Direct formula for area of a triangle formed by three lines, given their equations in the cartesian plane.

Question

I read this formula in some book but it didn't provide a proof so I thought someone on this website could figure it out. What it says is: If we consider 3 non-concurrent, non parallel lines represented by the equations : $$a_1x+b_1y+c_1=0$$ $$a_2x+b_2y+c_2=0$$ $$a_3x+b_3y+c_3=0$$ Then the area of the triangle that these lines will enclose is given by the magnitude of : $$\frac{det\begin{bmatrix}a_1 & b_1 & c_1\\a_2 & b_2 & c_2\\a_3 & b_3 & c_3\end{bmatrix}^2}{2C_1C_2C_3}$$ Where $C_1,C_2,C_3$ are the co-factors of $c_1,c_2,c_3$ respectively in the above matrix.

What I'm wondering is, where did this come from? And why isn't it famous? Earlier we had to calculate areas by finding the vertices and all but this does it in a minute or so and thus deserves more familiarity.

Answer 1

Clearly, we can scale the coefficients of a given linear equation by any (non-zero) constant and the result is unchanged. Therefore, by dividing-through by $\sqrt{a_i^2+b_i^2}$, we may assume our equations are in "normal form":

$$\begin{align} x \cos\theta + y \sin\theta - p &= 0 \\ x \cos\phi + y \sin\phi - q &= 0 \\ x \cos\psi + y \sin\psi - r &= 0 \end{align}$$

with $\theta$, $\phi$, $\psi$ and $p$, $q$, $r$ (and $A$, $B$, $C$ and $a$, $b$, $c$) as in the figure:

Then

$$C_1 = \left|\begin{array}{cc} \cos\phi & \sin\phi \\ \cos\psi & \sin\psi \end{array} \right| = \sin\psi\cos\phi - \cos\psi\sin\phi = \sin(\psi-\phi) = \sin \angle ROQ = \sin A$$ Likewise, $$C_2 = \sin B \qquad C_3 = \sin C$$

Moreover, $$D := \left|\begin{array}{ccc} \cos\theta & \sin\theta & - p \\ \cos\phi & \sin\phi & - q \\ \cos\psi & \sin\psi & - r \end{array}\right| = - \left( p C_1 + q C_2 + r C_3 \right) = - \left(\;p \sin A + q \sin B + r \sin C\;\right)$$

Writing $d$ for the circumdiameter of the triangle, the Law of Sines tells us that $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = d$$

Therefore,

$$\begin{align} D &= - \left( \frac{ap}{d} + \frac{bq}{d} + \frac{cr}{d} \right) \\[4pt] &= -\frac{1}{d}\left(\;ap + b q + c r\;\right) \\[4pt] &= -\frac{1}{d}\left(\;2|\triangle COB| + 2|\triangle AOC| + 2|\triangle BOA| \;\right) \\[4pt] &= -\frac{2\;|\triangle ABC|}{d} \end{align}$$

Also, $$C_1 C_2 C_3 = \sin A \sin B \sin C = \frac{a}{d}\frac{b}{d}\sin C= \frac{2\;|\triangle ABC|}{d^2}$$

Finally: $$\frac{D^2}{2C_1C_2C_3} = \frac{4\;|\triangle ABC|^2/d^2}{4\;|\triangle ABC|/d^2} = |\triangle ABC|$$

Answer 2

Let $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ be the vertices of the triangle. Then the three non-concurrent, non parallel lines can be represented by: $$(y_2-y_1)(x-x_1)-(x_2-x_1)(y-y_1)=0$$ $$(y_3-y_2)(x-x_2)-(x_3-x_2)(y-y_2)=0$$ $$(y_1-y_3)(x-x_3)-(x_1-x_3)(y-y_3)=0$$ The coefficients in the OP's question are thus represented by: $$ a_1 = (y_2-y_1) \quad ; \quad b_1 = -(x_2-x_1) \quad ; \quad c_1 = -(y_2-y_1)x_1+(x_2-x_1)y_1 \\ a_2 = (y_3-y_2) \quad ; \quad b_2 = -(x_3-x_2) \quad ; \quad c_2 = -(y_3-y_2)x_2+(x_3-x_2)y_2 \\ a_3 = (y_1-y_3) \quad ; \quad b_3 = -(x_1-x_3) \quad ; \quad c_3 = -(y_1-y_3)x_3+(x_1-x_3)y_3 $$ Now straightforward calculation should reveal that: $$ \frac{det\begin{bmatrix}a_1 & b_1 & c_1\\a_2 & b_2 & c_2\\a_3 & b_3 & c_3\end{bmatrix}^2}{2C_1C_2C_3} =\frac{1}{2} det\begin{bmatrix}(x_2-x_1) & (y_2-y_1)\\(x_3-x_1) & (y_3-y_1)\end{bmatrix} $$ Where the latter (half) determinant certainly represents the area of the triangle:

The algebra is somewhat tedious. Therefore I've invoked MAPLE to save time and effort:

A := array([[a1,b1,c1],[a2,b2,c2],[a3,b3,c3]]);
C1 := det(array([[a2,b2],[a3,b3]]));
C2 := -det(array([[a1,b1],[a3,b3]]));
C3 := det(array([[a1,b1],[a2,b2]]));
B := array([[x2-x1,y2-y1],[x3-x1,y3-y1]]);
verify(det(A)^2/(2*C1*C2*C3),det(B)/2,equal);
                        true

This completes the proof.

Answer 3

equations in the reduced form

$y=ax+b$

$y=cx+d$

$y=ex+f$

solving by the Cramer rule we get the A, B and C points

calculating the area of the ABC triangle

$ΔABC=\frac{1}{2}\left| \begin{array}{} \frac{b-d}{c-a} & \frac{bc-ad}{c-a} & 1 \\ \frac{d-f}{e-c} & \frac{de-cf}{e-c} & 1 \\ \frac{f-b}{a-e} & \frac{af-be}{a-e} & 1 \\ \end{array} \right| $

$ΔABC=\frac{1}{2(c-a)(e-c)(a-e)}\left| \begin{array}{} b-d & bc-ad & c-a \\ d-f & de-ef & e-c \\ f-b & af-be & a-e \\ \end{array} \right| $

$ΔABC=\frac{1}{2(c-a)(e-c)(a-e)}\left [ a(d-f)+c(f-b)+e(b-d) \right ] ^{2}$

$ΔABC=\frac{1}{2}\frac{\left|\begin{array}{} a & b & 1 \\ c & d & 1\\ e & f & 1 \\ \end{array} \right| ^{2}}{\left| \begin{array}{} a & a^{2} & 1 \\ c & c^2 & 1 \\ e & e^2 & 1 \\ \end{array} \right| }$

$a=\frac{-a_1}{b_1}$

$b=\frac{-c_1}{b_1}$

$c=\frac{-a_2}{b_2}$

$d=\frac{-c_2}{b_2}$

$e=\frac{-a_3}{b_3}$

$f=\frac{-c_3}{b_3}$

$ΔABC=|\frac{1}{2}\frac{\left| \begin{array}{} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2\\ a_3 & b_3 & c_3 \\ \end{array} \right| ^2}{\left| \begin{array}{} a_1^2 & b_1^2 & a_1b_1 \\ a_2^2 & b_2^2 & a_2b_2 \\ a_3^2 & b_3^2 & a_3b_3 \\ \end{array} \right| }|$

$ΔABC=|\frac{1}{2}\frac{Δ^2}{δ}|$

Answer 4

Let a₁x + b₁y + c₁ = O be the equation of the first side.( We will represent this by L₁=O )

Similarly a₂x + b₂y + c₂ = O and a₃x + b₃y + c₃ = O be the equation of 2^nd and 3^rd side respectively ,represented by L₂ and L₃ .

Let (x₁ , y₁) , (x₂ , y₂) , (x₃, y₃) be the vertices of the Triangle

So The Area of the Triangle will be the Determinant $\Delta$_o
\begin{align*} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \\ \end{vmatrix}. \end{align*} Consider another Determinant $\Delta$₁ \begin{align*} \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}. \end{align*}

Let Determinant $\Delta$₂ be $\Delta$_o $\Delta$₁ which is equal to the Determinant \begin{align*} \begin{vmatrix} a_1x_1+b_1y_1+c_1 & a_2x_1+b_2y_1+c_2 & a_3x_1+b_3y_1+c_3 \\ a_1x_2+b_1y_2+c_1 & a_2x_2+b_2x_2+c_2 & a_3x_2+b_3y_2+c_3 \\ a_1x_3+b_1y_3+c_3 & a_2x_3+b_3x_3+c_3 & a_3x_3+b_3x_3+c_3 \end{vmatrix}. \end{align*}
But (x₁ ,y₁) lies on L₂ and L₃

So a₁x₂ + b₁y₂+ c₁ = O and a₁x₃ + b₁y₃+ c₁ = O Similarly

a₂x₁ + b₂y₁+ c₂ = O a₂x₃ + b₂y₃+ c₂ = O and a₃x₁ + b₃y₁+ c₃ = O , a₃x₂ + b₃y₂+ c₃ = O

So the Determinant reduces to

\begin{align*} \begin{vmatrix} a_1x_1+b_1y_1+c_1 & 0 & 0 \\ 0 & a_2x_2+b_2x_2+c_2 & 0 \\ 0 & 0 & a_3x_3+b_3x_3+c_3 \end{vmatrix}. \end{align*} So $\Delta$₂ = L₁(x₁,y₁)L₂(x₂,y₂)L₃(x₃,y₃)

Now we shall prove that

a₁x₁+b₁y₁+c₁ = K₁(Let) = $\Delta$₁/a₃b₂-a₂b₃

Now we observe that (x₁,y₁) is the solution of the system of equations

a₁x₁+b₁y₁+c₁-K₁ = O a₂x₁+b₂y₁+c₂=O a₃x₁+b₃y₁+c₃=O

So \begin{align*} \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1-K_1 & c_2 & c_3 \end{vmatrix}. \end{align*} =0=$\Delta$₁ - $\Delta$₃ Where $\Delta$₃ = \begin{align*} \begin {vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ K & 0 & 0 \end{vmatrix}. \end{align*}

So $\Delta$₃ = K₁(a₃b₂-₂b₃)

So 0 = $\Delta$₁ - K₁(a₃b₂-₂b₃)

Therefore K₁= $\Delta$₁/a₃b₂-a₂b₃

So K₁ =L₁(x₁,y₁) = $\Delta$₁/a₃b₂-a₂b₃

Similarly K₂ =L₂(x₂,y₂) = $\Delta$₁/a₃b₁-a₁b₃

and

K₃ =L₃(x₃,y₃) = $\Delta$₁/a₁b₂-a₂b₁

Since C₁= a₃b₂-a₂b₃ and C₂= a₃b₁-a₁b₃ and C₃= a₁b₂-a₂b₁

So $\Delta$_o$\Delta$₁=L₁(x₁,y₁) L₂(x₂,y₂) L₃(x₃,y₃)= K₁K₂K₃ = $\Delta$³/C₁C₂C₃

And Therefore the area of the triangle is

$\Delta$²/C₁C₂C₃

Answer 5

Here is a proof of a special case. Let the three lines be $L_1,L_2,L_3$. A special case is $L_1$ and $L_2$ as the $x$- and $y$-axes respectively, and $L_3$ of the form $\dfrac xa+\dfrac yb=1$. Then the three lines form a right triangle with area $\dfrac{1}{2} ab$. On the other hand, the formula above gives an area of $$ \frac{1}{2 C_1 C_2 C_3}\left|\begin{matrix}1&0&0\\0&1&0\\a^{-1} & b^{-1} & -1\end{matrix}\right|^2=\frac{(-1)^2}{2(-a^{-1})(b^{-1})(1)}=-\frac12 ab$$ Since the triangle has a negative orientation (the order of the three lines is clockwise around the triangle) this matches the result above.

This may seem a good deal weaker than the general case. But we can map any generic set of lines to the set here by the following transformations: Translate one of the intersections to the origin, rotate the triangle into the upper half-plane such that one of its lines becomes the $x$-axis, and finally do a horizontal shear to make one of the remaining lines the $y$-axis. So all that should remain is to show that the above formula is preserved by these transformations---which, I'll confess, I don't know how to do off the top of my head. So as yet this is an incomplete proof of the general case.

Answer 6

Following is a derivation of the formula using homogeneous coordinates.

For any point $p = (u,v)$ on the plane, we will use $\vec{p} = (x,y,z) = (zu,zv, z)$ to denote any choice of its homogeneous coordinates. For any line $\ell = \{ (u,v) : au+bv+c = 0 \}$, we will associate it with a vector $\vec{\ell} = (a,b,c)$ as use it as a homogeneous coordinates of $\ell$.

In terms of homogeneous coordinates, the condition that point $p$ lies on line $\ell$ can be expressed as

$$au + bv + c = 0 \quad\iff\quad ax + by + cz = 0 \quad\iff\quad \vec{\ell}\cdot \vec{p} = 0$$

Let $\ell_1, \ell_2, \ell_3$ be $3$ lines on the plane forming the sides of a triangle $T$ in counterclockwise order.

Let $p_1 = (u_1,v_1)$ be the intersection of $\ell_2$ and $\ell_3$, we have $$p_1 \in \ell_2 \cap \ell_3 \quad\implies\quad \vec{\ell}_2\cdot \vec{p}_1 = \vec{\ell}_3 \cdot \vec{p}_1 = 0 \quad\implies\quad \vec{p}_1 \propto \vec{\ell}_2 \times \vec{\ell}_3$$ Let $p_2 = (u_2,v_2)$ be the intersection of $\ell_3$ and $\ell_1$, we have $$p_2 \in \ell_3 \cap \ell_1 \quad\implies\quad \vec{\ell}_3\cdot \vec{p}_2 = \vec{\ell}_1 \cdot \vec{p}_2 = 0 \quad\implies\quad \vec{p}_2 \propto \vec{\ell}_3 \times \vec{\ell}_1$$ Let $p_3 = (u_3,v_3)$ be the intersection of $\ell_1$ and $\ell_2$. we have $$p_3 \in \ell_1 \cap \ell_2 \quad\implies\quad \vec{\ell}_1\cdot \vec{p}_3 = \vec{\ell}_2 \cdot \vec{p}_3 = 0 \quad\implies\quad \vec{p}_3 \propto \vec{\ell}_1 \times \vec{\ell}_2 $$

Since $\vec{p}_1,\vec{p}_2,\vec{p}_3$ are homogeneous coordinates and rescaling them doesn't change the underlying points $p_1, p_2, p_3$, we can just pick $$ \vec{p}_1 = \vec{\ell}_2 \times \vec{\ell}_3,\quad \vec{p}_2 = \vec{\ell}_3 \times \vec{\ell}_1\quad\text{ and }\quad \vec{p}_3 = \vec{\ell}_1 \times \vec{\ell}_2 $$

In terms of them, the area of triangle $T$ is given by $$\verb/Area/(T) = \frac12 \left| \begin{matrix} u_1 & u_2 & u_3\\ v_1 & v_2 & v_3\\ 1 & 1 & 1\\ \end{matrix}\right| = \frac{1}{2z_1z_2z_3} \left|\begin{matrix} x_1 & x_2 & x_3\\ y_1 & y_2 & y_3\\ z_1 & z_2 & z_3\\ \end{matrix}\right| = \frac{ \vec{p}_1\cdot (\vec{p_2} \times \vec{p}_3) }{ 2 (\vec{p}_1\cdot \hat{z})(\vec{p}_2\cdot \hat{z})(\vec{p}_3\cdot \hat{z}) } $$

Notice $$\require{cancel} \vec{p}_2 \times \vec{p}_3 = (\vec{\ell}_3 \times \vec{\ell}_1) \times (\vec{\ell}_1 \times \vec{\ell}_2) = ((\vec{\ell}_3 \times \vec{\ell}_1) \cdot \vec{\ell}_2) \vec{\ell}_1 - \color{red}{\cancelto{0}{\color{gray}{((\vec{\ell}_3 \times \vec{\ell}_1) \cdot \vec{\ell}_1)}}} \vec{\ell}_2 = \Delta \vec{\ell}_1 $$ where $\Delta = \vec{\ell}_1\cdot (\vec{\ell}_2 \times \vec{\ell}_3) = \left|\begin{matrix} a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3\\ \end{matrix}\right|$, we find $$\vec{p}_1\cdot(\vec{p}_2\times\vec{p}_3) = \Delta( \vec{\ell}_2 \times \vec{\ell}_3 ) \cdot \vec{\ell}_1 = \Delta^2 $$ Together with $$\vec{p}_1 \cdot \hat{z} = (\vec{\ell}_2 \times \vec{\ell}_3 ) \cdot \hat{z} = \left|\begin{matrix} a_2 & a_3 & 0\\ b_2 & b_3 & 0\\ c_2 & c_3 & 1\\ \end{matrix}\right| = C_1 \stackrel{def}{=} \left|\begin{matrix} a_2 & a_3\\ b_2 & b_3\\ \end{matrix}\right| $$ and similar relations $$\vec{p}_2\cdot\hat{z} = C_2 \stackrel{def}{=} \left|\begin{matrix} a_3 & a_1\\ b_3 & b_1\\ \end{matrix}\right| \quad\text{ and }\quad \vec{p}_3\cdot\hat{z} = C_3 \stackrel{def}{=} \left|\begin{matrix} a_1 & a_2\\ b_1 & b_2\\ \end{matrix}\right|$$ we obtain $$\verb/Area/(T) = \frac{\Delta^2}{2C_1C_2C_3} = \frac{\left|\begin{matrix} a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3\\ \end{matrix}\right|^2 }{2C_1C_2C_3} $$

Answer 7

Let a₁x + b₁y + c₁ = O be the equation of the first side.( We will represent this by L₁=O )

Similarly a₂x + b₂y + c₂ = O and a₃x + b₃y + c₃ = O be the equation of 2^nd and 3^rd side respectively ,represented by L₂ and L₃ .

Let (x₁ , y₁) , (x₂ , y₂) , (x₃, y₃) be the vertices of the Triangle

So The Area of the Triangle will be the determinant $\Delta$_o
\begin{align*} \begin{vmatrix} x1 & y1 & 1 \\ x2 & y2 & 1 \\ x3 & y3 & 1 \\ \end{vmatrix}. \end{align*} Consider another determinant $\Delta$₁ \begin{align*} \begin{vmatrix} a1 & a2 & a3 \\ b1 & b2 & b3 \\ c1 & c2 & c3 \end{vmatrix}. \end{align*}

Let determinant $\Delta$₂ be $\Delta$_o $\Delta$₁ which is equal to the determinant \begin{align*} \begin{vmatrix} a1x1+b1y1+c1 & a2x1+b2y1+c2 & a3x1+b3y1+c3 \\ a1x2+b1y2+c1 & a2x2+b2x2+c2 & a3x2+by2+c3 \\ a1x3+b1y3+c3 & a2x3+b3x3+c3 & a3x3+b3x3+c3 \end{vmatrix}. \end{align*}
But (x₁ ,y₁) lies on L₂ and L₃

So a₁x₂ + b₁y₂+ c₁ = O and a₁x₃ + b₁y₃+ c₁ = O Similarly

a₂x₁ + b₂y₁+ c₂ = O a₂x₃ + b₂y₃+ c₂ = O and a₃x₁ + b₃y₁+ c₃ = O , a₃x₂ + b₃y₂+ c₃ = O

So the determinant reduces to

\begin{align*} \begin{vmatrix} a1x1+b1y1+c1 & 0 & 0 \\ 0 & a2x2+b2x2+c2 & 0 \\ 0 & 0 & a3x3+b3x3+c3 \end{vmatrix}. \end{align*} So $\Delta$₂ = L₁(x₁,y₁)L₂(x₂,y₂)L₃(x₃,y₃)

Now we shall prove that

a₁x₁+b₁y₁+c₁ = K₁(Let) = $\Delta$₁/a₃b₂-a₂b₃

Now we observe that (x₁,y₁) is the solution of the system of equations

a₁x₁+b₁y₁+c₁-K₁ = O a₂x₁+b₂y₁+c₂=O a₃x₁+b₃y₁+c₃=O

So \begin{align*} \begin{vmatrix} a1 & a2 & a3 \\ b1 & b2 & b3 \\ c1-K1 & c2 & c3 \end{vmatrix}. \end{align*} =0=$\Delta$₁ - $\Delta$₃ Where $\Delta$₃ = \begin{align*} \begin {vmatrix} a1 & a2 & a3 \\ b1 & b2 & b3 \\ K & 0 & 0 \end{vmatrix}. \end{align*}

So $\Delta$₃ = K₁(a₃b₂-₂b₃)

So O = $\Delta$₁ - K₁(a₃b₂-₂b₃)

Therefore K₁= $\Delta$₁/a₃b₂-a₂b₃

So K₁ =L₁(x₁,y₁) = $\Delta$₁/a₃b₂-a₂b₃

Similarly K₂ =L₂(x₂,y₂) = $\Delta$₁/a₃b₁-a₁b₃

and

K₃ =L₃(x₃,y₃) = $\Delta$₁/a₁b₂-a₂b₁

Since C₁= a₃b₂-a₂b₃ and C₂= a₃b₁-a₁b₃ and C₃= a₁b₂-a₂b₁

So $\Delta$_o$\Delta$₁=L₁(x₁,y₁) L₂(x₂,y₂) L₃(x₃,y₃)= K₁K₂K₃ = $\Delta$³/C₁C₂C₃

And Therefore the area of the triangle is

$\Delta$²/C₁C₂C₃