CUET Preparation Today
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) : |
0 ±1 ±2 $\frac{3}{2}$ |
±2 |
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 $R_2$ $\left|\frac{3k}{2}\right|=3⇒|k|=2$ $k=±2$ |
