Calculating a intersection of a triangle
This is going to be hard without a diagram......how can I add one?
I have two points "y1" (left) and "y2" (right)
Lets say their values are 1 and 10 respectively.
These are y coordinates...heights.
The two points are spaced n world units apart, lets say the variable "units" holds this and its value is 10.
x coordinate...
given another point "x", which is the distance from point "y1" that we wish to intersect. lets say its value is 4.
x coordinate...
given this informate i need to calculate y3, which is the height of the intersection of x.
which given all the above would in this case be 4.
The problem is I cannot for the life of me come up with a simple eqaution to calculate y3.
Currently im doing the following..
y3 = y1 + ((y2 - y1)/(units)*x)
which given the above would calculate y3 = 4.6
Any ideas....I have a diagram if it helps..
September 26, 2002 06:43 AM
As I understand you, you have two points, P1 = (x1, y1) and P2 = (x2, y2), and you want to determine a third point P3 = (x3, y3) lying on the straight line between P1 and P2. If this is how you mean, your formula is correct and y3 = y1 + x3*(y2 - y1)/(x2 - x1).
September 26, 2002 06:51 AM
Sorry, formula should be y3 = y1 + (x3 - x1)*(y2 - y1)/(x2 - x1).
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