# Calculating the distance from a line to a point in 2D

I have a simple setup: How do I calculate the orange distance?

It's the distance between the block on the left to the white line in the center (X & Z axes combined).

I don't have the intersection point. My only knowledge of the white line is that it goes through the point (0, 0, 0), running between (10, 0, -10) and (-10, 0, 10) My problem is that I need to calculate this distance without the depth (Y value). Only on two planes combined: X & Z

• Maybe you can use Vector3.Project with normal = new Vector3( 1, 0, 1 ) ? Jul 3 '17 at 13:19
• How do you define the white line? Do you know any point on the white line, and the direction the line is pointing? Or do you define it in some other way? Jul 3 '17 at 14:42
• Hey Gregory, I've edited the question :) Jul 3 '17 at 15:11
• Edited it again, I've added the image from the wrong angle. Now it's ok. Jul 3 '17 at 15:14

I think you should be able to get the position vector using Vector3.Project Vector3 vector = yourGameObject.GetComponent<Transform>().position ;
vector.y = 0 ;
Vector3 onNormal = new Vector3( 1, 0, 1 ) ;
Vector3 projectionOnWHiteAxis = Vector3.Project( vector, onNormal ) ;
float distance = (projectionOnWHiteAxis - vector).magnitude ;


You can rotate the point by -45 degrees, so that the x component becomes the distance.

If you always want to find the distance between the point and the XZ axes, then you can just precalculate the values:

| cos -45° -sin -45° |   | √2/2, -√2/2 |
| sin -45   cos -45° | = | √2/2,  √2/2 |


If we take the vector (x; y) the result is:

(√2/2 * x - √2/2 * y; √2/2 * x + √2/2 * y)


Then the distance is abs(√2/2 * x - √2/2 * y)

• Yeah, pythagoras would be nice, If I new the intersection point of lines white and orange. My only info is that it's the X & Z axes combination. Jul 3 '17 at 12:56
• @Spectre It wasn't clear that you didn't know where the white line was, let me try to figure it out Jul 3 '17 at 13:00
• @Spectre Here you go Jul 3 '17 at 13:15
• @Bálint don't forget to put the absolute value to your result, otherwise you might be getting negative values depending on (x,y). Also you might want to use 1/sqrt(2) instead of sqrt(2)/2. Mathematically is the same, but less operations ;) Jul 7 '17 at 12:39
• @Turms sqrt(2) / 2 is actually less CPU intensive, than 1/sqrt(2), because in the latter you need to calculate the square root of 2, then divide 1 with the result (division is hard), while the former only needs to calculate square root of 2 then use a bit shift. You're right about the absolute value though Jul 7 '17 at 13:33
float SignedDistanceFromLine(
Vector2 sourcePoint,
Vector2 lineDirection,
Vector2 pointOnLine = default(Vector2))
{
// Translate the whole situation to the origin.
sourcePoint -= pointOnLine;

// Construct a unit vector pointing 90 degrees right of our lineDirection.
Vector2 perpendicular = (new Vector2(
lineDirection.y,
-lineDirection.x)).normalized;

// Extract the component of our offset pointing in this perpendicular direction.
return Vector2.Dot(sourcePoint, perpendicular);
}


This returns a signed distance, with positive values signalling that the sourcePoint is to the right of the line, and negative values to the left.

You can use this to get the positive distance like so:

Vector2 objectPosition = new Vector2(
objectTransform.position.x,
objectTransform.position.z);
float distance = Mathf.Abs(SignedDistanceFromLine(objectPosition, new Vector2 (1, -1)));