How to find the equation of the line....?

Question

the points were (-1/2,-1/2) (11,5) so i got the slope of 5.5/11.5 but they have to stay in fraction form.....someone help im so confused

Answer 1

11 - (-1/2) = 22/2 - (-1)/2 = 23/2
5 - (-1/2) = 10/2 - (-1)/2 = 11/2

So the slope is (23/2) / (11/2)
= (23/2) * (2/11)
= 23/11.

So the line has equation
y - (-1/2) = (23/11) (x - (-1/2))
<=> y + 1/2 = (23/11)x + (23/11)(1/2)
<=> y = (23/11) x + 23/22 - 1/2
<=> y = (23/11) x + 6/11

(note that 23/22 - 1/2 = 23/22 - 11/22 = 12/22 = 6/11.)

Answer 2

for slop:
(y2 - y1) / (x2 - x1) = m
(5 + 1/2) / (11 + 1/2) = m
5.5/11.5 = m or m = 55 / 115 OR m = 11 / 23

to find constant of line:
y = mx + c
put values of m and y n x for point (11, 5)
5 = 11/23 * 11 + c
5 - 121/23 = c
c = -0.261
c = -261/1000 = -6/23

to find equation of the line:
putting values of m and c.
y = mx + c
y = (11/23) x - (6/23)
23y = 11x - 6

Answer 3

slope of the given line, m
(y2 - y1)/(x2 - x1) = 5 -(-1/2)/(11- (-1/2) =
= (5 + 1/2)/(11 + 1/2) =
= (11/2)/(23/2) = 11/23
The point - slope form of a line , y - y1 = m(x - x1)
y - y1 = 11/23(x - x1)
This passes through points (-1/2, -1/2) and (11, 5)
substituting coordinates of any one point
(y - 5) = 11/23(x - 11) =
23(y - 5) = 11(x - 11)
= 23y - 115 = 11x - 121
= 11x - 23y - 121 + 115 = 0
= 11x - 23y - 6 = 0

Answer 4

The equation of a line is: y = mx + b
with m being the slope and you can fill in the x and y from the points to find b.