Question

The area of the triangle with vertices $A(3,7),B(-5,2)$ and $C(2,5)$ is denoted by $\Delta$. If ${\Delta }_A,{\Delta }_B,{\Delta }_C$ denote the areas of the triangles with vertices $OBC,AOC$ and $ABO$ respectively, $O$ being the origin, then

• A. ${\Delta }_A+{\Delta }_B=\Delta +{\Delta }_C$
• B. ${\Delta }_A+{\Delta }_B={\Delta }_C-\Delta$
• C. ${\Delta }_A+{\Delta }_B=2{\Delta }_C$
• D. ${\Delta }_A+{\Delta }_B+{\Delta }_C=2\Delta$

Answer 1

The correct option is B ${\Delta }_A+{\Delta }_B={\Delta }_C-\Delta$

Given: $A(3,7)\equiv (x_1,y_1),B(-5,2)\equiv (x_2,y_2)$ and $C(2,5)\equiv (x_3,y_3)$

$\Delta =\frac{1}{2}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$

$\Delta =\frac{1}{2}\left|\begin{array}{ccc}3& 7& 1\\ -5& 2& 1\\ 2& 5& 1\end{array}\right|=\left|\frac{1}{2}(3(-3)-7(-7)+1(-29))\right|=\frac{11}{2}$

here $(x_1,y_1)\equiv (0,0)$

${\Delta }_A=\frac{1}{2}\left|\begin{array}{ccc}0& 0& 1\\ -5& 2& 1\\ 2& 5& 1\end{array}\right|=\left|\frac{1}{2}\left(1(-29)\right)\right|=\frac{29}{2}$

here $(x_2,y_2)\equiv (0,0)$

${\Delta }_B=\frac{1}{2}\left|\begin{array}{ccc}3& 7& 1\\ 0& 0& 1\\ 2& 5& 1\end{array}\right|=\left|\frac{1}{2}\left(1\left(1\right)\right)\right|=\frac{1}{2}$

here $(x_3,y_3)\equiv (0,0)$

${\Delta }_C=\frac{1}{2}\left|\begin{array}{ccc}3& 7& 1\\ -5& 2& 1\\ 0& 0& 1\end{array}\right|=\left|\frac{1}{2}\left(1\left(41\right)\right)\right|=\frac{41}{2}$
Hence
${\Delta }_A+{\Delta }_B={\Delta }_C-\Delta$