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
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