Vertices on lines (Posted on 2012-04-25)

	Submitted by Bractals
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Let the vertices be the points (c,-a), (0,0), and (d,b) and s the side length of the triangle. Then s2 = (c-0)2 + (-a-0)2 = c2 + a2 s2 = (d-0)2 + (b-0)2 = d2 + b2 s2 = (c-d)2 + (-a-b)2 = (c-d)2 + (a+b)2 = c2-2cd+d2+a2+2ab+b2 = (s2-a2)-2cd+(s2-a2)+a2+2ab+b2 2cd = s2+2ab 4c2d2 = s4+4abs2+4a2b2 4(s2-a2)(s2-b2) = s4+4abs2+4a2b2 3s4 = 4(a2+ab+b2)s2 s2 = 4(a2+ab+b2)/3 Area(Δ) = s2√3/4 = (a2+ab+b2)/√3 QED

	Subject	Author	Date
	answer	Dej Mar	2012-04-25 14:28:36