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# Right triangle PQR is to be constructed in the xy-plane so

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Director
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Right triangle PQR is to be constructed in the xy-plane so [#permalink]

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26 May 2006, 20:06
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

248. Right triangle PQR is to be constructed in the xy-plane so that the right angle is at P and PR is parallel to the x-axis. The x and y coordinates of P, Q and R are to be integers that satisfy the inequalities -4 <= x <= 5 and
6 <=y <= 16. How many different triangles with these properties could be constructed?

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26 May 2006, 21:51
here is what I am thinking

this kind of triangle would have coordintes

p = x1,y1
q = x1, y2
r = x2 , y1

where x1 <> x2 and
y1 <> y2

range of x = 10 ,
range of Y = 11

x1, y1 could be selected in 10 * 11
after that x1 , y2 can be selected in 1 * 10
after that the third point x2 , y1 can be selected in 9 * 1 ways
so total number of ways are 10* 11* 10 * 9 = 9900

Could you please post OA .
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26 May 2006, 22:54
I think it will be 11 * 10 * 9 = 990

there are 10 possible values of x so it has to 10 * 9 since P and R can't occupy same point.
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27 May 2006, 03:44
Can you please explain how did you get 10C2*11*2?
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27 May 2006, 05:37
248. Right triangle PQR is to be constructed in the xy-plane so that the right angle is at P and PR is parallel to the x-axis. The x and y coordinates of P, Q and R are to be integers that satisfy the inequalities -4 <= x <= 5 and
6 <=y <= 16. How many different triangles with these properties could be constructed?

Possible values of x =10
possible values of y = 11

total points = 110

you need three distinct points to create a triangle

so 110C3 = 110x109x36 - (10x10C3 + 11x11C3)
Re: Trangles + Coordinates   [#permalink] 27 May 2006, 05:37
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