How to find the equation of a line.
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Transcript of How to find the equation of a line.
How to find
the equation
of a line.
Equations of lines start with y =
Equations of lines start with y =
Next comes the slope (slopes are fractions)
Equations of lines start with y =
Next comes the slope (slopes are fractions)
Last is the y-intercept.
(the point where the line crosses the y-axis)
Name this line’s equation.
.A.B
.A.B
This line
is going uphill
so the slope is
positive.
.A.B
To get from point
A to point B, go
up 2 and right 2
so this line’s slope
is or 1.2
2
This line
crosses at (0,2)
so the y-intercept
is = to 2.
.A.B
.A.B
This line’s equation is
y = slope = 1 and y –int = 2
.A.B
y = 1slope = 1 and y –int = 2
.A.B
y = 1xslope = 1 and y –int = 2
.A.B
This line’s equation is
y = 1x +2slope = 1 and y –int = 2
Name this line’s equation.
.C
.D
.C
.D
This line
is going downhill
so the slope is
going to be
negative.
.C
.D
To get from point
C to point D, go
down 6 and right 3
so this line’s slope
is or -2.3
6
.C
.D
This line
crosses at (0,4)
so the y-intercept
is = to 4.
.C
.D
We have all the parts
to name this line.
.C
.D
y =
slope = -2 and y-int = 4
.C
.D
y = -2
slope = -2 and y-int = 4
.C
.D
y = -2x
slope = -2 and y-int = 4
.C
.D
The line’s equation is
slope = -2 and y-int = 4
y = -2x + 4
Name this line’s equation.
. J
. K
. J
. KThis line is going
uphill so the slope
will be positive.
. J
. KTo get from point J
to point K, go
up 5 and right 3
so this line’s slope
is .3
5
. J
. KThis line
crosses at (0,-4)
so the y-intercept
is = to -4.
. J
. KWe have all the parts
to name this line.
. J
. Ky =
Slope = and y-int = -4
3
5
3
5
. J
. Ky = x
Slope = and y-int = -4
3
5
3
5
. J
. Ky = x - 4
Slope = and y-int = -4
3
5
3
5
The line’s equation is
Name this lines equation.
. L . M
. L . MThis line is going
uphill so the slope
will be positive.
. L . MTo get from point L
to point M, go
up 1 and right 2
so this line’s slope
is .2
1
. L . MThis line
crosses at (0,2)
so the y-intercept
is = to 2.
. L . MWe have all the parts
to name this line.
. L . M y =Slope = and y-int = 2
2
1
. L . M y =Slope = and y-int = 2
2
12
1
. L . M y = xSlope = and y-int = 2
2
12
1
. L . M y = x + 2Slope = and y-int = 2
2
12
1
The line’s equation is
Name this line’s equation.
. Q . R
. Q . R
Slope =5
1
. Q . R
Slope =
y- int = -3
5
1
. Q . R
The line’s equation is
y = x - 35
1
Name this line’s equation.
. S. T
. S. T
Slope = 3
2
. S. T
Slope =
y-int = 23
2
. S. T
The line’s equation is
y = x +23
2
Name this line’s equation.
. U
. V
. U
. V
Slope = 1
6
. U
. V
Slope =
y-int = +5 1
6
. U
. V
The line’s equation is
y = -6 x + 5
. W
. X
Name this line’s equation.
. W
. X
slope = 5
7
. W
. X
slope =
y-int = 4 5
7
. W
. X
The line’s equation is
y = x + 45
7
. Y
. Z
Name this line’s equation.
. Y
. Z
Slope = 3
4
. Y
. Z
Slope =
y-int = 0 3
4
. Y
. Z
The line’s equation is
y = x + 03
4