# Straight line graphs *help* please. (Further Maths)

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#1
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)
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1 month ago
#2
I don’t do further maths but I can help.

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1 month ago
#3
Then find what the equation will be
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1 month ago
#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)
I never got taught this way, I always use y= mx + c
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1 month ago
#5
So (y2-y1) / (x2 - x1) for the gradient then you can do the rest
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#6
(Original post by CaptainDuckie)
So (y2-y1) / (x2 - x1) for the gradient then you can do the rest
Right!! ty
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1 month ago
#7
(Original post by Alice.gates)
I never got taught this way, I always use y= mx + c
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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1 month ago
#8
(Original post by kittymityyy)
Right!! ty
Tell me what you get, (I can’t show you worked solutions or else I’ll get banned) sorry
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#9
(Original post by kittymityyy)
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
so the gradient is -11/7, right?
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#10
(Original post by CaptainDuckie)
Tell me what you get, (I can’t show you worked solutions or else I’ll get banned) sorry
thank you for the help and i got -11/7? is that right? for the gradient
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1 month ago
#11
(Original post by kittymityyy)
so the gradient is -11/7, right?
Not -11/7, do it again
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#12
(Original post by CaptainDuckie)
Not -11/7, do it again
isn't it (-6-5)/(10-3)
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1 month ago
#13
(Original post by kittymityyy)
isn't it (-6-5)/(10-3)
No

y1 is (-5)
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1 month ago
#14
(Original post by kittymityyy)
isn't it (-6-5)/(10-3)
Made a mistake with the negatives. it's - 6 - (-5) which is - 6 + 5, = -1
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#15
(Original post by CaptainDuckie)
No

y1 is (-5)
so -6-(-5)/(10-3) = -1/7?
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1 month ago
#16
(Original post by kittymityyy)
so -6-(-5)/(10-3) = -1/7?
Yes

So substitue into y- y1 = m( x - x1)
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#17
(Original post by CaptainDuckie)
Yes

So substitue into y- y1 = m( x - x1)
oh ok!
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1 month ago
#18
(Original post by kittymityyy)
oh ok!
No problem, you can probably leave it in that form in the exam unless told otherwise...

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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#19
(Original post by kittymityyy)
oh ok!
is it y = -1/7x - 32/7
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1 month ago
#20
And don’t get tripped up by the negatives - - makes a plus, rookie mistakes that the examiners love
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