Question

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

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

Answer 1

The correct option is D $\frac{1}{8}$

Given, $\left|\begin{array}{ccc}2a& x_1& y_1\\ 2b& x_2& y_2\\ 2c& x_3& y_3\end{array}\right|=\frac{abc}{2}$

Area of triangle formed by joining the given points is

$\Delta =\frac{1}{2}\left|\begin{array}{ccc}\frac{x_1}{a}& \frac{y_1}{a}& 1\\ \frac{x_2}{b}& \frac{y_2}{b}& 1\\ \frac{x_3}{c}& \frac{y_3}{c}& 1\end{array}\right|\Delta =\frac{1}{2abc}\left|\begin{array}{ccc}{x}_1& y_1& a\\ x_2& y_2& b\\ x_3& y_3& c\end{array}\right|\Delta =\frac{1}{4abc}\left|\begin{array}{ccc}{x}_1& y_1& 2a\\ x_2& y_2& 2b\\ x_3& y_3& 2c\end{array}\right|$

Interchanging the colums

$\Delta =\frac{1}{4abc}\left|\begin{array}{ccc}2a& x_1& y_1\\ 2b& x_2& y_2\\ 2c& x_3& y_3\end{array}\right|$

Substituting the value of determinant

$\Delta =\frac{1}{4abc}\times \frac{abc}{2}=\frac{1}{8}$

So, option D is correct.