Question

If $\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|=\left|\begin{array}{ccc}1& 1& 1\\ b_1& b_2& b_3\\ a_1& a_2& a_3\end{array}\right|$ then the two triangles whose vertices are $(x_1,y_1),(x_2,y_2),((x_3,y_3)$ and $(a_1,b_1),(a_2,b_2),(a_{13},b_3),$ are

• A. congruent
• B. similar
• C. equal in area
• D. none of these

Answer 1

The correct option is C equal in area

If $(x_1,y_1),(x_2,y_2),((x_3,y_3)$ are the vertices of triangle , then its area is

$A_1=\frac{1}{2}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$
If $(a_1,b_1),(a_2,b_2),(a_3,b_3)$ are the vertices of triangle , then its area is

$A_2=\frac{1}{2}\left|\begin{array}{ccc}{a}_1& b_1& 1\\ a_2& b_2& 1\\ a_3& b_3& 1\end{array}\right|$
$A_2=\frac{1}{2}\left|\begin{array}{ccc}{a}_1& a_2& a_3\\ b_1& b_2& b_3\\ 1& 1& 1\end{array}\right|(\because |A|=|A^T|)$
$A_2=-\frac{1}{2}\left|\begin{array}{ccc}1& 1& 1\\ b_1& b_2& b_3\\ a_1& a_2& a_3\end{array}\right|$

Since, area is positive,

$A_2=\frac{1}{2}\left|\begin{array}{ccc}1& 1& 1\\ b_1& b_2& b_3\\ a_1& a_2& a_3\end{array}\right|$
Given, $\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|=\left|\begin{array}{ccc}1& 1& 1\\ b_1& b_2& b_3\\ a_1& a_2& a_3\end{array}\right|$
$\Rightarrow A_1=A_2$