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Geometry and discrete question

 
 
cadre
 
Reply Wed 15 Mar, 2006 10:53 pm
My class was given this question in our review sheet for an upcoming exam. This was really the only one I couldn't really solve. It seems simple enough, but i'm not sure where to start or what equations to use :s It seems that it would fall under the 'Thinking and Inquiry' section of exam because we haven't seem anything like it before. The question is as follows-

The line segment joining A(2,3) to B(9,2) is the hypotenuse of a right triangle. The third vertex C, lies on the line with these parametric equations: y=8-t x=2+2t
Determine the coordinates of C. Illustrate with a diagram.

I can solve this easily with by using a protractor and that you can easily solve for the vector equation of the line, being (x,y)=(2,8)+t[2,-1]. The answers are (6,6) and (8,5), but our teacher wants a mathematical solution. Since the angle ACB has to be 90 degrees do you have to use the cos rule with vector magnitudes? I'm really confused :s
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Type: Discussion • Score: 1 • Views: 804 • Replies: 5
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markr
 
  1  
Reply Thu 16 Mar, 2006 12:34 am
C must be such that the slope of AC is the opposite of the inverse of the slope of BC (requirement for them to be perpendicular).

So,
(x-2)/(y-3) = -(y-2)/(x-9)

Substitute 8-t for y and 2+2t for x and solve for t.
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engineer
 
  1  
Reply Thu 16 Mar, 2006 12:08 pm
You can also say that a^2 + b^2 = c^2. You know that c^2 = (9-2)^2 + (2-3)^2 = 50. That equals (x-2)^2 + (y-3)^2 + (x-9)^2 + (y-2)^2. Like above, substitute in for x and y and solve for t.
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cadre
 
  1  
Reply Tue 11 Apr, 2006 01:53 pm
Sorry I forgot to thank you guys Wink I have another question now too. :p

It is as follows.

Vectors 'a' and 'b' are drawin tail-to-tail. Vector 'c' is the reflection of 'a' in the line containing vector 'b'. Express 'c' as a linear combination of 'a' and 'b'.

Okay, I've tried showing it through the use of the projection of a vector formula but I keep getting stuck. Could any of you please offer some guidance or a suggestion on where to start?
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dimatt
 
  1  
Reply Wed 12 Apr, 2006 03:40 pm
^ I had this same problem in class almost a month ago. Sorry i really cant remember much from it now though. I can only recall using the orthogonal through the process, i think :s Sorry.
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cadre
 
  1  
Reply Fri 14 Apr, 2006 09:01 pm
The answer to my new question is apparently c=2((a.b/b.b)b)-a if anyone could help me or suggest me a place to start at. My teacher said something about orthogonal and parallel vectors but I have no idea how to work those in Sad
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