mantu
9x-16x-49=0 -7x=49 -x=7 x=7
mantu
what's the formula
Modress
-x=7
Modress
new member
siame
what is trigonometry
deals with circles, angles, and triangles. Usually in the form of Soh cah toa or sine, cosine, and tangent
Thomas
solve for me this equational y=2-x
what are you solving for
Alex
solve x
Rubben
you would move everything to the other side leaving x by itself. subtract 2 and divide -1.
Nikki
then I got x=-2
Rubben
it will b -y+2=x
Alex
goodness. I'm sorry. I will let Alex take the wheel.
Nikki
ouky thanks braa
Rubben
I think he drive me safe
Rubben
how to get 8 trigonometric function of tanA=0.5, given SinA=5/13? Can you help me?m
More example of algebra and trigo
What is Indices
If one side only of a triangle is given is it possible to solve for the unkown two sides?
cool
Rubben
kya
Khushnama
please I need help in maths
Okey tell me, what's your problem is?
Navin