Cisz Posted November 16, 2008 MIN_ALPHA = 75 MAX_ALPHA = 100 AREA = 512 Distance = X MaxDistance = AREA / 2 = 256 Point A (0, 75) Point B (256, 100) Point C (200, ???) A => d = 75 y = mx + 75 y - 75 = mx (y - 75)/x = m m = (y - 75)/x <= insert Point B m = (100 - 75)/256 m = 25/256 = 0,09765625 y = (25/256)*x + 75 <= insert Point C y = (25/256)*200 + 75 y = 94,53125 Agreed? Go to top Share this post Link to post
echinodermata Posted November 17, 2008 So let me reword (or rather, word) this: A triangle ABC in the coordinate plane has vertices A(0, 75) , B(256, 100), C (200, y). Angle A is between 75 and 100 degrees. The area of the triangle is 512. Find y. Comments: You attempted to find C by finding the intersection of x=200 and some line y=mx+b through point A. However, by saying that point B was on the line, you have found the intersection of line AB with the line x=200, which is not point C. Go to top Share this post Link to post
Cisz Posted November 17, 2008 AREA is not the area of a triangle, but a length, the diameter of a region. At least that's what Uber told me. All three points are part of a line, and it's about interpolating an alpha value with given extremes on both borders of the region in question. Go to top Share this post Link to post
echinodermata Posted November 18, 2008 Please do excuse me, then, for my interpolation of your words. Go to top Share this post Link to post