Asked by Bright ulubi
Find the equation of a straight line whuch passes through the points (2,3) and (-1,4)
Answers
Answered by
Reiny
slope = (4-3)/(-1-2) = -1/3
using that slope and the point (2,3)
y-3 = (-1/3)(x-2)
arrange into whatever version of line you learned.
Sub (-1,4) into your equation to make sure it works
using that slope and the point (2,3)
y-3 = (-1/3)(x-2)
arrange into whatever version of line you learned.
Sub (-1,4) into your equation to make sure it works
Answered by
collins
y=mx+c......
m=y2-y1/x2-x1
m=4-3/-1-2
m=-1/3
at any point i pick
p(2,3)
y=mx+c
3=(-1)2/3+c
-9=-2/3+c
-9+2=3c
-9+2/3=c
c=-7/3
now
y=mx+c.......crack out the equation
m=y2-y1/x2-x1
m=4-3/-1-2
m=-1/3
at any point i pick
p(2,3)
y=mx+c
3=(-1)2/3+c
-9=-2/3+c
-9+2=3c
-9+2/3=c
c=-7/3
now
y=mx+c.......crack out the equation
Answered by
Steve
not quite. You got m right, but
3 = (-1/3)(2)+c
3 = -2/3 + c
3 + 2/3 = c
11/3 = c
y = -1/3 x + 11/3
why did you change 9 to -9? It didn't change sides.
If you check Reiny's equation
y-3 = (-1/3)(x-2)
y-3 = -1/3 x + 2/3
y = -1/3 x + 11/3
You may want to work on logs and advanced algebra, but maybe you should work on the basic skills some more for now...
3 = (-1/3)(2)+c
3 = -2/3 + c
3 + 2/3 = c
11/3 = c
y = -1/3 x + 11/3
why did you change 9 to -9? It didn't change sides.
If you check Reiny's equation
y-3 = (-1/3)(x-2)
y-3 = -1/3 x + 2/3
y = -1/3 x + 11/3
You may want to work on logs and advanced algebra, but maybe you should work on the basic skills some more for now...
Answered by
collins
i just did i without solving ob paper it was a typo okay
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