Question

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

Answer 1

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

Answer 2

Btw, you plug in (8, -8) for their corresponding variable. 8 is the x-coordinate and -8 is the y-coordinate.

Answer 3

ok thank you that makes it much easier to understand i was doing it the hard way.

Answer 4

You're welcome

Answer 5

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)

Answer 6

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

Answer 7

ok so I added wrong thank you again

Answer 8

You're welcome. Just remember to find the common denominator

Answer 9

my problem was im so tired i was adding instead of subtracting or vice versa lol

Answer 10

lol