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

Concept:
The area of a triangle with vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \) is given by: \[ \text{Area} = \frac{1}{2} \left| \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \right| \] 
Step 1: Set up the area determinant for the given vertices.
\[ \text{Area} = \frac{1}{2} \left| \begin{vmatrix} \dfrac{x_1}{a} & \dfrac{y_1}{a} & 1 \\ \dfrac{x_2}{b} & \dfrac{y_2}{b} & 1 \\ \dfrac{x_3}{c} & \dfrac{y_3}{c} & 1 \end{vmatrix} \right| \] 
Step 2: Factor out constants from the rows.
To remove the denominators, factor out \( \dfrac{1}{a} \) from \( R_1 \), \( \dfrac{1}{b} \) from \( R_2 \), and \( \dfrac{1}{c} \) from \( R_3 \): \[ \text{Area} = \frac{1}{2abc} \left| \begin{vmatrix} x_1 & y_1 & a \\ x_2 & y_2 & b \\ x_3 & y_3 & c \end{vmatrix} \right| \] 
Step 3: Relate to the given determinant.
Swap columns to match the given form (one swap changes the sign, but we take absolute value): \[ \text{Area} = \frac{1}{2abc} \left| \begin{vmatrix} a & x_1 & y_1 \\ b & x_2 & y_2 \\ c & x_3 & y_3 \end{vmatrix} \right| \] The given determinant is: \[ \begin{vmatrix} 2a & x_1 & y_1 \\ 2b & x_2 & y_2 \\ 2c & x_3 & y_3 \end{vmatrix} = 2 \begin{vmatrix} a & x_1 & y_1 \\ b & x_2 & y_2 \\ c & x_3 & y_3 \end{vmatrix} = \frac{abc}{2} \] Therefore, \[ \begin{vmatrix} a & x_1 & y_1 \\ b & x_2 & y_2 \\ c & x_3 & y_3 \end{vmatrix} = \frac{abc}{4} \] 
Step 4: Calculate final area.
\[ \text{Area} = \frac{1}{2abc} \times \frac{abc}{4} = \frac{1}{8} \]