Find the last co-ordinate of isosceles triangle given coordinates of base and altitude


I have no clue about trigonometry, despite learning it in school way back when, and I figure this should be pretty straightforward, but trawling through tons of trig stuff on the web makes my head hurt 🙂 So maybe someone could help me…

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

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

Does anyone have a simple function for this??

Best Solution

construct a normal to the vector (x1,y1)->(x2,y2). place it at the midpoint ((x1+x2)/2,(y1+y2)/2) and go out a distance h.

the normal will look like (-(y2-y1),x2-x1). make this a unit vector (

add h times this unit vector to the midpoint.

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