I have 2 points P1
and P2
. I need to find the P3
, in order that
- all points to be on the same line;
P3
should be at the distanced
from theP2
(away fromP1
)
I started a complicated system apparently hardly to resolve...
P开发者_JS百科S.
Vectorial answers is cool, but I use C# and don't know how to add vectors over there.
P3 = P2 + d * ±(P2 - P1) / |P2 - P1|
EDIT:
Because shopping is easy:
mag = sqrt((P2x - P1x) ** 2 + (P2y - P1y) ** 2)
P3x = P2x + d * (P2x - P1x) / mag
P3y = P2y + d * (P2y - P1y) / mag
I have translated the code to Objective C
float distanceFromPx2toP3 = 1300.0;
float mag = sqrt(pow((px2.x - px1.x),2) + pow((px2.y - px1.y),2));
float P3x = px2.x + distanceFromPx2toP3 * (px2.x - px1.x) / mag;
float P3y = px2.y + distanceFromPx2toP3 * (px2.y - px1.y) / mag;
CGPoint P3 = CGPointMake(P3x, P3y);
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