Let on the x-y plane, the distance between the points $A(x_1,y_1)$ and $B(x_2,y_2)$ be $d$. Another point $P(a,b)$ satisfies the equations $x_1a+y_1b=1$ and $x_2a+y_2b=1$ and the distance between point $P$ from the origin,$O(0,0)$ is $p$. Find the area of triangle $\Delta OAB$.
My solution: Define a non-singular square matrix $$A=\begin{bmatrix} x_{1} & y_{1} \\ x_{2} & y_{2} \end{bmatrix}$$.
The required area, $\Delta = \frac{1}{2} \begin{vmatrix} x_{1} & y_{1} \\ x_{2} & y_{2} \end{vmatrix} = \frac{1}{2} \det(A)$
The given condition of point $P$ can be written as
$$\begin{bmatrix} x_{1} & y_{1} \\ x_{2} & y_{2} \end{bmatrix} \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$$
$$\implies \begin{bmatrix} a \\ b \end{bmatrix} = A^{-1} \begin{bmatrix} 1 \\ 1 \end{bmatrix}$$ (on pre multiplication of both sides with $A^{-1}$)
The inverse of matrix $A$,
$A^{-1} = \frac{1}{2 \Delta} \begin{bmatrix} y_{2} & -y_{1} \\ -x_{2} & x_{1} \end{bmatrix}$
$$\implies \begin{bmatrix} a \\ b \end{bmatrix} = \frac{\begin{bmatrix} y_{2}-y_{1} \\ -x_{2}+x_{1} \end{bmatrix}}{2 \Delta}$$.
Comparing element wise and taking sum of their squares,
$p^2 = \frac{d^2}{4 \Delta^2} \implies \Delta = \frac{d}{2p}$
Is there any other way to solve this question?
Points $A$ and $B$ lie on the line $\ell \ldots ax+by-1=0$. The distance of $\ell$ from the origin is $$d(\ell,0) = \frac{|a\cdot 0+b\cdot 0-1|}{\sqrt{a^2+b^2}}=\frac1p$$ since $\sqrt{a^2+b^2} = d(P,O) = p$. The area of your triangle is $$\text{area} = \frac12 \,\text{base}\cdot \text{height} = \frac12 d(A,B)\cdot d(\ell,O) = \frac12 \cdot d \cdot \frac1p = \frac{d}{2p}.$$