The area (in sq. units) of the triangle whose vertices are (0, 0), (a, 0), (0, b), is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{1}{2}|ab|\)
Using the determinant formula for the area of a triangle:
\( \displaystyle \Delta = \frac{1}{2}\left|\begin{vmatrix}0&0&1\\a&0&1\\0&b&1\end{vmatrix}\right| \)
Expanding along the first row, only the third entry contributes:
\( \displaystyle \begin{vmatrix}0&0&1\\a&0&1\\0&b&1\end{vmatrix} = 1 \times (ab - 0) = ab \)
\( \displaystyle \Delta = \frac{1}{2}|ab| \)
The modulus is needed because a or b may be negative, while an area is positive. (The triangle is right-angled at the origin, with legs \(|a|\) and \(|b|\).)