Let a_{1}x + b_{1}y + c_{1} = O be the equation of the first side.( We will represent this by L_{1}=O )

Similarly a_{2}x + b_{2}y + c_{2} = O
and a_{3}x + b_{3}y + c_{3} = O be the equation of 2^{nd} and 3^{rd} side respectively ,represented by L_{2} and L_{3} .

Let (x_{1} , y_{1}) , (x_{2} , y_{2}) , (x_{3}, y_{3}) 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 (x_{1} ,y_{1}) lies on L_{2} and L_{3}

So a_{1}x_{2} + b_{1}y_{2}+ c_{1} = O
and a_{1}x_{3} + b_{1}y_{3}+ c_{1} = O
Similarly

a_{2}x_{1} + b_{2}y_{1}+ c_{2} = O
a_{2}x_{3} + b_{2}y_{3}+ c_{2} = O
and
a_{3}x_{1} + b_{3}y_{1}+ c_{3} = O
, a_{3}x_{2} + b_{3}y_{2}+ c_{3} = 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} = L_{1}(x_{1},y_{1})L_{2}(x_{2},y_{2})L_{3}(x_{3},y_{3})

Now we shall prove that

a_{1}x_{1}+b_{1}y_{1}+c_{1} = K_{1}(Let) = $\Delta$_{1}/a_{3}b_{2}-a_{2}b_{3}

Now we observe that (x_{1},y_{1}) is the solution of the system of equations

a_{1}x_{1}+b_{1}y_{1}+c_{1}-K_{1} = O
a_{2}x_{1}+b_{2}y_{1}+c_{2}=O
a_{3}x_{1}+b_{3}y_{1}+c_{3}=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} = K_{1}(a_{3}b_{2}-_{2}b_{3})

So 0 = $\Delta$_{1} - K_{1}(a_{3}b_{2}-_{2}b_{3})

Therefore K_{1}= $\Delta$_{1}/a_{3}b_{2}-a_{2}b_{3}

So K_{1} =L_{1}(x_{1},y_{1}) = $\Delta$_{1}/a_{3}b_{2}-a_{2}b_{3}

Similarly K_{2} =L_{2}(x_{2},y_{2}) = $\Delta$_{1}/a_{3}b_{1}-a_{1}b_{3}

and

K_{3} =L_{3}(x_{3},y_{3}) = $\Delta$_{1}/a_{1}b_{2}-a_{2}b_{1}

Since C_{1}= a_{3}b_{2}-a_{2}b_{3}
and C_{2}= a_{3}b_{1}-a_{1}b_{3}
and C_{3}= a_{1}b_{2}-a_{2}b_{1}

So $\Delta$_{o}$\Delta$_{1}=L_{1}(x_{1},y_{1}) L_{2}(x_{2},y_{2}) L_{3}(x_{3},y_{3})= K_{1}K_{2}K_{3} = $\Delta$^{3}/C_{1}C_{2}C_{3}

And Therefore the area of the triangle is

**$\Delta$**^{2}/C_{1}C_{2}C_{3}