# Lines

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## Line equations

A line can be defined using any two points lying on the line. Let’s call the points on our line \(a\) and \(b\). If we draw a line through these two points, we end up with our original line.

Now let’s build a vector from these points going from \(a\) to \(b\) called \(\vec{ab}\).

\[\vec{ab}=b-a\]Our line can now be defined using the point \(a\) and the vector \(\vec{ab}\).

\[p=a+s*\vec{ab}\]For every scalar s the point p will be a point on our line.

This is the vector equation of a line. However most people might be more acquainted with the parametric equation of a line

\[y=cx+d\]Where \(d\) is the offset from the x-origin and \(c\) is the slope (amount of \(y\) units for one \(x\) unit). Both equations describe the same function and can be derived from each other. We can write the parametric equation as two functions of \(t\) as follows

\[x=t\] \[y=ct+d\]The function of x can be written as a line equation as well by using an offset of 0 and a slope of one.

\[x=1t+0\] \[y=at+b\]If we write this in vector form we get

\[(x,y)=(1,c)t+(0,d)\]Which is our vector equation of a line, where the point of the line is chosen where the line crosses the x axis at 0 thus \((0,d)\), and the direction vector is divided by its x coordinate to make the x coordinate 1 thus \((1,c)\).

### Line intersection

Given two lines through a, b and c, d respectively.

If these lines intersect, then there is a point p for which there is both a scalar t and a scalar s for which the following equations are equal to p.

\[p=a+s*\vec{ab}\] \[p=c+t*\vec{cd}\]Since p is the same in this case we can write

\[a+s*\vec{ab}=c+t*\vec{cd}\]If we subtract a from both sides we get

\[a-a+s*\vec{ab}=c-a+t*\vec{cd}\]The point \(a\) subtracted from the point \(c\) gives us the vector \(\vec{ac}\)

\[s*\vec{ab}=\vec{ac}+t*\vec{cd}\]Now we are going to use the cross product with \(\vec{cd}\) to get rid of the term \(t*\vec{cd}\), as \(\vec{cd}\times\vec{cd}\) is 0.

\[s*\vec{ab}\times\vec{cd}=\vec{ac}\times\vec{cd}+t*\vec{cd}\times\vec{cd}\]Erasing the 0 term gives us

\[s*\vec{ab}\times\vec{cd}=\vec{ac}\times\vec{cd}\]This means s is

\[\frac{\vec{ac}\times\vec{cd}}{\vec{ab}\times\vec{cd}}=s\]Before we go on, let’s note that \(\vec{ab}\times\vec{cd}\) can be 0, and we shouldn’t divide by 0. This cross product is zero when our lines are parallel and thus have either no intersection, or are collinear and are thus intersecting everywhere. Remember, the cross product gives us the sine of the angle (multiplied with a factor if the vectors are not normalized). The sine is 0 in the cases of 0 and 180 degrees. If there’s an angle of 0 or 180 degrees between the lines, they are definitely parallel, or collinear.

To know whether the lines are collinear, we can look whether \(\vec{ac}\times\vec{cd}\) is 0. because in that case, all points lie on the same line.

Now that we know s, we can calculate the point p from the formula

\[p=a+s*\vec{ab}\]This gives us the intersection point. Note that if the lines are collinear, we still return nil, as we can’t tell what point to return when everything collides.

```
function intersectLines(ax, ay, bx, by, cx, cy, dx, dy)
local abx, aby = bx - ax, by - ay -- ab
local cdx, cdy = dx - cx, dy - cy -- cd
local s = cross(abx, aby, cdx, cdy) -- ab x cd
if s < 0.00001 and s > -0.00001 then
-- Collinear if ac x cd == 0, parallel otherwise
return nil
end
s = cross(cx - ax, cy - ay, cdx, cdy) / s -- ac x cd / ab x cd
return ax + s * abx, ay + s * aby -- a + s * ab
end
```

### Point on line

Let’s say we have the line through points a and b, and we need to know whether c is on the same line. Remember how we tested for parallel and collinear lines using the cross product? That is exactly what we need now. If we build a vector from a to c called ac, and cross it with the vector ab, the result is 0 if c lies on ab.

```
function pointOnLine(ax, ay, bx, by, cx, cy)
local abx, aby = bx - ax, by - ay -- ab
local acx, acy = cx - ax, cy - ay -- ac
local t = cross(abx, aby, acx, acy) -- ab x ac
return t < 0.00001 and t > -0.00001
end
```

## Line segments

Sometimes our lines don’t go to infinity in both directions, but have given boundaries. Line segments are not much different in usage compared to lines, except for some extra tests to make sure we are actually still on the line segment. Given a segment between points \(a\) and \(b\), we saw we could define the line as

\[p=a+s*\vec{ab}\]We can observe that if \(s\) is 0, \(p\) is equal to \(a\), while if \(s\) is 1, \(p\) is equal to \(b\). Any value of \(s\) not between 0 and 1 gives a point outside of the line segment.

### Line segment intersection

#### Line segment and line

Given what we know from line intersection, intersecting a line segment with another line is not so hard. We only need to be sure that the intersection point is actually within the segment(s). This is simple if we have s for the point on the segment. If \(s\) is outside the interval [0,1], the point is outside the range of the segment. We can use this test to know whether our intersection point lies on the segment.

```
function intersectSegmentLine(ax, ay, bx, by, cx, cy, dx, dy)
local abx, aby = bx - ax, by - ay -- ab
local cdx, cdy = dx - cx, dy - cy -- cd
local s = cross(abx, aby, cdx, cdy) -- ab x cd
if s < 0.00001 and s > -0.00001 then
-- Collinear if ac x cd == 0, parallel otherwise
return nil
end
s = cross(cx - ax, cy - ay, cdx, cdy) / s -- ac x cd / ab x cd
if s < 0 or s > 1 then
return nil
end
return ax + s * abx, ay + s * aby -- a + s * ab
end
```

#### Line segment and line segment

To know whether our intersection point lies on the second segment, we need to calculate \(t\) like we calculated \(s\) before.

\[a+s*\vec{ab}=c+t*\vec{cd}\]If we subtract c from both sides we get

\[a-c+s*\vec{ab}=c-c+t*\vec{cd}\]The point \(c\) subtracted from \(a\) gives us the vector \(\vec{ca}\)

\[\vec{ca}+s*\vec{ab}=t*\vec{cd}\]Now we use the cross product with \(\vec{ab}\) to get rid of the term \(s*\vec{ab}\), as \(\vec{ab}\times\vec{ab}\) is 0.

\[\vec{ca}\times\vec{ab}+s*\vec{ab}\times\vec{ab}=t*\vec{cd}\times\vec{ab}\]Erasing the 0 term gives us

\[\vec{ca}\times\vec{ab}=t*\vec{cd}\times\vec{ab}\]This means that t is

\[t=\frac{\vec{ca}\times\vec{ab}}{\vec{cd}\times\vec{ab}}\]Now we already have \(\vec{ab}\times\vec{cd}\), and we know that \(\vec{cd}\times\vec{ab}=-\vec{ab}\times\vec{cd}\) because switching vectors in a cross product gives use the cross product of the other angle, and \(sin(\alpha)=-sin(-\alpha)\). We also have \(\vec{ac}\), while here we require \(\vec{ca}\). If we invert the vector \(\vec{ca}\) to the vector \(\vec{ac}\) in \(\vec{ca}\times\vec{ab}\) however, we invert the sign once more, as \(\vec{ca}\times\vec{ab}=-\vec{ac}\times\vec{ab}\). So we get

\[t=\frac{\vec{ca}\times\vec{ab}}{\vec{cd}\times\vec{ab}}=\frac{-\vec{ac}\times\vec{ab}}{-\vec{ab}\times\vec{cd}}=\frac{\vec{ac}\times\vec{ab}}{\vec{ab}\times\vec{cd}}\]Using this we can efficiently write the tests to see if our point lies within the other segment using already calculated values.

```
function intersectSegments(ax, ay, bx, by, cx, cy, dx, dy)
local abx, aby = bx - ax, by - ay -- ab
local cdx, cdy = dx - cx, dy - cy -- cd
local abxcd = cross(abx, aby, cdx, cdy) -- ab x cd
if abxcd < 0.00001 and abxcd > -0.00001 then
-- Collinear if ac x cd == 0, parallel otherwise
return nil
end
local acx, acy = cx - ax, cy - ay
s = cross(acx, acy, cdx, cdy) / abxcd -- ac x cd / ab x cd
if s < 0 or s > 1 then
return nil
end
local t = cross(acx, acy, abx, aby) / abxcd -- ac x ab / ab x cd
if t < 0 or t > 1 then
return nil
end
return ax + s * abx, ay + s * aby -- a + s * ab
end
```

### Point on line segment

We first need to test whether the point lies on the line using the cross product as we did before for infinite lines.

If the point is on the line, we need to test whether it lies between the segment’s boundaries. We can do this second step partly by calculating the dot product, like when we did vector projection. If the dot product is smaller than 0, the point lies before the start of the segment, as the angle is greater than 90 degrees in that case.

To be sure that the point doesn’t lie past the end of the segment, we can look whether the dot product of \(\vec{ac}\) with itself is smaller than the dot product of \(\vec{ab}\) with itself, as the dot product gives us the length squared.

```
function pointOnSegment(ax, ay, bx, by, cx, cy)
local abx, aby = bx - ax, by - ay -- ab
local acx, acy = cx - ax, cy - ay -- ac
local t = cross(abx, aby, acx, acy) -- ab x ac
if t > 0.00001 or t < -0.00001 then
return false
end
t = dot(abx, aby, acx, acy) -- ab . ac
if t < 0 then
return false
end
t = dot(acx, acy, acx, acy) -- ac . ac
return t <= dot(abx, aby, abx, aby) -- ab . ab
end
```

## Rays

A ray is a segment which is bounded at one side, the origin, while it stretches to infinity one the other side. Rays have various usages. As rays of light to determine highlights for example. The dot product of the ray with a surface normal gives a simple but fast diffuse shading. Or to determine where shadows fall by determining walls whose normals point towards the light and projecting those walls further along the ray. Sometimes a line segment is used as a ray, like when determining visibility from the player’s position when there is a “fog of war”, as the ray needs a limited range in that case.

### Ray intersection

Intersecting a ray with a line is similar to a segment with a line except that we only need to test for t < 0, as the ray is only bounded at the origin.

```
function intersectRayLine(ax, ay, bx, by, cx, cy, dx, dy)
local abx, aby = bx - ax, by - ay -- ab
local cdx, cdy = dx - cx, dy - cy -- cd
local s = cross(abx, aby, cdx, cdy) -- ab x cd
if s < 0.00001 and s > -0.00001 then
-- Collinear if ac x cd == 0, parallel otherwise
return nil
end
s = cross(cx - ax, cy - ay, cdx, cdy) / s -- ac x cd / ab x cd
if s < 0 then
return nil
end
return ax + s * abx, ay + s * aby -- a + s * ab
end
```

### Point on ray

Also here we just need to test whether the point is not before the origin of the ray.

```
function pointOnRay(ax, ay, bx, by, cx, cy)
local abx, aby = bx - ax, by - ay -- ab
local acx, acy = cx - ax, cy - ay -- ac
local t = cross(abx, aby, acx, acy) -- ab x ac
if t > 0.00001 or t < -0.00001 then
return false
end
return dot(abx, aby, acx, acy) >= 0 -- ab . ac
end
```

## Subdivision

### Subdivision of a line segment

Dividing a line segment into smaller segments, or finding the middle of a line segment is very straightforward when we use the vector equation. Given a segment from a to b, thus using the equation \(a+s*(b-a)\), we find the middle at s=0.5. Similarly we can get the points at ⅓rd or ⅔rd or any other possible fraction. It’s a simple linear interpolation between two points, since that is what the vector equation of a line defines.

Note that if specifically the middle of the line is needed, the equation can be simplified as

\[a+0.5*(b-a)=a+0.5*b-0.5a=0.5*a+0.5*b=0.5*(a+b)\]So we only need to add the points and divide the result by 2.

### Cutting a line segment interactively

Splitting a line in two lines is easy enough, but how do we know where the user wants it cut? There are several solutions to this.

We can use vector projection to find the closest line as well as the point on that line closest to where the user clicked. The squared distance to the line can be the deciding factor whether to cut or not, whether the user clicked near enough.

Alternatively we can let the user do a cut gesture, which represents a line segment. Now we can check which line segments are being intersected by the cut segment as well as calculate the intersection points where the cut was made.

## A note on axis aligned lines

When lines are axis aligned, quite some simplifications can be done.

### Horizontal lines

Remember that the cross product is defined as \(a_x*b_y-a_y*b_x\). For a horizontal line we can choose \(c\) and \(d\) so that $\vec{cd}$ is equal to \(\langle -1,0\rangle\) which makes \(\vec{ab}\times\vec{cd}=ab_y\).

Similarly, \(\vec{ac}\times\vec{cd}\) becomes \(ac_y\). So we can write a specific function for horizontal lines as follows.

```
function intersectSegmentHorizontalLine(ax, ay, bx, by, y)
local abx, aby = bx - ax, by - ay -- ab
-- cd is (-1,0)
-- ab x cd is aby
if aby < 0.00001 and aby > -0.00001 then
-- Collinear if ac x cd == 0, y-ay in this case
-- parallel otherwise
return nil
end
local s = (y-ay) / aby -- ac x cd / ab x cd
if s < 0 or s > 1 then
return nil
end
return ax + s * abx, ay + s * aby -- a + s * ab
end
```

Can we simplify this? What does \(s<0\) and \(s>0\) mean in this case? well, let’s find out. The inequality

\[\frac{y-a_y}{b_y-a_y}<0\]has two solutions, depending whether \(b_y-a_y\) is positive or negative.

If \(b_y-a_y\) is positive, thus \(b_y>a_y\), multiplying both sides with \(b_y-a_y\) keeps the inequality as $<$, thus

\[y-a_y < 0\]or

\[y < a_y\]If \(b_y-a_y\) is negative, thus \(b_y<a_y\), we need to flip the inequality when multiplying, so we get

\[y-a_y > 0\]or

\[y>a_y\]For the other inequality

\[\frac{y-a_y}{b_y-a_y}>1\]we can do the same. If \(b_y-a_y\) is positive, thus \(b_y>a_y\), we get

\[y-a_y>b_y-a_y\]Or, if we add \(a_y\) on both sides

\[y>b_y\]Similarly when \(b_y-a_y\) is negative, thus \(b_y<a_y\) we get

\[y<b_y\]So to summarize, we have that if \(b_y>a_y\), we don’t have an intersection if \(y < a_y\) or \(y>b_y\). When \(b_y<a_y\), we don’t have an intersection if \(y>a_y\) or \(y<b_y\). In code this would be expressed as for example

```
if y < min(ay,by) or y > max(ay,by) then
return nil
end
```

Thus we don’t have an intersection if the segment lies entirely above or under the horizontal line at y. Whether you use s < 0 or s > 1 or this min-max test won’t make a difference, but since you already need to calculate s to get the intersection, testing s with 0 and 1 will be faster than doing 2 extra comparisons (and potential function calls). But I hope this analysis shows what exactly is being tested.

If the test, whichever we choose, indicates an intersection, we can calculate the intersection point, which was

\[i_x=a_x+s*(b_x-a_x)\\i_y=a_y+s*(b_y-a_y)\]if we fill in s we get

\[i_x=a_x+\frac{y-a_y}{b_y-a_y}*(b_x-a_x)\\i_y=a_y+\frac{y-a_y}{b_y-a_y}*(b_y-a_y)\]For \(i_y\) we can remove the division by \(b_y-a_y\) since we multiply by \(b_y-a_y\) as well. Then we can erase \(a_y\) as well as it is negated by \(-a_y\). So \(i_y=y\) as we would expect of an intersection with an horizontal line at y.

For \(i_x\) we see that we actually compute the slope of the line as \(\frac{dx}{dy}\) which we then multiply with the distance of \(a_y\) to the line, thus getting the distance of \(a_x\) to the line. Adding \(a_x\) and its distance to the line gives us the intersection coordinate \(i_x\).

So finally we can write our simplified function as

```
function intersectSegmentHorizontalLine(ax, ay, bx, by, y)
local abx, aby = bx - ax, by - ay -- ab
-- cd is (-1,0)
-- ab x cd is aby
if aby < 0.00001 and aby > -0.00001 then
-- Collinear if ac x cd == 0, y-ay in this case
-- parallel otherwise
return nil
end
local s = (y-ay) / aby -- ac x cd / ab x cd
if s < 0 or s > 1 then
return nil
end
return ax + s * abx, y -- a + s * ab
end
```

### Vertical lines

For vertical lines we can choose our vector as \(\langle 0,1\rangle\) which gives us a similar solution.

```
function intersectSegmentVerticalLine(ax, ay, bx, by, x)
local abx, aby = bx - ax, by - ay -- ab
-- cd is (0,1)
-- ab x cd is abx
if abx < 0.00001 and abx > -0.00001 then
-- Collinear if ac x cd == 0, x-ax in this case
-- parallel otherwise
return nil
end
local s = (x-ax) / abx -- ac x cd / ab x cd
if s < 0 or s > 1 then
return nil
end
return ax, ay + s * aby -- a + s * ab
end
```

## horizontal line segments

While lines can be useful to test against screen borders, in real situations it is more useful if we can test against line segments, like those of walls or boxes. We can do this like we did for lines, but in this case we need to pay attention to the fact that the length of our direction vector (-1, 0) doesn’t match our actual line segment length of \(l=abs(x1-x2)\).

What actually happens when the direction vector of the horizontal segment has length 1 instead of the actual length \(l\)? Since the cross product is a product, and one of the factors is \(l\) times smaller, the product itself will be \(l\) times smaller. Thus every cross product involving \(cd\), which has been replaced with (-1,0) will be \(\lvert cd\rvert\) times smaller.

For \(s\) this gives no change, as for \(s\), \(cd\) features in both the numerator and denominator and it cancels itself out.

For \(t\) however, \(cd\) is only present in the denominator. Since the denominator is \(\lvert cd\rvert\) times smaller than it should be, \(t\) will be \(\lvert cd\rvert\) times greater than we expect it to be. Thus we need to adapt our test of \(t>1\) to \(t>\lvert cd\rvert\).

To get the length of \(cd\) we could subtract d from c and take the absolute value. However we need to know which of c or d is greater than the other because we decided to use the vector (-1,0) as approximation of \(cd\) and thus the direction of \(cd\) needs to match this. This is important when calculating \(ac\). Once we know which is greater, an absolute value is no longer needed, we just subtract the smaller one from the greater one.

```
function intersectSegmentHorizontalSegment(ax, ay, bx, by, x1, x2, y)
local abx, aby = bx - ax, by - ay -- ab
-- cd is (-1,0)
-- ab x cd is aby
if aby < 0.00001 and aby > -0.00001 then
-- Collinear if ac x cd == 0, y-ay == 0 in this case
-- parallel otherwise
return nil
end
local s = (y - ay) / aby -- ac x cd / ab x cd
if s < 0 or s > 1 then
return nil
end
if x1 > x2 then x1, x2 = x2, x1 end
local t = cross(x2-ax, y-ay, abx, aby) / aby -- ac x ab / ab x cd
if t < 0 or t > (x2-x1) then
return nil
end
return ax + s * abx, y -- a + s * ab
end
```

## Vertical line segments

Vertical line segments are almost the same, with the exception that, because we choose our vector as (0,1), the direction of \(cd\) is opposite to that of our horizontal segment. Thus we need to use the smallest of c and d to build \(ac\).

```
function intersectSegmentVerticalSegment(ax, ay, bx, by, x, y1, y2)
local abx, aby = bx - ax, by - ay -- ab
-- cd is (-1,0)
-- ab x cd is abx
if abx < 0.00001 and abx > -0.00001 then
-- Collinear if ac x cd == 0, x-ax == 0 in this case
-- parallel otherwise
return nil
end
local s = (x - ax) / abx -- ac x cd / ab x cd
if s < 0 or s > 1 then
return nil
end
if y1 > y2 then y1, y2 = y2, y1 end
local t = cross(x-ax, y1-ay, abx, aby) / abx -- ac x ab / ab x cd
if t < 0 or t > (y2-y1) then
return nil
end
return x, ay + s * aby -- a + s * ab
end
```

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