Getting a Point That's 200 Pixels Up a Line
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I have two coordinate pairs (A and B, for example). How can I find the point that is 1) colinear to A and B, and 2) 200 pixels away from A?
What do you mean by "200 pixels". I assume that you're not primarily interested in straight horizontal nor straight vertical lines. Do you mean like: What is the Y-coordinate of the point that intersects the line at X-coordinate 200? For this you can use the standard line equation:
y = k * x + m
where slope: k = (By - Ay) / (Bx - Ax)
displacement: m = Ay - k * Ax
The terminology I use is probably lousy, but hope you get the picture. It should be in any math book.
[edited by - CWizard on September 15, 2002 6:24:39 PM]
y = k * x + m
where slope: k = (By - Ay) / (Bx - Ax)
displacement: m = Ay - k * Ax
The terminology I use is probably lousy, but hope you get the picture. It should be in any math book.
[edited by - CWizard on September 15, 2002 6:24:39 PM]
But in what unit? What would you say the distance is between (0,0)-(14,6)? If your answer is about 15.23, we can use something like this: (pseudo-code)
// Ax,Ay = First point
// Bx,By = Second point
// d = Distance from first point
// Angle of line
V = atan2((By - Ay), (Bx - Ax));
// Point d units from A
Px = Ax + cos(V);
Py = Ay + sin(V);
Something like that. Anyone can corret me if I''m wrong.
// Ax,Ay = First point
// Bx,By = Second point
// d = Distance from first point
// Angle of line
V = atan2((By - Ay), (Bx - Ax));
// Point d units from A
Px = Ax + cos(V);
Py = Ay + sin(V);
Something like that. Anyone can corret me if I''m wrong.
Just for fun I coded a function that do this:
#include <math.h> void PointOnLine(int Ax, int Ay, int Bx, int By, int *Cx, int *Cy, int d){ float fAngle = atan2f((float)(By - Ay), (float)(Bx - Ax)); *Cx = Ax + (int)cosf(fAngle); *Cy = Ay + (int)sinf(fAngle);} int cxPoint, cyPoint; PointOnLine(100, 50, 800, 450, &cxPoint, &cyPoint, 200);
Damn, made an error in the above code. Here's a fixed one, also including an alternative method:
Not tested, though.
[edited by - CWizard on September 15, 2002 8:15:39 PM]
#include <math.h> void PointOnLine(int Ax, int Ay, int Bx, int By, int *Cx, int *Cy, int d){ float fAngle = atan2f((float)(By - Ay), (float)(Bx - Ax)); *Cx = Ax + (int)(cosf(fAngle) * (float)d); *Cy = Ay + (int)(sinf(fAngle) * (float)d);} void PointOnLine2(int Ax, int Ay, int Bx, int By, int *Cx, int *Cy, int d){ int Vx = Bx - Ax; int Vy = By - Ay; float fLength = sqrtf((float)(Vx * Vx + Vy * Vy)); *Cx = Ax + ((float)Vx / fLength * (float)d); *Cy = Ay + ((float)Vy / fLength * (float)d);} int cxPoint, cyPoint; PointOnLine(100, 50, 800, 450, &cxPoint, &cyPoint, 200); [edited by - CWizard on September 15, 2002 8:15:39 PM]
quote:Original post by rjahrman
I mean the point where the distance between that point and A is 200.
That is a circle around A with a radius of 200.
A point cannot be colinear.
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