Finding a point along a line a certain distance away from another point!

Question 1

Let’s say you have two points, $(x_0, y_0)$ and $(x_1, y_1)$ .

The gradient of the line between them is:

$m = (y_1 - y_0)/(x_1 - x_0)$

And therefore the equation of the line between them is:

$y = m (x - x_0) + y_0$

Now, since I want another point along this line, but a distance $d$ away from $(x_0, y_0)$ , I will get an equation of a circle with radius $d$ with a center $(x_0, y_0)$ then find the point of intersection between the circle equation and the line equation.

Circle Equation w/ radius $d$ :

$(x - x_0)^2 + (y - y_0)^2 = d^2$

Now, if I replace $y$ in the circle equation with $m(x - x_0) + y_0$ I get:

$(x - x_0)^2 + m^2(x - x_0)^2 = d^2$

I factor is out and simplify it and I get:

$x = x_0 \pm d/ \sqrt{1 + m^2}$

However, upon testing this equation out it seems that it does not work! Is there an obvious error that I have made in my theoretical side or have I just been fluffing up my calculations?

Question 2

Another way, using vectors:

Let $\mathbf v = (x_1,y_1)-(x_0,y_0)$ . Normalize this to $\mathbf u = \frac{\mathbf v}{||\mathbf v||}$ .

The point along your line at a distance $d$ from $(x_0,y_0)$ is then $(x_0,y_0)+d\mathbf u$ , if you want it in the direction of $(x_1,y_1)$ , or $(x_0,y_0)-d\mathbf u$ , if you want it in the opposite direction. One advantage of doing the calculation this way is that you won’t run into a problem with division by zero in the case that $x_0 = x_1$ .

Finding a point along a line a certain distance away from another point!

Answer

Leave a Comment Cancel reply