Q.If $\begin{vmatrix} 2a & x_1 & y_1 \\ 2b & x_2 & y_2 \\ 2c & x_3 & y_3 \end{vmatrix} = \dfrac{abc}{2} \neq 0$, then the area of the triangle whose vertices are $\left(\dfrac{x_1}{a}, \dfrac{y_1}{a}\right),\ \left(\dfrac{x_2}{b}, \dfrac{y_2}{b}\right),\ \left(\dfrac{x_3}{c}, \dfrac{y_3}{c}\right)$ is

This tests area of a triangle via determinants and row operations.

Step 1. Area $= \tfrac{1}{2}\left|\,M\,\right|$ where $M = \begin{vmatrix} x_1/a & y_1/a & 1 \\ x_2/b & y_2/b & 1 \\ x_3/c & y_3/c & 1 \end{vmatrix}$.

Step 2. Multiply row 1 by $a$, row 2 by $b$, row 3 by $c$; this multiplies the determinant by $abc$: $abc\,M = \begin{vmatrix} x_1 & y_1 & a \\ x_2 & y_2 & b \\ x_3 & y_3 & c \end{vmatrix}$. A cyclic column shift turns this into $\begin{vmatrix} a & x_1 & y_1 \\ b & x_2 & y_2 \\ c & x_3 & y_3 \end{vmatrix}$ (even permutation, sign unchanged). …