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Find the last coordinate of an isosceles triangle with given base and height coordinates

I have no idea about trigonometry even though he is in school when I think it should be pretty easy but trawling through tons of triggers on the internet makes my head ache :) So maybe someone can help me ...

The title explains what exactly I want to do, I have a line: x1, y1 and x2, y2 and want the function to find x3, y3 to complete an isosceles triangle given the height.

To be clear, the line x1, y2 -> x2, y2 will be the base and it will not be aligned to any axis (it will be at a random angle.)

Does anyone have a simple function for this?

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construct a normal to the vector (x1, y1) → (x2, y2). place it in the middle of ((x1 + x2) / 2, (y1 + y2) / 2) and go out to distance h.

the normal will look like (- (y2-y1), x2-x1). make it a unit vector ( http://en.wikipedia.org/wiki/Unit_vector ).



add h times this unit vector to the midpoint.

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The third point is at the perpendicular bisector of your base and is at a distance altitude

from the line.

  • Calculate the midpoint of the datum by averaging the x and y coordinates.
  • Calculate the slope of your height: -dx / dy (perpendicular to dy / dx). You now have your line (point and slope).
    • y - my = -dx / dy * (x - mx)
  • Replace the variables in the distance formula: d = sqrt (dx ^ 2 + dy ^ 2)
    • d = sqrt ((x - mx) ^ 2 + (y - my) ^ 2)
    • d = sqrt ((x - mx) ^ 2 + (-dx / dy * (x - mx)) ^ 2)
    • d ^ 2 = (x - mx) ^ 2 + (-dx / dy * (x - mx)) ^ 2
    • d ^ 2 - (x - mx) ^ 2 = (-dx / dy * (x - mx)) ^ 2
    • ± sqrt (d ^ 2 - (x - mx) ^ 2) = -dx / dy * (x - mx)
    • ± sqrt (d ^ 2 - (x - mx) ^ 2) * dy / dx = x - mx
    • ± sqrt (d ^ 2 - (x - mx) ^ 2) * dy / dx + mx = x
    • x = ± sqrt (d ^ 2 - (x - mx) ^ 2) * dy / dx + mx
  • Calculate the other variable (y here) using your linear equation (from # 2).
  • You now have two points; choose what you want ...

In pseudocode:



dx = x1 - x2
midpoint = ((x1 + x2) / 2, (y1 + y2) / 2)
slope = -dx / (y1 - y2)
x = sqrt(altitude*altitude - dx*dx) / slope + midpoint.x
y = slope * (x - midpoint.x) + midpoint.y

This is probably not the most optimal method. Not sure if it works. XD

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Al I remember that an isosceles triangle will have sides of equal length and equal angles at the base. If you have the height, then you have the end coordinate, because that will be the intersection point, right?

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