dungtd wrote:
How many triangles with positive area can be drawn on the coordinate plane such that the vertices have integer coordinates (x,y) satisfying 1≤x≤3 and 1≤y≤3?
(A) 72
(B) 76
(C) 78
(D) 80
(E) 84
Thanks for your kind help!
It would be better if you draw it while reading this explanation. With the restriction given (1≤x≤3 and 1≤y≤3) we get 9 points, from which we can form the triangle: (1,1), (1,2), (1,3), (2,1)...
From this 9 points any three (9C3) will form the triangle BUT THE SETS of three points which are collinear.
We'll have 8 sets of collinear points of three:
3 horizontal {(1,1),(2,1),(3,1)}; {(1,2)(2,2)(3,2)}; ...
3 vertical
2 diagonal {(1,1)(2,2)(3,3)}; {(1,3)(2,2)(3,1)};
So the final answer would be; 9C3-8=84-8=76.
Answer: B.
Similar problems with different solutions:
arithmetic-og-question-88380.html700-question-94644.htmltough-problem-88958.htmlHope it helps.
I got stuck when trying to understand the set of three points which are collinear to make a triangle.
By the way, I like your avatar. It reminds me about my ancestors in my country.