Showing posts with label Analytic Geometry. Show all posts
Showing posts with label Analytic Geometry. Show all posts

Friday, May 17, 2013

A line segment has a midpoint of (–3, 2) and an endpoint of (–6, 4). What is the other endpoint of the line segment?

The midpoint coordinates are: 

%0D%0Ax%5Bm%5D=%28x%5B1%5D%2Bx%5B2%5D%29%2F2%0D%0A 


%0D%0Ay%5Bm%5D=%28y%5B1%5D%2By%5B2%5D%29%2F2%0D%0A 

Substituting... 




%0D%0A-3=%28-6%2Bx%5B2%5D%29%2F2%0D%0A 


%0D%0A2=%284%2By%5B2%5D%29%2F2%0D%0A 

After simplifying we get (0,0) 

:)

find the distance from (6,-4) to the line defined by y=2x-6


Typically we ask for the shortest distance because the distance from the line varies. So here we go... 

There is a line perpendicular to our line that passes through (6,-4). Since it is perpendicular it has the slope -1/2 

%0D%0Agraph%28300%2C300%2C-10%2C10%2C-10%2C10%2C2%2Ax-6%2C2%2Ax-16%2C-.5%2Ax-1%29%0D%0A 


%0D%0Ay-%28-4%29=%28-1%2F2%29%28x-6%29%0D%0A 

After some simplifying we can get 

%0D%0Ay=%28-1%2F2%29x-1%0D%0A 



Notice that this new line must intersect our original at a point exactly parallel to (6,-4) (why?) 

Therefore if we find the point of intersection we can use that point to find our distance 


y=2x-6=%28-1%2F2%29x-1 
and this leads us to the x-coordinate which is x=2 and so y=2(2)-6=-2 

Next we find the distance between (6,-4) and (2,-2) 


d=sqrt%28%286-2%29%5E2%2B%28-4-%28-2%29%29%5E2%29 
and this gives us 

d=2sqrt%285%29%0D%0A 



Actually in precalculus there is a formula: 
 



:)