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 $(\frac{x_1}{a},\frac{y_1}{a}),(\frac{x_2}{b},\frac{y_2}{b}),(\frac{x_3}{c},\frac{y_3}{c})$ is

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

Answer 1

Answer: (D)

$\frac{1}{8}$

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

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 given by,

$\frac{1}{2}\left|\begin{array}{ccc}\frac{x_1}{a}& \frac{y_1}{a}& 1\\ \frac{x_2}{b}& \frac{y_1}{b}& 1\\ \frac{x_3}{c}& \frac{y_3}{c}& 1\end{array}\right|=\frac{1}{2abc}\left|\begin{array}{ccc}{x}_1& y_1& a\\ x_2& y_2& b\\ x_3& y_3& c\end{array}\right|$, take $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ common from first, second and third row respectively

$=\frac{1}{4abc}\left|\begin{array}{ccc}{x}_1& y_1& 2a\\ x_2& y_2& 2b\\ x_3& y_3& 2c\end{array}\right|$, take $\frac{1}{2}$ common from third column

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

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