I have 2 points P1
and P2
. I need to find the P3
, in order that
P3
should be at the distance d
from the P2
(away from P1
)I started a complicated system apparently hardly to resolve...
PS.
Vectorial answers is cool, but I use C# and don't know how to add vectors over there.
To find points on the line y = mx + b, choose x and solve the equation for y, or. choose y and solve for x.
Three coordinate axes are given, each perpendicular to the other two at the origin, the point at which they cross. They are usually labeled x, y, and z.
A simple way is to: find the vectors connecting the point to each of the triangle's three vertices and sum the angles between those vectors. If the sum of the angles is 2*pi then the point is inside the triangle.
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);
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With