Area of △ABC = \(\frac 12\begin{vmatrix} 2&3&1\\5&7&1\\-3&-1&1\end{vmatrix}\) \(\begin{pmatrix} \because A = (2,3),\\B=(5,7)\text{ &}\\C = (-3, -1)\end{pmatrix}\)
\(= \frac 12 (2(7 + 1) - 3(5 + 3) + 1(-5+ 21))\)
\(= \frac 12 (16 - 24 + 16)\)
\(= 4\,sq.\,units\)
After shift the origin (0, 0) to (-1, 3)
Points A, B & C are interchanged into
\(A(1, 6), B (4, 10) , C(-4, 2)\)
\(\therefore \) Area of new △ABC = \(\frac 12 \begin{vmatrix}1&6&1\\4&10&1\\-4&2&1\end{vmatrix}\)
\(= \frac 12 (1 (10 - 2) - 6 (4 + 4) + 1(8 + 40))\)
\(= \frac 12 (8 - 48 + 48)\)
\(= 4\,sq.\,units\)
Hence, area of triangle remains unchanged.
Hence Proved.