the points (-2,4), (0,2) and (3,a-1) are on the line. find a?

Answer 1

m = (2 - 4) / (0 + 2)
m = - 2 / 2
m = - 1

-1 = (a - 3) / (3 - 0)
- 1 = (a - 3) / 3
- 3 = a - 3
a = 0

Answer 2

First, find the gradient or slope of the line. We take point (-2 , 4) and point (0 , 2) as they have x and y values.

Gradient,
m = (2 - 4) / (0 - (-2))
m = -2 / 2
m = -1

Using the same concept, we substitute the value of gradient into the following equation,
m = [ (a - 1) - 2 ] / [ 3 - 0 ]
Since the points are on a line, they share a same gradient,
-1 = [ (a - 1) - 2 ] / 3
(a - 1) - 2 = -3
a - 1 = -1
a = 0

Answer 3

Gradient of (-2,4), (0,2)=(4-2)/(-2-0)
=-1
so gardient of (0,2) and (3,a-1) =-1
(a-1-2)/(3)=-1
a-3=-3
a=0

Answer 4

y=(4-2)/(-2-0)x + 2
y= -x +2 so
a-1 = -3 + 2
a = 0

Answer 5

a=0

1st find slope

4 - 2 -2
____ = __ = -1
-2 - 0 2

slope=-1

since you know slope

find y of 3rd point

y=mx+b
y=-1(3)+2
y=-3+2
y=-1

find a
y=a-1
-1=a-1
a=-1+1
a=0

Answer 6

find the line equation with the first two... and then place the third to find a

Answer 7

(4-2)=m(-2-0)
2=-2m
-2/2=m
m=-1

(a-1)-2=-1(3-0)
a-3=-3
a=3-3
a=0

Bye.

SM