To prove that the point does not lie in the triangle plane, it is possible simple check of all possible situations especially as there is not a lot of them. It is only not necessary to forget that it is possible to come also to an event opposite, that is to a case when the point is internal for the set triangle.
Instruction
1. Before looking for the solution of an objective, the reader should most make the decision on accessory of the parties of a triangle. To consider their points external for a triangle or not. At this stage we consider that it is the area closed and therefore it includes the borders. For simplicity consider "a flat case", but do not forget also about spatial generalization. Therefore the standard equations for straight lines of the plane of a type of y=kx+b, it is not necessary to use, on extremely measure at the beginning of the decision.
2. Choose a way of a task for the parties of a triangle. Judging by problem statement, it has no basic value. Therefore consider that coordinates of its tops of A (xa, ya), B (xb are given, to yb), C (xc, yc) (see fig. 1.). Find the directing vectors of the parties of a triangle of AB= {xb-xa, yb-ya }, BC= {xc-xb, yc-yb }, AC= {xc-xa, yc-ya } and write down the initial equations of the straight lines supporting these parties. For AB – (x-xa) / (xb-xa)= (y-ya)/(yb-ya). For BC – (x-xb) / (xc-xb)= (y-yb)/(yc-ya). For AC – (x-xa) / (xc-xa)= (y-ya)/(yc-ya). According to the drawing draw horizontal and vertical lines which can be written down as x=xc, x = xa, x=xb, y=yc, y=ya, y=yb. It will allow to reduce number of calculations to a minimum. Further follow the offered algorithm. In the drawing the set M point (xo, yo) is placed in the most "adverse" place.
3. Following along an axis of 0th, check performance of inequality of xcxoxb. If it is not executed, then the point already lies outside a triangle as "not inside" is and is "outside". If inequality is executed, then further check justice of xc
4. Check inequality performance usuoua. If it is not fair, then the point does not lie in a triangle. Otherwise find ordinate of the straight line containing AB. u1=u (xo)= [(yb-ya) (xo-xa)] / (xb-xa) of +ya. Also arrive with straight line ordinate for BC. u2= at (ho)= [(ys-yb) (xo-xb)] / (xc-xb) of +yc. Make inequality of y2≤yo≤y1. Its performance allows to make the conclusion that the set point is in a triangle. If this inequality is false, then it lies out of its limits, in particular according to the drawing.