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Finding Equation of Lines
Equations
y  mx  b
Slope-intercept form of a line
Point-slope form on a line
y  y1  m( x  x1 )
y  y1
how to find slope given two points ( x1 , y1 )( x2 , y2 )
slope  m  2
x2  x1
Slope
Slope is the slant of the line. The greater the number the greater the slant. A
vertical line has no slope. A horizontal line has slope of 0. The slope is often referred
has rise over run.
rise movement in the y direction

run movement in the x direction
2
So if you are given a slope of 2. Where 2 is equal to , to move to the next point you
1
would move 1 in the x direction (right) and 2 in the y direction (up). If you are given a
2
slope of -2 which is equal to
, you will move 1 in the x direction (right) and 2 in the
1
negative y direction (down).
To find the slope given two points you need to use the slope equation,
y  y1
, and fill in the variables and do the math.
slope  m  2
x2  x1
Example 1: Find the slope of a line that passes through (1,2) and (6,8). First label the
number and determine what values the variables are equal to.
(1,2) = ( x1 , y1 ) and (6,8) = ( x2 , y2 )
so
x1  1
y1  2
x2  6
y2  8
82 6 3
 
62 4 2
so the slope is equal
up 3.
3
. This means to move to the next point move right 2 and
2
Example 2: Find the slope of the line that passes through (-2,4) and (8, 8).
First label the number and determine what values the variables are equal to.
(-2,4) = ( x1 , y1 ) and (8,8) = ( x2 , y2 )
so
x1  2
y1  4
x2  8
y2  8
84
84 4 2



8  (2) 8  2 10 5
2
. This means to move to the next point move right 5 and up 2.
5
Notice that in the denominator if the second term is negative the minus and the negative
become one plus sign. So it ends up being 8 + 2.
so the slope is equal
Finding Equations of Lines
The end result the equation should be in the slope-intercept form, y=mx+b. The
information given in the problem is a clue to what line equation you should start with. If
given the y intercept and slope, then use the slope-intercept formula (y=mx+b). If you
are given a point on the line and the slope, then start out with the point-slope formula
y  y1  m( x  x1 ) , plug in the values you know and then transform it in to the y=mx+b
form. The point given will go into x1 and y1 . If you are given two points you need to
y  y1
first find the slope using: slope  m  2
. So now you have a point and a slope so
x2  x1
use the point-slope equation, plug in the numbers, and transform it into the y=mx+b form.
Here are a few examples.
Example 1: Find the equation of a line that has a y intercept at (0,5), and slope -2.
We are given the y intercept and the slope so we can use the slope-intercept
formula (y=mx+b)
m = -1
b =5
so y = -1x + 5
y = -x + 5
no need to have the 1 there!
Example 2: Find the equation of a line that goes through (-5,-1), and slope -3.
We are given a point and the slope, so we can use the point-slope formula.
y  y1  m( x  x1 )
y1  1
m  3
x1  5
y  (1)  3( x  (5))
y  1  3( x  5)
y  1  3x  15
y  3x  16
To check you answer you can plug the point in the equation and see if it works
out.
 1  3(5)  16
 1  15  16
 1  1
Example 3: Find the equation of a line that passes through (-5,6) and (-3,-4).
We are given two points, so first we need to find the slope.
y  y1
slope  m  2
x2  x1
x1  5
y1  6
x2  3
y 2  4
46
 4  6  10


 5
 3  (5)  3  5
2
Now we have the slope and a point, so you can use the point-slope formula.
y  y1  m( x  x1 )
y1  6
m  5
x1  5
y  6  5( x  (5))
y  6  5( x  5)
y  6  5 x  25
y  5 x  19
To check you answer you can plug the point in the equation and see if it works
out.
 4  5(3)  19
 4  15  19
 4  4