What are the values of the trigonometric ratios for this triangle?
2 answers:
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Answer:
(-3, 5), (-1, -1), (5, -3)
Step-by-step explanation:
Each pair of vertices can be one of the diagonals. Then the missing point will be found at the coordinates that are the sum of those, less the coordinates of the third point.
Given points are ...
A(-2, 2), B(1, 1), C(2, -2)
For AB a diagonal, D1 is ...
A+B-C = (-2+1-2, 2+1-(-2)) = (-3, 5)
For AC a diagonal, D2 is ...
A+C-B = (-2+2-1, 2-2-1) = (-1, -1)
For BC a diagonal, D3 is ...
B+C-A = (1+2-(-2), 1-2-2) = (5, -3)
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For a lot of parallelogram problems I find it easiest to work with the fact that the diagonals bisect each other. This means they both have the same midpoint, so for quadrilateral ABCD, we have (A+C)/2 = (B+D)/2. Multiplying this by 2 gives the equation we used above, A+C = B+D, so D=A+C-B. Remember, in ABCD, AC and BD are the diagonals.
