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

Question

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?

Answer 1

Let me explain the answer in a simple way.

Start point - $(x_0, y_0)$

End point - $(x_1, y_1)$

We need to find a point $(x_t, y_t)$ at a distance $d_t$ from start point towards end point.

The distance between Start and End point is given by $d = \sqrt{(x_1-x_0)^2+(y_1-y_0)^2}$

Let the ratio of distances, $t=d_t/d$

Then the point $(x_t, y_t) =(((1-t)x_0+tx_1), ((1-t)y_0+ty_1))$

When $0<t<1$, the point is on the line.

When $t<0$, the point is outside the line near to $(x_0,y_0)$.

When $t>1$, the point is outside the line near to $(x_1,y_1)$.

Answer 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 $x_0 = x_1$.

Answer 3

You can very easily find it with trigonometry!!

Let's say Xa and Xb are the two points of your line, and D is the distance between them. And you are looking to find Xc which is D2 away from Xa (as the diagram bellow):

You can easily find D:

The formulas that you can find Xa, Xb, Xc, D and D2 are:

But SINa-b and SINa-c share the same the same corner, so they are equal:

Since you know the distance (D2) between Xa and Xc that you are looking for, you can easily solve the following:

In conclusion by solving the formula for D and the last one you are done. (You need one for the Y as well, just replace in the last one, X with Y )

Hope it helps!!

Answer 4

The easy way in rectangular coordinate systems is to use the vector formula
P = d(B - A) + A
where
A is the starting point (x₀, y₀) of the line segment
B is the end point (x₁, y₁)
d is the distance from starting point A to the desired collinear point
P is the desired collinear point

Answer 5

I think your method is working. Here is the python code for the test and output of the test is linked below,

import numpy as np
import matplotlib.pyplot as plt
def f(x, m, c):
    y = m*x + c
    return y
def X(x0,d,m):
    x1 = x0 + (d/np.sqrt(1+m2))
    x2 = x0 - (d/np.sqrt(1+m2))
    return np.array([x1, x2])
x = np.arange(1,5,0.5)
m = 1
c = 0.5
plt.plot(x, f(x, m, c), "-")
x0 = 3
d = np.array([0.25, 0.5, 0.75, 1])
X = X(x0,d,m)
print(type(X))
plt.plot(X, f(X,m, c), "o")
plt.show()

Output of the above code:[1] https://i.stack.imgur.com/n0XJH.png

Answer 6

I think you need to check $x_0 > x_1$ when you try to calculate $x$ (last equation in your calculation) then you determine it will be $(+)$ or $(-)$ in your equation.

Answer 7

There is an other way to solve this problem.

The function $\rho:\mathbb R \to \mathbb R^2$ defined by

$$\rho(t) = (1-t)(x_0,y_0) + t(x_1,y_1)$$

describes the line through the points $(x_0,y_0)$ and $(x_1,y_1)$ with $\rho(0) = (x_0, y_0)$ and $\rho(1)=(x_1,y_1)$. If you let $$D = \sqrt{(x_1-x_0)^2 + (y_1-y_0)^2},$$ then it is fairly easy to show that

$$|\rho(t) - \rho(s)| = D|t-s|$$

So, if you want to find the points a distance of $d$ from $(x_0,y_0)$, then you need to solve $$d = |\rho(t) - \rho(0)| = D|t-0|$$

You get $ t = \pm \dfrac dD$. Hence the points are

$$\rho\left( \pm \dfrac dD \right) = \left\{ \begin{array}{c} \left( 1-\dfrac dD \right)(x_0,y_0) + \dfrac dD (x_1,y_1) \\ \left( 1+\dfrac dD \right)(x_0,y_0) - \dfrac dD (x_1,y_1) \end{array} \right.$$

Answer 8

Simply put, for desired distance d, x of this point is:

$$ x_1 + d \cdot \cos(\arctan(a)) \iff a = (\frac{y_2-y_1}{x_2-x_1}) $$

THIS IS HOW:

You have given two points:

$$ P = (x_p, y_p) \\ Q = (x_q, y_q) $$

and line between them defined with the function f(x):

$$ m = \frac{y_q - y_p}{x_q - x_p} \\\ \\ f(x) = m(x - x_p) + y_p $$

or:

$$ a = \frac{y_q-y_p}{x_q-x_p} \\\ \\ b = \frac{x_{q}y_{p} - x_{p}y_{q}}{x_q-x_p} \\\ \\ f(x) = ax + b $$

You're looking for a point R between P and Q.
Note that we can "render" a triangle with the points P and R and use it to get a width of a side that we have to know. We know one side - it's desired distance, and the angle that equals $\arctan(a)$. Let's see all this stuff in the picture:

We need to calculate x' and then our point R is $(x_p+x', f(x_p+x'))$.
Example calculations for desired distance 3:

• we have two points:

$$ P = (0, 2) \\ Q = (4, 4) $$

• and the line between them:

$$ f(x) = \frac{1}{2}x + 2 $$

• our distance is 3:

$$ d = 3 $$

• we're getting an angle:

$$ \alpha = \arctan(\frac{1}{2}) = 0.4637 $$

• now we can get a distance along X-axis:

$$ x' = d \cdot cos(\alpha) = 3 \cdot 0.8944 = 2.6832 $$

so x of our point R is:

$$ x_r = x_p + x' = 0 + 2.6832 = 2.6832 $$

and our point is at:

$$ (x_r, f(x_r)) = (2.6832, 3.3416) $$

---

Example implementation in JavaScript:

function findPointByDistance(p1, p2, dist) {
    const [x1,y1] = p1;
    const [x2,y2] = p2;
const m = (y2-y1)/(x2-x1);
const f = x => m*(x-x1) + y1;

const alpha = Math.atan(m);
const xVector = dist * Math.cos(alpha);
const foundX = x1 + xVector;

return 
    foundX,
    f(foundX)
].map(v => +v.toFixed(2));

}

const P = [0,2];
const Q = [4,4];
const d = 3;

findPointByDistance(P, Q, d);
// [ 2.68, 3.34 ]