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Mar 15th, 2006, 01:40 PM
#1
Thread Starter
New Member
in the form of y = m x + c.
Hi All!
I am trying to figure out how to solve problems like the one below. Any guidence would be appreciated.
Work out the equation of the straight line through (3, -7) and (6, -16) in the form y = m x + c.
m = (-7) - (-16) = 9 = -3
__________ ___
(3) - (6) -3
m = -3
y = mx + c
-7 = (-3)(3) + c
-7 = -9 + c
c = -9 - (-7)
c = -2
y = (-3)x + -2
That is what I have done so far, am I doing it correctly ?
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Mar 15th, 2006, 01:54 PM
#2
Addicted Member
Re: in the form of y = m x + c.
Very close except c should be 2.
-7 = (-3)(3) + c
c = -7 - (-9)
c = 2
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Mar 15th, 2006, 02:23 PM
#3
Thread Starter
New Member
Re: in the form of y = m x + c.
ah silly mistake! thank you for your input. So the answer is correct then ? Its for homework you see.
Thanking you
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Mar 21st, 2006, 10:36 AM
#4
Junior Member
Re: in the form of y = m x + c.
i remember the two point form of line equation. if a line passes thru (x1,y1),(x2,y2) the eq will be
(y-y1)/(y2-y1)=(x-x1)/(x2-x1)
hen u can convert it to y=mx+c form
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