# How do I determine if a point is within a rhombus?

I know the coordinates of the 4 rhombus' vertices. I also have the coordinates of another arbitrary point (the result of a click on the screen).

How do I determine if that point is within the rhombus?

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Are you working in $\mathbb R^2$? (Also, since this is really an implementation problem, you might want to try this question on some of the programming-oriented stack-exchanges instead.) –  Potato Feb 23 '13 at 23:45

The interior of a square, rhombus, rectangle, parallelogram, are given in $(x,y)$ coordinates by inequalities. One pair of parallel edges gives $$E \leq A x + B y \leq F,$$ the other pair of parallel edges $$G \leq C x + D y \leq H,$$ for constants $A,B,C,D,E,F,G,H.$ If all four inequalities are strict, the new point is strictly inside the figure, if one of the four indicated inequalities is actually an equation the point is on an edge.

Meanwhile, the four boundary equations such as $Ax+By=E$ are just the lines between consectutive vertices of the figure. You need to exercise care in choosing $\pm$ signs throughout, as writing it this way necessitates $E<F$ and $G <H.$

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Let's give the names $A$, $B$, $C$, $D$ to the four vertices of the rhombus. Let $Q$ be the center of the rhombus. You can calculate $Q$ as either $\tfrac{1}{2}(A+C)$ or $\tfrac{1}{2}(B+D)$. Also, let $a = \tfrac{1}{2}\Vert C - A \Vert$ and $b = \tfrac{1}{2}(\Vert D - B \Vert$. We're going to work in a coordinate system that has its origin at $Q$, its $x$-axis along the line $AC$, and its $y$-axis along the line $BD$. Let $U$ and $V$ denote unit vectors along these $x$ and $y$ axes respectively. In fact, $U = (C - A)/2a$ and $V = (D - B)/2b$. All of this is shown in the figure below:

In our chosen coordinate system, $C$ has coordinates $(a,0)$ and $D$ has coordinates $(0,b)$, so the line $DC$ has equation

$$\frac{x}{a} + \frac{y}{b} = 1$$

Using symmetry, the other three sides have similar eqautions. I think it's fairly clear that the point $(x,y)$ is inside the rhombus if (and only if):

$$\frac{\vert x \vert}{a} + \frac{\vert y \vert}{b} \le 1$$

In your code, you should first pre-compute (once):

Q = 0.5*(A + C);           // center point
a = 0.5*distance(A,C);     // half-width (in the x-direction)
b = 0.5*distance(B,D);     // half-height (y-direction)
U = (C - A)/(2*a);         // unit vector in x-direction
V = (D - B)/(2*b);         // unit vector in y-direction


Then, given a point $P$, the inclusion test is as follows:

W = P - Q;
xabs = abs(W*U);    // here W*U is the dot product of W and U
yabs = abs(W*V);    // here W*V is the dot product of W and V
if (xabs/a + yabs/b <= 1) then inside = true;


So, pretty efficient -- you can decide inclusion just by doing a vector subtraction and two dot products. All of the above assumes you have functions for doing basic vector arithmetic, of course (addition, subtraction, dot products).

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Here is one way of doing it, perhaps not the most elegant. First note that the rhombus is the intersection of 4 half planes determined by the lines that bound the rhombus. If you know the 4 points then you can figure out the 4 equations of the lines that bound the rhombus and therefore the 4 inequalities that define the desired half-planes. Therefore a point $p$ will be in the rhombus iff it is in all of the half-planes that determine the rhombus.

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I would do this with some vector math. Take two edges of your rhombus and consider them unit vectors of a custom coordinate system. Transform the point to this system. If the point has its x coordinate between 0 and 1 as well as its y coordinate between 0 and 1, it is inside.

This should work well for any true rhombus. (Including exactly vertical and horizontal edges, but excluding things with an area of zero)

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Here is a single equation that gives you an idea of when a point lies inside , outside or on the boundary of a regular octagon of radius 15. You can easily write in your test a meaningful and useful expression .( I have written out the equation for the boundary of an octagon, inside use the equation in the form of an inequality , and the same for points that are exterior of the octagon with the sense of the inequality turned around. ) Hope this helps.

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Oops ! My bad, next time I'll read the question before replying !!! –  Alan Jun 29 at 17:19