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The area of the triangle with vertices (1, 4), (2, 7) and (4, 13) is : |
0 1 2 3 |
0 |
The correct answer is Option (1) → 0 area = $\frac{1}{2}\begin{vmatrix}1&4&1\\2&7&1\\4&13&1\end{vmatrix}$ $R_3→R_3-R_2,R_2→R_2-R_1$ $\frac{1}{2}\begin{vmatrix}1&4&1\\1&3&0\\2&6&0\end{vmatrix}=\frac{6-6}{2}=0$ |
