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I have the points (5,1) and (-3, 4) I have to put them in Slope Intercept, Point-Slope, and Standard form. I got Point-Slope. ( y - 1 = 3/8(x - 5) but I can't figure out the other two for the life of me! Anyone?

Linear Algebra
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Point slope form is: \[y-y_{1}=m(x-x_{1})\] Standard form is: \[Ax+By=C\] Slope intercept form is: \[y=mx+b\]
I know, I tried to work them out but the 3/8 is throwing me off. I've been working on this for over two hours now. :/ I used Khanacademy but his problems all came out even, which this won't and I'm scared to veer too far away from what I know that I'll mess it up.
i need help can u help me

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Other answers:

Depends on what you need help with?
Shouldn't your slope be negative?
I get y-1=-3/8(x-5)
It might be wrong? That was just the slope I came up with, both ways.
To find the slope intercept form multiply -3/8 through (x-5) -3/8(x) is easy to figure out what -3/8(5) is, put the 5 in a fraction form then multiply straight across. \[(\frac{ -3 }{ 8 })(\frac{ 5 }{ 1 })=?\]
You should get -3/8x-(-15/8) or -3/8x+15/8 now add the one from the left side to the right side 1+15/8 find a common denominator first then add 8/8+15/8= 23/8 y=-3/8x +23/8 slope intercept form to get standard form just add 3/8x to the left side 3/8x +y = 23/8
Thank you soo much!
But wait, if you add 3/8x to one side don't you have to do it to the other as well?
It seems like you can just leave it 3/8x + y = 23/8
adding 3/8x to the left will cancel it out on the right
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