Question

If the area of a triangle with vertices (-3,0), (3, 0) and (0, k) is 9 sq. units, then, the value of k will be:

(a) 9
(b) 3
(c) -9
(c) 6

Answer 1

(b) We know that, area of a triangle with vertices $(x_1,y_1)$, $(x_2,y_2)$ and $(x_3,y_3)$ is given by
$\Delta =\frac{1}{2}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$
$\therefore \Delta =\frac{1}{2}\left|\begin{array}{ccc}-3& 0& 1\\ 3& 0& 1\\ 0& k& 1\end{array}\right|$
Expanding along $R_1$,
$9=\frac{1}{2}[-3(-k)-0+1(3k\left)\right]\Rightarrow 18=3k+3k=6k\therefore k=\frac{18}{6}=3$