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You can find if a point is in a triangle using barycentric coordinates.

If you have a triangle with vertices $A$, $B$ & $C$, & a point P in the plane of the triangle, you simply need to find:

$ AreaABC = \frac{ \left| \overline{AB} \times \overline{AC} \right| }{ 2 } $

$ \alpha = \frac{ \left| \overline{PB} \times \overline{PC} \right| }{ 2AreaABC } $

$ \beta = \frac{ \left| \overline{PC} \times \overline{PA} \right| }{ 2AreaABC } $

$ \gamma = 1 − \alpha − \beta $

Here $\alpha$ is the ratio of the area of a subtriangle $PBC$ over the area of the whole triangle $ABC$, as shown in this image from Peter Shirley's book:

If ALL of the following 4 restrictions are met:

• $ 0 \le \alpha \le 1 $
• $ 0 \le \beta \le 1 $
• $ 0 \le \gamma \le 1 $
• $\alpha + \beta + \gamma = 1$

then the point P is inside the triangle.

If ANY of $\alpha$,$\beta$,$\gamma$ are outside those ranges, or if the sum of $ \alpha + \beta + \gamma \ne 1 $ then the point P is not inside the triangle.

Notice how we exploit the 4th condition ($\alpha + \beta + \gamma = 1$) in finding $\gamma$.

Even though we find $\gamma$ that way, we still must check that $ 0 \le \alpha \le 1 $ and $ 0 \le \beta \le 1 $ and $ 0 \le \gamma \le 1 $ for the point to be inside the triangle.

$\gamma$ can easily be outside $[0, 1]$. Say $\alpha=0.99$ and $\beta=0.85$. Then $\gamma = 1 - 0.99 - 0.85 = -0.84$, and the point that got us those barycentric coordinates would be outside the triangle.

Things to observe:

• When one of ($\alpha$, $\beta$, $\gamma$) is 1 and the other two are 0, then the point P is exactly at a vertex of the triangle.

• When one of ($\alpha$, $\beta$, $\gamma$) is 0, and the other 2 coordinates are between 0 and 1, the point P is on an edge of the triangle.

You can find if a point is in a triangle using barycentric coordinates.

If you have a triangle with vertices $A$, $B$ & $C$, & a point P in the plane of the triangle, you simply need to find:

$AreaABC = \frac{ \left| \overline{AB} \times \overline{AC} \right| }{ 2 }$

$\alpha = \frac{ \left| \overline{PB} \times \overline{PC} \right| }{ 2AreaABC }$

$\beta = \frac{ \left| \overline{PC} \times \overline{PA} \right| }{ 2AreaABC }$

$\gamma = 1 − \alpha − \beta$

Here $\alpha$ is the ratio of the area of a subtriangle $PBC$ over the area of the whole triangle $ABC$, as shown in this image from Peter Shirley's book:

If ALL of the following 4 restrictions are met:

• $ 0 \le \alpha \le 1 $
• $ 0 \le \beta \le 1 $
• $ 0 \le \gamma \le 1 $
• $\alpha + \beta + \gamma = 1$

then the point P is inside the triangle.

If ANY of $\alpha$,$\beta$,$\gamma$ are outside those ranges, or if the sum of $ \alpha + \beta + \gamma \ne 1 $ then the point P is not inside the triangle.

Notice how we exploit the 4th condition ($\alpha + \beta + \gamma = 1$) in finding $\gamma$.

Even though we find $\gamma$ that way, we still must check that $ 0 \le \alpha \le 1 $ and $ 0 \le \beta \le 1 $ and $ 0 \le \gamma \le 1 $ for the point to be inside the triangle.

$\gamma$ can easily be outside $[0, 1]$. Say $\alpha=0.99$ and $\beta=0.85$. Then $\gamma = 1 - 0.99 - 0.85 = -0.84$, and the point that got us those barycentric coordinates would be outside the triangle.

Things to observe:

• When one of ($\alpha$, $\beta$, $\gamma$) is 1 and the other two are 0, then the point P is exactly at a vertex of the triangle.

• When one of ($\alpha$, $\beta$, $\gamma$) is 0, and the other 2 coordinates are between 0 and 1, the point P is on an edge of the triangle.

A common technique in a computer program is to use barycentric coordinates.

Barycentric coordinates are a lot easier to find than any web resources indicate, so I'm not linking to them.

The easiest way to obtain barycentric coordinates of a point P, given a triangle with vertices described by the vectors A, B, C is likely this method:

$ AreaABC = \frac{ \left| \overline{AB} \times \overline{AC} \right| }{ 2 } $

$ \alpha = \frac{ \left| \overline{PB} \times \overline{PC} \right| }{ 2AreaABC } $

$ \beta = \frac{ \left| \overline{PC} \times \overline{PA} \right| }{ 2AreaABC } $

$ \gamma = 1 − \alpha − \beta $

Here $\alpha$ is the ratio of the area of a subtriangle PBC over the area of the whole triangle ABC, as shown in this image from Peter Shirley's book:

If ALL of the following 4 restrictions are met:

• $ 0 \le \alpha \le 1 $
• $ 0 \le \beta \le 1 $
• $ 0 \le \gamma \le 1 $
• $\alpha + \beta + \gamma = 1$

then the point P is inside the triangle.

If ANY of $\alpha$,$\beta$,$\gamma$ are outside those ranges, or if the sum of $ \alpha + \beta + \gamma \ne 1 $ then the point P is not inside the triangle.

Note also when one of $\alpha$,$\beta$,$\gamma$ is 0, and the other 2 coordinates are between 0 and 1, the point P is on an edge of the triangle.

When one of $\alpha$,$\beta$,$\gamma$ is 1 and the other two are 0, then the point P is exactly at a vertex of the triangle.

Of course, these computations assume P is already in the plane of the triangle. If P is not in the plane of the triangle, then you should project it there first, before computing the barycentric coordinates.

You can find if a point is in a triangle using barycentric coordinates.

If you have a triangle with vertices $A$, $B$ & $C$, & a point P in the plane of the triangle, you simply need to find:

$ AreaABC = \frac{ \left| \overline{AB} \times \overline{AC} \right| }{ 2 } $

$ \alpha = \frac{ \left| \overline{PB} \times \overline{PC} \right| }{ 2AreaABC } $

$ \beta = \frac{ \left| \overline{PC} \times \overline{PA} \right| }{ 2AreaABC } $

$ \gamma = 1 − \alpha − \beta $

Here $\alpha$ is the ratio of the area of a subtriangle $PBC$ over the area of the whole triangle $ABC$, as shown in this image from Peter Shirley's book:

If ALL of the following 4 restrictions are met:

• $ 0 \le \alpha \le 1 $
• $ 0 \le \beta \le 1 $
• $ 0 \le \gamma \le 1 $
• $\alpha + \beta + \gamma = 1$

then the point P is inside the triangle.

If ANY of $\alpha$,$\beta$,$\gamma$ are outside those ranges, or if the sum of $ \alpha + \beta + \gamma \ne 1 $ then the point P is not inside the triangle.

Notice how we exploit the 4th condition ($\alpha + \beta + \gamma = 1$) in finding $\gamma$.

Even though we find $\gamma$ that way, we still must check that $ 0 \le \alpha \le 1 $ and $ 0 \le \beta \le 1 $ and $ 0 \le \gamma \le 1 $ for the point to be inside the triangle.

$\gamma$ can easily be outside $[0, 1]$. Say $\alpha=0.99$ and $\beta=0.85$. Then $\gamma = 1 - 0.99 - 0.85 = -0.84$, and the point that got us those barycentric coordinates would be outside the triangle.

Things to observe:

• When one of ($\alpha$, $\beta$, $\gamma$) is 1 and the other two are 0, then the point P is exactly at a vertex of the triangle.

• When one of ($\alpha$, $\beta$, $\gamma$) is 0, and the other 2 coordinates are between 0 and 1, the point P is on an edge of the triangle.

A common technique in a computer program is to use barycentric coordinates.

Barycentric coordinates are a lot easier to find than any web resources indicate, so I'm not linking to them.

The easiest way to obtain barycentric coordinates of a point P, given a triangle with vertices described by the vectors A, B, C is likely this method:

$ AreaABC = \frac{ \left| \overline{AB} \times \overline{AC} \right| }{ 2 } $

$ \alpha = \frac{ \left| \overline{PB} \times \overline{PC} \right| }{ 2AreaABC } $

$ \beta = \frac{ \left| \overline{PC} \times \overline{PA} \right| }{ 2AreaABC } $

$ \gamma= \frac{ \left| \overline{PB} \times \overline{PA} \right| }{ 2AreaABC } $

Here $\alpha$ is the ratio of the area of a subtriangle PBC over the area of the whole triangle ABC, as shown in this image from Peter Shirley's book:

If ALL of the following 4 restrictions are met:

• $ 0 \le \alpha \le 1 $
• $ 0 \le \beta \le 1 $
• $ 0 \le \gamma \le 1 $
• $\alpha + \beta + \gamma = 1$

then the point P is inside the triangle.

If ANY of $\alpha$,$\beta$,$\gamma$ are outside those ranges, or if the sum of $ \alpha + \beta + \gamma \ne 1 $ then the point P is not inside the triangle.

Note also when one of $\alpha$,$\beta$,$\gamma$ is 0, and the other 2 coordinates are between 0 and 1, the point P is on an edge of the triangle.

When one of $\alpha$,$\beta$,$\gamma$ is 1 and the other two are 0, then the point P is exactly at a vertex of the triangle.

Of course, these computations assume P is already in the plane of the triangle. If P is not in the plane of the triangle, then you should project it there first, before computing the barycentric coordinates.

A common technique in a computer program is to use barycentric coordinates.

Barycentric coordinates are a lot easier to find than any web resources indicate, so I'm not linking to them.

The easiest way to obtain barycentric coordinates of a point P, given a triangle with vertices described by the vectors A, B, C is likely this method:

$ AreaABC = \frac{ \left| \overline{AB} \times \overline{AC} \right| }{ 2 } $

$ \alpha = \frac{ \left| \overline{PB} \times \overline{PC} \right| }{ 2AreaABC } $

$ \beta = \frac{ \left| \overline{PC} \times \overline{PA} \right| }{ 2AreaABC } $

$ \gamma = 1 − \alpha − \beta $

Here $\alpha$ is the ratio of the area of a subtriangle PBC over the area of the whole triangle ABC, as shown in this image from Peter Shirley's book:

If ALL of the following 4 restrictions are met:

• $ 0 \le \alpha \le 1 $
• $ 0 \le \beta \le 1 $
• $ 0 \le \gamma \le 1 $
• $\alpha + \beta + \gamma = 1$

then the point P is inside the triangle.

If ANY of $\alpha$,$\beta$,$\gamma$ are outside those ranges, or if the sum of $ \alpha + \beta + \gamma \ne 1 $ then the point P is not inside the triangle.

Note also when one of $\alpha$,$\beta$,$\gamma$ is 0, and the other 2 coordinates are between 0 and 1, the point P is on an edge of the triangle.

When one of $\alpha$,$\beta$,$\gamma$ is 1 and the other two are 0, then the point P is exactly at a vertex of the triangle.

Of course, these computations assume P is already in the plane of the triangle. If P is not in the plane of the triangle, then you should project it there first, before computing the barycentric coordinates.