# Do you know how to find the equation of lines?

1. May 4, 2010

### crouch88

1. The problem statement, all variables and given/known data

3. The attempt at a solution

Points: (0, 0); ( [x/2], [x√3]/2 ); (x, 0)

For points (0, 0) and ( [x/2], [x√3]/2 )

Gradient = m = {[x√3]/2 - 0}/ {[x/2] - 0} = √3

m = (y-0)/(x-0)
√3 = y/x
y = x√3 <--- eqn. 1

For points ( [x/2], [x√3]/2 ) and (x, 0)

Gradient = m = {0 - (x√3)/2}/ {x-(x/2)} = -√3

m = [y - {(x√3)/2)] / [ x- (x/2) ]
-√3 = [y - {(x√3)/2)] / [ x- (x/2) ]
y = [(-x√3)/2] + [(x√3)/2]
y = 0 <--- eqn. 2

For points (0, 0) and (x, 0)

Gradient = m = (0 - 0)/ (x-0) = 0

m = (y - 0) / (x-0)
0 = 0​

If you went through all of the above you'll notice that I got two y=0 equations...Where did I mess up?

Last edited by a moderator: Apr 25, 2017
2. May 4, 2010

### Staff: Mentor

In the second pair of points. You found the slope to be -sqrt(3), but substituted +sqrt(3) when you found the equation.

Last edited by a moderator: Apr 25, 2017
3. May 4, 2010

### The Chaz

Plugged in the wrong slope in the second equation. It should be negative.

4. May 4, 2010

### The Chaz

Mark, you genius!

5. May 4, 2010

### crouch88

Sorry guys! That was a typo! I basically just copied the whole response off my copy. I did take -√3 as gradient for the second pair of equations.

6. May 4, 2010

### The Chaz

Oh, now I see. The problem is the redefinition of "x". As used in the coordinates, it is a constant. When used in "y=mx+b", it is a variable.

Let's use the points (0,0) (a,0) and... whatever that third one was, but with "a" instead of "x". You'll get the right answer.

7. May 4, 2010

### crouch88

Okay, now the first gradient becomes (-a√3) / (2x-a)... thanks Chaz! I hope this works.

Last edited: May 4, 2010
8. May 4, 2010

### Staff: Mentor

Well, we both spotted it at the same time.