Do you know how to find the equation of lines?

[crouch88]

Homework Statement

http://img1.uploadscreenshot.com/images/orig/5/12317385174-orig.jpg"

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?

 

[Mark44]

Homework Statement

http://img1.uploadscreenshot.com/images/orig/5/12317385174-orig.jpg"

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?

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

 

[The Chaz]

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

 

[The Chaz]

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

Mark, you genius!

 

[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.

 

[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.

 

[crouch88]

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

 

[Mark44]

Mark, you genius!

Well, we both spotted it at the same time.