[2010-01] finding the point to make an equilaterial trangle with a line segment, i was figuring out for fractical triangle khoh curve or whatever

gistfile1.txt

		"""
			The equation of the line segment near A is
			
			y = a.y + (x - a.x) * slopeA
			
			The equation of the line segment near B is
			
			y = b.y + (x - b.x) * slopeB
			
			Isolate x in the system;
			
			a.y + (x - a.x) * slopeA = b.y + (x - b.x) * slopeB
			(x - a.x) * slopeA - (x - b.x) * slopeB = b.y - a.y
			x * slopeA - a.x * slopeA - x * slopeB + b.x * slopeB = b.y - a.y
			x * slopeA - x * slopeB = b.y - a.y - b.x * slopeB + a.x * slopeA
			x (slopeA - slopeB) = b.y - a.y - b.x * slopeB + a.x * slopeA
			x = (b.y - a.y - b.x * slopeB + a.x * slopeA) / (slopeA - slopeB)
			
			There.
		"""
		
		x = (b.y - a.y - b.x * slopeB + a.x * slopeA) / (slopeA - slopeB)
		y = a.y + (x - a.x) * slopeA

gistfile2.sc

def equalaterize(a, b):
	"""Given two points return the third point
	needed to form an equalaterial triangle.
	
	Two possible points fit this criteria, returned is
	the one that would be above the two points if they
	were drawn on a vertical line with a to the left."""
	
	distance = a.distanceFrom(b)
	
	dX = b.x - a.x
	dY = b.y - a.y
	
	if dY:
		lineSlope = (b.x - a.x) / (b.y - a.y)
		lineAngle = math.asin(lineSlope)
	else:
		log.info("Points are vertically aligned, angle is 0°.")
		lineAngle = 0
	
	angleA = (lineAngle + 60) % 360
	if angleA: slopeA = math.sin(angleA)
	
	angleB = (lineAngle - 60) % 360
	if angleB: slopeB = math.sin(angleB)
	
	if angleA == 0:
		return Point(a.x, b.y + dX * slopeB)
	elif angleB == 0:
		return Point(b.x, a.y - dX * slopeA)
	else:
		"""
			The equation of the line segment near A is
			
			y = a.y + (x - a.x) * slopeA
			
			The equation of the line segment near B is
			
			y = b.y + (x - b.x) * slopeB
			
			Isolate x in the system;
			
			a.y + (x - a.x) * slopeA = b.y + (x - b.x) * slopeB
			(x - a.x) * slopeA - (x - b.x) * slopeB = b.y - a.y
			x * slopeA - a.x * slopeA - x * slopeB + b.x * slopeB = b.y - a.y
			x * slopeA - x * slopeB = b.y - a.y - b.x * slopeB + a.x * slopeA
			x (slopeA - slopeB) = b.y - a.y - b.x * slopeB + a.x * slopeA
			x = (b.y - a.y - b.x * slopeB + a.x * slopeA) / (slopeA - slopeB)
			
			There.
		"""
		
		x = (b.y - a.y - b.x * slopeB + a.x * slopeA) / (slopeA - slopeB)
		y = a.y + (x - a.x) * slopeA
		return Point(x, y)