# Shortest distance along line within a certain distance of point

I have two points, $$(x_0, y_0)$$ and $$(x_1, y_1)$$. The point $$(x_0, y_0)$$ is on a line, which has angle $$\theta$$, relative to the X-axis.

Using these values, I want to be able to calculate the shortest distance, $$d$$, along the line (relative to $$(x_0, y_0)$$) which is less than 2 units away from $$(x_1, y_1)$$.

I suspect there may be something to do with slope and arctangent, but I can't make heads or tails of how I should attempt this (It's for a video game, in case that matters).

Sorry if this is a dupe, or if my math-speak is a bit too much programmer and a bit too little mathematician.

• Angle $\theta$ with respect to what? Commented Apr 1, 2019 at 2:53
• @MichaelBiro X-axis, sorry Commented Apr 1, 2019 at 2:54
• I want to be able to calculate the _shortest_ distance, d, along the line Commented Apr 1, 2019 at 2:55
• It’s either $0$ if the distance between the two points is $\le2$ or the distance to the nearest intersection of the line with a circle of radius $2$.
– amd
Commented Apr 1, 2019 at 3:12
• I added an image, maybe you can refine it or replace it with more information. Commented Apr 1, 2019 at 3:17

I'm assuming you want a programmatic approach instead of a purely analytic one, so I would attack this problem using standard transformations.

First, translate so $$(x_0,y_0)$$ lands on the origin - this corresponds to subtracting $$(x_0,y_0)$$ from each point.

Second, rotate by $$-\theta$$ so that the line $$L$$ lands on the $$x$$-axis - this corresponds to multiplying by the rotation matrix $$\begin{bmatrix} \cos \theta & \sin \theta\\-\sin \theta & \cos \theta\end{bmatrix}$$

Now, $$(x_1, y_1)$$ has been transformed to $$(x_1^\prime , y_1^\prime)$$, and we want to find the point on the $$x$$-axis within a distance of $$2$$ that has the smallest $$x$$ coordinate. Assuming $$y_1^\prime \leq 2$$:

If $${x^\prime_1}^2 + {y^\prime_1}^2 \leq 4$$ then $$(0,0)$$ is the nearest point.

Otherwise:

If $$x^\prime_1 > 0$$, then $$(x^\prime_1 - \sqrt{2^2 - {y_1^\prime}^2},0)$$ is the nearest point.

If $$x^\prime_1 < 0$$, then $$(x^\prime_1 + \sqrt{2^2 - {y_1^\prime}^2},0)$$ is the nearest point.

Now, just reverse the transformations to find the coordinates you need.

• Thanks! This is just what I need. Commented Apr 1, 2019 at 3:51

I'll propose the example as I first understood it:

The line through x0,y0 does not hit x1,y1. Also there is a point-of-origin of x,y. So there is a line from x,y to x0,y0 as the first line and a line from x,y to x1,y1 as the second line. Then there is an angle between the two lines such that Sin(angle) * distance-of-second-line = right-angle-distance off the first line. Also, Tan(angle) * unknown-distance = previously-calculated-right-angle-distance then calculates the distance from x,y to the right-angle-offset on the first line.

I calculate problems like this with a land surveying system of working with bearing-directions and distances. For instance an angle to the first line of 10 degrees off the x-axis is a bearing of N80E.

The direction and distance from x,y to x1,y1 is an "inverse" of the second line:

Direction = InvTan((x1 - x) / (y1 - y)). If (x1 - x) is positive that is E otherwise W. If (y1 - y) is positive that is N otherwise S. The expected result in this example is a direction of N_angle_E.

Distance = Square Root of ((x1 - x)^2 + (y1 - y)^2) .