If K is such that the area of triangle ABC is 3, where A(1, 3), B(0, 0) and C(K, 0), then the value(s) of K is/(are) :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → ±2
area = $\frac{1}{2}\begin{vmatrix}1&3&1\\0&0&1\\k&0&1\end{vmatrix}=3$
expanding along $R_2$
$\left|\frac{3k}{2}\right|=3⇒|k|=2$
$k=±2$