Given two points, find a third point on the line - c++

Question

I have two points A (x1,y1) and B (x2,y2) that are given as an input to the program. I have to find a third point C that lies on the line AB and is at a distance 10 away from the point A.
I can easily get the slope of the line but that doesn't give me the full equation for the line. Even if I get the full equation, I am not sure using this equation, how would I find out a point that is x distance away from A.
Any suggestions on how to approach this?

Answer 1

There are always two points on each line:
get the vector from A to B (subtract the coordinates)
normalize the vector (divide by its length; pythagorean theorem)
multiply the vector by 10 or -10
add the vector to A to get C
Note that if A==B, the line is not defined, and this algorithm causes a division by zero. You may want to add a test for equality at the beginning.

Answer 2

You can use the sine or the cosine (times 10) of the angle of the line to get the horizontal or vertical distance of the point that is a distance of 10 from a given point. A shortcut is to use the horizontal or vertical distance divided by the direct distance between the points to get the sine or cosine.

Answer 3

You can do it using vectors like this:
Let D = the difference between B and A (D = B - A)
Then any point on the line can be described by this formula:
point = A + Dt
where t is a real number.
So just plug in any value for t to get another point. For example if you let t == 1 then the equation above reduces to point = B. If you let t = 0 then it reduces to point = A. So you can see that you can use this to find a point between A and B simply by let t range from 0 to 1. Additionally if you let t > 1, you will find a point past B.

Answer 4

You can see from the image that your given points are x1,y1 and x2,y2. You need to find an intermediate point at a distance 'R' from point x1,y1.
All you need to do is to find θ using
Tan θ = (y2-y1)/(x2-x1)
Then you can get the intermediate point as (R * cos θ),(R * Sin θ)
I have drawn this assuming positive slope.
Going on similar lines you can seek a solution for other special cases lile:
i. Horizontal line
ii. Vertical line
iii. Negative slope
Hope it clarifies.

Answer 5

I have done the calculation in Andengine using a Sprite object. I have two Array List x coordinates and y coordinates. Here i am just calculating using the last two values from these two array list to calculate the third point 800 pixel distant from Your point B. you can modify it using different values other than 800. Hope it will work.The coordinate system here is a little different where (0,0) on the top left corner of the screen. Thanks
private void addExtraCoordinate(CarSprite s) {
int x0, y0, x1, y1;
float x = 0f, y = 0f;
x0 = Math.round(xCoordinates.get(xCoordinates.size() - 2));
x1 = Math.round(xCoordinates.get(xCoordinates.size() - 1));
y0 = Math.round(yCoordinates.get(yCoordinates.size() - 2)) * (-1);
y1 = Math.round(yCoordinates.get(yCoordinates.size() - 1)) * (-1);
if (x1 == x0 && y1 == y0) {
return;
} else if (y1 == y0 && x1 != x0) {
if (x1 > x0) {
x = (float) x1 + 800f;
} else
x = (float) x1 - 800f;
y = Math.round(yCoordinates.get(yCoordinates.size() - 1));
} else if (y1 != y0 && x1 == x0) {
if (y1 > y0) {
y = (float) Math.abs(y1) - 800f;
} else
y = (float) Math.abs(y1) + 800f;
x = Math.round(xCoordinates.get(xCoordinates.size() - 1));
} else {
float m = (float) (yCoordinates.get(yCoordinates.size() - 1) * (-1) - yCoordinates
.get(yCoordinates.size() - 2) * (-1))
/ (float) (xCoordinates.get(xCoordinates.size() - 1) - xCoordinates
.get(xCoordinates.size() - 2));
if (x1 > x0) {
x = (float) ((float) x1 + 800f / (float) Math
.sqrt((double) ((double) 1f + (double) (m * m))));
} else
x = (float) ((float) x1 - 800f / (float) Math
.sqrt((double) ((double) 1f + (double) (m * m))));
if (y0 > y1) {
y = (float) ((float) Math.abs(y1) + 800f / (float) Math
.sqrt((double) (((double) 1f / (double) (m * m)) + (double) 1f)));
} else
y = (float) ((float) Math.abs(y1) - 800f / (float) Math
.sqrt((double) (((double) 1f / (double) (m * m)) + (double) 1f)));
}
xCoordinates.add(x);
yCoordinates.add(y);
}