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 $\left[\frac{x_1}{a},\frac{y_1}{a}\right]$, $\left[\frac{x_2}{b},\frac{y_2}{b}\right]$ and $\left[\frac{x_3}{c},\frac{y_3}{c}\right]$ is

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

Answer 1

The correct option is D

$\frac{1}{8}$

Explanation for the correct option

It is given that, $\left|\begin{array}{ccc}2a& x_1& y_1\\ 2b& x_2& y_2\\ 2c& x_3& y_3\end{array}\right|=\frac{abc}{2}\ne 0$

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

The vertices of the triangle are $\left[\frac{x_1}{a},\frac{y_1}{a}\right]$, $\left[\frac{x_2}{b},\frac{y_2}{b}\right]$ and $\left[\frac{x_3}{c},\frac{y_3}{c}\right]$.

The area of the triangle can be given by, $A=\frac{1}{2}\left|\left|\begin{array}{ccc}\frac{x_1}{a}& \frac{y_1}{a}& 1\\ \frac{x_2}{b}& \frac{y_2}{b}& 1\\ \frac{x_3}{c}& \frac{y_3}{c}& 1\end{array}\right|\right|$

$$\begin{gathered} \Rightarrow A=\frac{1}{2}\left|\frac{1}{a}\times \frac{1}{b}\times \frac{1}{c}\left|\begin{array}{ccc}{x}_1& y_1& a\\ x_2& y_2& b\\ x_3& y_3& c\end{array}\right|\right| \\ \Rightarrow A=\frac{1}{2abc}\left|\begin{array}{ccc}{x}_1& y_1& a\\ x_2& y_2& b\\ x_3& y_3& c\end{array}\right| \\ \Rightarrow A=\frac{1}{2abc}\times \frac{abc}{4} \\ \Rightarrow A=\frac{1}{8} \end{gathered}$$

Therefore, the area of the triangle is $\frac{1}{8}$.

Hence, option(D) is the correct option i.e. $\frac{1}{8}$