Question

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

• A. \( \frac{1}{4} abc \)
• B. \( \frac{1}{8} abc \)
• C. \( \frac{1}{4} \)
• D. \( \frac{1}{8} \)
• E. \( \frac{1}{12} \)

Hint:

Factoring out $1/a, 1/b, 1/c$ from the rows of the area determinant effectively multiplies the final answer by $1/abc$. This is a powerful way to simplify complex-looking coordinate geometry problems.

Answer 1

The Correct Option is D