Question

If $\left|\begin{array}{c}2a{x}_1{y}_1\\ 2b{x}_2{y}_2\\ 2c{x}_3{y}_3\end{array}\right|=\frac{abc}{2}\ne 0$, then the area of the triangle whose vertices are$(\frac{x_1}{a},\frac{y_1}{a}),(\frac{x_2}{b},\frac{y_2}{b}),(\frac{x_3}{c},\frac{y_3}{c})$ is

• A. $\frac{1}{4}abc$
• B. $\frac{1}{8}abc$
• C. $\frac{1}{4}$
• D. $\frac{1}{8}$

Answer 1

The correct option is D

$\frac{1}{8}$

$2abc\left|\begin{array}{ccc}1& \frac{x_1}{a}& \frac{y_1}{a}\\ 1& \frac{x_2}{b}& \frac{y_2}{a}\\ 1& \frac{x_3}{c}& \frac{y_3}{c}\end{array}\right|=2abc\left(area\right)=\frac{abc}{2}\therefore Area=\frac{1}{8}$