Emra
Find The equestion in slope intercept form for the line contaning the two points. (1,1 ) and ( 5,4 )
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Mrs.Math101
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|dw:1326768472676:dw|
Mrs.Math101
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you know how to use the equation?
Callum29
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that gives the gradient of the line
Emra
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Do You Start With Y1 Or Y2
Callum29
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call x_1= 1, y_1 =1, x_2=5, y_2 = 4
Mrs.Math101
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|dw:1326768658229:dw|
Callum29
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the gradient, m is 3/4
Callum29
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now you want the equation in the form: y=mx + c
Mrs.Math101
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y=mx+B
MohamedDada
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the answer is y = 3/5x -1
MohamedDada
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i did all my work the answer doesnt show up
Mrs.Math101
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do you get emra?
MohamedDada
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no i didnt get emra
she's hard to catch o.O
Callum29
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you know m = 3/4. Take the first coordinate pair (1,1). Set y=1 and x=1. Then:
\[1=3/4 + c\]
\[c=1/4\]
You can check this value by using the other coordinate pair (5,4)
\[4=15/4 + c\]
\[c=(16-15)/4 = 1/4\]
Therefore the equation of your line is
\[y=\frac{3}{4}x + \frac{1}{4}\]
Callum29
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it's not y=3/5x - 1 though...
MohamedDada
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thats how we learned it
MohamedDada
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oh my bad
MohamedDada
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i see my mistake
MohamedDada
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yeah its y = 3/4 x + 1/4
Callum29
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you forgot to minus the 1 from the 5 didn't you?
MohamedDada
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no i wrote the slope wrong -.-
Emra
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imm confused alot know
Emra
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5-1 is 4 right
Callum29
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yeah ^^ i remember when I was doing all that geometry was fun, I went to classes drunk for the lols
MohamedDada
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lol
Callum29
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5-1 is 4, bravo, bravo
Emra
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heheheheh
Emra
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How would you do this one (8,6 ( and (-3,6)
Callum29
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y_2 - y_1 = 6 - 6 = 0
x_2 - x_1 = -3 - 8 = -11
Emra
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0
--
11
Emra
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then what do i do
Callum29
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it's zero yes, what does that mean about the line?
Callum29
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hint: m=0
Emra
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?/// thats what i dont get ohh m=0 know we have to see what b=
Emra
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??
Callum29
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go back to the general equation of a line:
\[y=mx+b\]
in this case, m=0 so it' just
\[y=b\]
This tells you that the line is a straight line forever, parallel to the x axis and intersecting the y axis at 6.
Now look at your two pairs of coordinates. y is 6 in both (it better be)!
So the equation of your line is......?
Emra
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6 ?
Emra
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6
--
6 = 0 ??
Callum29
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yes \[y=6\] is the equation of your line, it's a very boring graph if you ask me
Emra
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so what the equestions
y=0x
Callum29
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yes 0x=0 for any x
Callum29
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so it's not in the equation
Emra
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so whats the equestion then is it y=ox
Callum29
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remember x is a variable, and is allowed to take all values on the real line, but it obeys the law 0x=0
Emra
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kk is that the equestion
Emra
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k how would you do this one ( 100,-300 ) and (101,-299 )
Callum29
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do it how you think...
Emra
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-1
____
-1 = 1 ?
Emra
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H-E-L-P
Callum29
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y_2-y_1 = -299 - (-300) = -299 + 300 = 1
x_2 - x_1 = 101 - 100 = 1
so it is 1/1 = 1. Your way worked too which is a good thing!
Callum29
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so yes, you are right
Emra
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so what would the equestion bee y=-1x+ ??
Callum29
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use y=x+b and one of your coordinate pairs to solve the equation for b. (Gradient is 1 not -1 by the way).