# line segment

• Aug 6th 2010, 05:41 AM
sfspitfire23
line segment
Given two end points of a line segment as (5, t) and (-5, -t). If the length of the line segment is 2sqrt(34), find the slope of the line?
• Aug 6th 2010, 05:44 AM
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Quote:

Originally Posted by sfspitfire23
Given two end points of a line segment as (5, t) and (-5, -t). If the length of the line segment is 2sqrt(34), find the slope of the line?

The point goes through the origin, and you can simplify by saying the length of the segment from (0,0) to (5,t) is sqrt(34). Apply Pythagorean theorem to find t, then slope will be t/5.

Edit: There is not enough information. We do not know if t is positive or negative. So we get two possible answers, and can't know which one is true. (I guess just use the $\displaystyle \pm$ sign.)
• Aug 6th 2010, 05:47 AM
sfspitfire23
can I also write the equation of the line m=(y-y1)/(x-x1) using a point and the slope and then plug in the other point for values x and y to solve for t?
• Aug 6th 2010, 05:52 AM
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Quote:

Originally Posted by sfspitfire23
can I also write the equation of the line m=(y-y1)/(x-x1) using a point and the slope and then plug in the other point for values x and y to solve for t?

And where are you using the information of the length of the line segment?
• Aug 6th 2010, 05:53 AM
sfspitfire23
i wasn't ;-)
• Aug 6th 2010, 05:54 AM
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Quote:

Originally Posted by sfspitfire23
i wasn't ;-)

You can't solve the problem without using that information..

Also please note the edit in my first reply of the thread, regarding plus/minus.
• Aug 6th 2010, 06:08 AM
sfspitfire23
i get plus/minus sqrt(111)/5
• Aug 6th 2010, 06:11 AM
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Quote:

Originally Posted by sfspitfire23
i get plus/minus sqrt(111)/5

5^2 + t^2 = (sqrt(34))^2

25 + t^2 = 34

t^2 = 9

$\displaystyle t = \pm 3$

slope $\displaystyle m = \pm \frac{3}{5}$