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.

  • $\begingroup$ Hint: The intersection point of two lines in homogeneous coordinates is $P=(a_1,b_1,c_1) \times (a_2,b_2,c_2)$ where $\times$ is the vector cross product. So three lines can be transformed two three vertices. $\endgroup$ Commented Aug 18, 2014 at 13:50
  • $\begingroup$ That numerator is perplexing. Is it supposed to be a 6-by-6 matrix determinant? If so, it vanishes since the first and third rows are identical. So I'd check for transcription errors. $\endgroup$ Commented Aug 20, 2014 at 1:41
  • $\begingroup$ No it's not a 6x6 but it's a 3x3 matrix determinant in which the elements themselves are 2x2 determinants $\endgroup$
    – najayaz
    Commented Aug 20, 2014 at 8:29

6 Answers 6


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:

enter image description here


$$C_1 = \left|\begin{array}{cc} \cos\phi & \sin\phi \\ \cos\psi & \sin\psi \end{array} \right| = \sin\psi\cos\phi - \cos\psi\sin\phi = \sin(\psi-\phi) = \sin \angle ROQ = \sin A$$ 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 $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = d$$


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

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

  • 5
    $\begingroup$ Ah, good old Euclidian geometry! (+1) for an answer that leads to understanding where the formula comes from. $\endgroup$ Commented Aug 23, 2014 at 15:33
  • $\begingroup$ Can you explain where the sides( when you divided by √a^2+b^2 throughout) a/√a^2+b^2 OR b /√a^2+b^2 exist. this is really getting me into circles!!! $\endgroup$
    – Riya Verma
    Commented Nov 19, 2019 at 20:27
  • $\begingroup$ @RiyaVerma: Writing $ux+vy-w=0$ as $y=-\frac{u}{v}+\frac{w}{v}$ shows that $-u/v$ is the line's slope, so $+v/u$ is the perpendicular (aka, "normal") slope. Thus, $u$ & $v$ are the normal's "run" & "rise", so that $u=k\cos t$ & $v=k\sin t$, for some $k>0$, where $t$ is the angle the normal makes with the $x$-axis. But, it's not just "some $k$": since $u^2+v^2=k^2\cos^2t+k^2\sin^2t=k^2$, we have $k:=\sqrt{u^2+v^2}$. Dividing the original eqn by this $k$ reduces the $u$ and $v$ coefficients to $\cos t$ and $\sin t$. (I'll leave you to show that $w/k$ is the original line's distance from origin.) $\endgroup$
    – Blue
    Commented Nov 20, 2019 at 3:47
  • 1
    $\begingroup$ @harry: A proof author "should" take care to cover edge-cases and degeneracies and whatnot, at least off-the-page. Whether these aspects get included in the proof as presented depends upon the nature of the platform (eg, an informal StackExchange answer vs a formal academic journal), how well the author can expect the audience to fill-in any holes, etc. (A common cliché is to leave verifying such details as "an exercise to the reader".) But there are limits to how much hand-waving can be done, especially with deceptively-simple problems (eg, the Collatz conjecture). (continued) $\endgroup$
    – Blue
    Commented Sep 5, 2021 at 5:04
  • 1
    $\begingroup$ @harry: (continuing) Occasionally, holes in presented proofs are trickier to fill than expected. (See the history of Fermat's Last Theorem.) ... In any case, ideally, one presents a proof in a context that naturally covers the edge cases (eg, making allowances for signed distances, infinite values, etc), but there are reasonable trade-offs between bogging-down in details and providing a conceptual overview. As a practical matter: If/when one seeks to publish in a journal, the editors and peer-reviewers will help determine when a proof has too much, too little, or just-enough detail. $\endgroup$
    – Blue
    Commented Sep 5, 2021 at 5:13

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:
enter image description here
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]]);
This completes the proof.


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 $$\verb/Area/(T) = \frac{\Delta^2}{2C_1C_2C_3} = \frac{\left|\begin{matrix} a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3\\ \end{matrix}\right|^2 }{2C_1C_2C_3} $$


equations in the reduced form




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| $

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

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







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



  • 3
    $\begingroup$ This doesn't seem to answer the question, which is about the derivation of this formula. $\endgroup$
    – Xander Henderson
    Commented Feb 18, 2020 at 1:02

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 2nd and 3rd 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


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



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.


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