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Straight line graphs *help* please. (Further Maths)
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How would you solve the equation of the line AB in this case, using the
y-y1=m(x-x1) formula?
A(3,-5) B(10,-6)
y-y1=m(x-x1) formula?
A(3,-5) B(10,-6)
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#4
(Original post by kittymityyy)
How would you solve the equation of the line AB in this case, using the
y-y1=m(x-x1) formula?
A(3,-5) B(10,-6)
How would you solve the equation of the line AB in this case, using the
y-y1=m(x-x1) formula?
A(3,-5) B(10,-6)
Basically find the gradient
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(Original post by CaptainDuckie)
So (y2-y1) / (x2 - x1) for the gradient then you can do the rest
So (y2-y1) / (x2 - x1) for the gradient then you can do the rest
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#7
(Original post by Alice.gates)
I never got taught this way, I always use y= mx + c
Basically find the gradient
I never got taught this way, I always use y= mx + c
Basically find the gradient
Once you’ve found the m, which is the gradient
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#8
(Original post by kittymityyy)
Right!! ty
Right!! ty
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(Original post by kittymityyy)
Right!! ty
Right!! ty
(Original post by CaptainDuckie)
You can rearrange y- y1 = m(x-x1) into the form of y=mx+ c
Once you’ve found the m, which is the gradient
You can rearrange y- y1 = m(x-x1) into the form of y=mx+ c
Once you’ve found the m, which is the gradient
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(Original post by CaptainDuckie)
Tell me what you get, (I can’t show you worked solutions or else I’ll get banned) sorry
Tell me what you get, (I can’t show you worked solutions or else I’ll get banned) sorry
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#11
(Original post by kittymityyy)
so the gradient is -11/7, right?
so the gradient is -11/7, right?
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(Original post by CaptainDuckie)
Not -11/7, do it again
Not -11/7, do it again
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#14
(Original post by kittymityyy)
isn't it (-6-5)/(10-3)
isn't it (-6-5)/(10-3)
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#16
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#18
(Original post by kittymityyy)
oh ok!
oh ok!
Remember the gradient equation as (y2-y1)/(x2 -x1)
Memorise it. These questions are naturally reoccurring, (I’m assuming you do AS level Maths too)
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#20
And don’t get tripped up by the negatives - - makes a plus, rookie mistakes that the examiners love
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