An equilateral triangle has each side equal to a the co-ordinates of its vertices are \( \Large \left(x_{1},\ y_{1}\right),\ \left(x_{2},\ y_{2}\right)\ and\ \left(x_{3},\ y_{3}\right) \), then the square of determinant \( \begin{vmatrix} 
x_{1} & y_{1} & 1 \\ 
x_{2} & y_{2} & 1 \\ 
x_{3} & y_{3} & 1  
\end{vmatrix}  \) equals


A) \( \Large 3a^{4} \)

B) \( \Large \frac{3}{4}a^{4} \)

C) \( \Large 4a^{4} \)

D) none of these

Correct Answer:
B) \( \Large \frac{3}{4}a^{4} \)

Description for Correct answer:

Since, the triangle is an equilateral triangle

Therefore, Area of equilateral triangle = \( \Large \frac{\sqrt{3}}{4}a^{2} \) ...(i)

Also, if \( \Large \left(x_{1},y_{1}\right),\ \left(x_{2},y_{2}\right)\ and\ \left(x_{3}, y_{3}\right) \) are the vertices of A, then

Area of equilateral triangle = \( \frac{1}{2} \begin{vmatrix} 
x_{1} & y_{1} & 1 \\ 
x_{2} & y_{2} & 1 \\ 
x_{3} & y_{3} & 1  
\end{vmatrix}  \)...(ii)

From equations (i) and (ii) we get

\( \begin{vmatrix} 
x_{1} & y_{1} & 1 \\ 
x_{2} & y_{2} & 1 \\ 
x_{3} & y_{3} & 1  
\end{vmatrix} = \frac{\sqrt{3}}{2}a^{2} \)

= \( \Large
 \begin{vmatrix} 
x_{1} & y_{1} & 1 \\ 
x_{2} & y_{2} & 1 \\ 
x_{3} & y_{3} & 1  
\end{vmatrix} = \left(\frac{\sqrt{3}}{2}a^{2}\right)^{2}= \frac{3a^{4}}{4} \)


Part of solved Rectangular and Cartesian products questions and answers : >> Elementary Mathematics >> Rectangular and Cartesian products








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