find an equation of the line having the given slope and containing the given points m=2/3, (8,-8). i get this far and then always have trouble with the last part. y-y1=m(x-x1) y-(-8)=2/3(x-8) y-(-8)=2/3x-5(1/3) then for some reason or some how when I subtract the 8 from 5 (1/3) I don't do it right some how. can some one explain this to me
y = mx b where m is the slope and b is the y-intercept. y = 2/3x + b Then plug in (8, -8). -8 = (2/3)(8) + b -8 = 16/3 + b -8 - 16/3 = b -24/3 - 16/3 = b b = -40/3 Therefore, y = 2/3x - 40/3
Btw, you plug in (8, -8) for their corresponding variable. 8 is the x-coordinate and -8 is the y-coordinate.
ok thank you that makes it much easier to understand i was doing it the hard way.
I guess that I am really not getting this. my next problem is m=(-3/5), (1,-2). the answer i got is -2=(-3/5)1+b -2=(-3/5)+b -2+3/5=b -2 (3/5)=b is this correct so it would be y=-3/5x+(-13/5)
Now, once you get here -->-2+3/5=b, you're correct. -10/5 + 3/5 = b = -7/5. THerefore, you're answer is y = -3/5x - 7/5
ok so I added wrong thank you again
You're welcome. Just remember to find the common denominator
my problem was im so tired i was adding instead of subtracting or vice versa lol
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