Question

If $\begin{vmatrix}2a&x_{1}&y_{1}\\ 2b&x_{2}&y_{2}\\ 2c&x_{3}&y_{3}\end{vmatrix}=\frac{abc}{2}\ne0 ,$
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 $\left(\frac{x_{3}}{a},\frac{y_{3}}{a}\right)$ 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

Answer: (D)

Given, $\begin{vmatrix}2 a & x_{1} & y_{1} \\ 2 b & x_{2} & y_{2} \\ 2 c & x_{3} & y_{3}\end{vmatrix}=\frac{a b c}{2}$
$\Rightarrow 2\begin{vmatrix}a & x_{1} & y_{1} \\ b & x_{2} & y_{2} \\ C & x_{3} & y_{3}\end{vmatrix}=\frac{a b c}{2}$
$\Rightarrow \begin{vmatrix}a & x_{1} & y_{1} \\ b & x_{2} & y_{2} \\ C & x_{3} & Y_{3}\end{vmatrix} =\frac{a b c}{4} \,\,\,...(i)$
Given, vertices of a triangle are
$\left(\frac{x_{1}}{a}, \frac{y_{1}}{a}\right),\left(\frac{x_{2}}{b}, \frac{y_{2}}{b}\right) \text { and }\left(\frac{x_{3}}{c}, \frac{Y_{3}}{c}\right)$
$\therefore $ Area of triangle
$=\frac{1}{2}\begin{vmatrix}\frac{x_{1}}{a} & \frac{y_{1}}{a} & 1 \\ \frac{x_{2}}{b} & \frac{y_{2}}{b} & 1 \\ \frac{x_{3}}{c} & \frac{y_{3}}{c} & 1\end{vmatrix}=\frac{1}{2} \cdot \frac{1}{a} \cdot \frac{1}{b} \cdot \frac{1}{c}\begin{vmatrix}x_{1} & y_{1} & a \\ x_{2} & y_{2} & b \\ x_{3} & y_{3} & C\end{vmatrix}$
$=\frac{1}{2 a b c}\begin{vmatrix}a & x_{1} & y_{1} \\ b & x_{2} & y_{2} \\ c & x_{3} & y_{3}\end{vmatrix}=\frac{1}{2 a b c}\left(\frac{a b c}{4}\right)$ [from Eq. (i)]
$=\frac{1}{8}$