mreditor16
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thanks for the valuable contribution.I remember this question, reminded me of 2010's integration spam
Basically what Ruse did was a shorter variation of another question that asked you to prove it was a parallelogram/rhombus before that (I think it was an Independent Paper Question can't remember).
sounds quicker than what the others suggested - so i'll try that...Basically what Ruse did was a shorter variation of another question that asked you to prove it was a parallelogram/rhombus before that (I think it was an Independent Paper Question can't remember).
But essentially if i remember quickly, the fastest way to prove this was to prove rhombus (diagonals perpendicular to each other) followed by proving 1 pair of adjacent sides was perpendicular too, thus making a rhombus with one right angle a square.
ceebs, besides I'd have to set it up myself.Why not use complex numbers?
The tangents are perpendicular to each other (this is apparent when using point gradient to find their equations). Squares have diagonals that are perpendicular and bisect (I think), so you only need to prove that the midpoint is the same?
that only proves it's a rhombus though I think.The tangents are perpendicular to each other (this is apparent when using point gradient to find their equations). Squares have diagonals that are perpendicular and bisect (I think), so you only need to prove that the midpoint is the same?
Also, this midpoint has to be the point P.
Oh hmm, yeah true! My bad, it would probably be easiest to find the adjacent sides equal thenthat only proves it's a rhombus though I think.
James Ruse 2011 4U TrialEditor what paper was that? Which JRAHS Paper?
Thanks.James Ruse 2011 4U Trial
hmmmI just figured an easier way to prove, for anyone interested:
-Prove AC = BD (ie diagonals equal length)
-Prove m(AC) * m(BD) = -1 (ie perpindicular.)
Thus, Square.
...Well, if anyone's still wondering.