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

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

Answer 1

Answer: (D)

$\frac {1}{8}$

Answer 2

Area of triangle whose vertices are
$\left(\frac{x_{1}}{a}, \frac{y_{1}}{a}\right),\left(\frac{x_{2}}{b}, \frac{y_{2}}{b}\right), \left(\frac{x_{3}}{c}, \frac{y_{3}}{c}\right)$is
$\Delta=\frac{1}{2}\begin{vmatrix}\frac{x_{1}}{a}&\frac{y_{1}}{a}&1\\\ \frac{x_{2}}{b}&\frac{y_{2}}{b}&1\\\ \frac{x_{3}}{c}&\frac{y_{3}}{c}&1\end{vmatrix}$
On multiplying $_1, R_2$ and $R_3$ by a.b.c respectively, we get
$\Delta=\frac{1}{2\,abc}\begin{vmatrix}x_{1}&y_{1}&a\\\ x_{2}&y_{2}&b\\\ x_{3}&y_{3}&c\end{vmatrix}$
On multiplying $C_3$ by 2 we get
$\Delta=\frac{1}{4\,abc}\begin{vmatrix}x_{1}&y_{1}&2a\\\ x_{2}&y_{2}&2b\\\ x_{3}&y_{3}&2c\end{vmatrix}$
On applying $C_1$ $\leftrightarrow$ $C_3$ , by 2 we get
$\Delta=-\frac{1}{4\,abc}\begin{vmatrix}2a&y_{1}&x_{1}\\\ 2b&y_{2}&x_{2}\\\ 2c&y_{3}&x_{3}\end{vmatrix}$
On applying $C_2$ $\leftrightarrow$ $C_3$ , we get
$\Delta=-\frac{1}{4\,abc}\begin{vmatrix}2a&x_{1}&y_{1}\\\ 2b&x_{2}&y_{2}\\\ 2c&x_{3}&y_{3}\end{vmatrix}$
$=\frac{1}{4\,abc}. \frac{abc}{2}\left[\because\begin{vmatrix}2a&x_{1}&y_{1}\\\ 2b&x_{2}&y_{2}\\\ 2c&x_{3}&y_{3}\end{vmatrix}=\frac{abc}{2}\right]=\frac{1}{8}$