Question

If the area of the pentagon formed by the vertices $A(1,3),B(-2,5),C(-3,-1),D(0,-2)$ and $E=(2,t)$ is $\frac{45}{2}\text{sq. units}$, then possible value(s) of $t$ is/are

• A. $-1$
• B. $1$
• C. $-99$
• D. $89$

Answer 1

Answer: (A), (D)

$89$

Given vertices of the pentagon are $A(1,3),B(-2,5),C(-3,-1),D(0,-2)$ and $E=(2,t)$
We know that the area of pentagon $ABCDE$
$\frac{45}{2}=\frac{1}{2}\left|\begin{array}{cccccc}{x}_1& x_2& x_3& x_4& x_5& x_1\\ y_1& y_2& y_3& y_4& y_5& y_1\end{array}\right|\Rightarrow \frac{45}{2}=\frac{1}{2}\left|\begin{array}{cccccc}1& -2& -3& 0& 2& 1\\ 3& 5& -1& -2& t& 3\end{array}\right|\Rightarrow \frac{45}{2}=\frac{1}{2}|(5+6)+(2+15)+(6+0)+(0+4)+(6-t)|\Rightarrow 45=|44-t|\Rightarrow 44-t=\pm 45\Rightarrow -t=-44\pm 45\therefore t=-1,89$