1. ## equation of line

Can someone help me with this problem:
Find the equation of the line which passes through the points (1
,3)
and (
1, 1). Sketch the graph of the line.

Please go step by step. Thank You.

2. Originally Posted by Jea9
Can someone help me with this problem:
Find the equation of the line which passes through the points (1
,3)
and (
1, 1). Sketch the graph of the line.

Please go step by step. Thank You.

the equation of a line is of the form y = mx + b, where m is the slope (or gradient) and b is the y-intercept.

if m is not given, we can find it if two points are given in the following way.

let the points given be (x1,y1) and (x2,y2), then
m = (y2 - y1)/(x2 - x1)

after finding the value for m, we can use it and any one set of points to find b, by either plugging in all the values into y = mx + b (at this time, b will be the only thing unknown) or plugging the values into the point-slope form and solving for y.

point-slope form: y - y1 = m(x - x1)

so now let's do your problem.

the points given are (1,-3) and (-1,1)

=> m = (y2 - y1)/(x2 - x1) = (1 - (-3))/(-1 - 1) = 4/-2 = -2

using (x1,y1) = (-1,1), we have by the point-slope form:

y - y1 = m(x - x1)
=> y - 1 = -2(x - (-1)) = -2x - 2
=> y = -2x - 1

3. using (x1,y1) = (-1,1), we have by the point-slope form:

y - y1 = m(x - x1)
=> y - 1 = -2(x - (-1)) = -2x - 2
=> y = -2x - 1[/quote]

Wait a minute... i didnt understand this part fully... if its y-y1=m(x-x1) then wouldnt the minuses become pluses? and on this part "y - 1 = -2(x - (-1)) = -2x - 2" how did you get the -2 on the end?

RE:

y = -2x - 1

Sketch