Match List-I with List-II
| List-I (Vertices of a triangle) | List-II (Area in square units) |
|---|---|
| (A) \((0, 0), (4, 0), (0, 6)\) | (I) 0 |
| (B) \((1, 2), (3, 5), (5, 8)\) | (II) 5 |
| (C) \((-1, 0), (3, 0), (1, 5)\) | (III) 10 |
| (D) \((2, 1), (4, 3), (1, 5)\) | (IV) 12 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(IV), (B)-(I), (C)-(III), (D)-(II)
The area of a triangle is the absolute value of \(\displaystyle \frac{1}{2}\begin{vmatrix}x_1 & y_1 & 1\\ x_2 & y_2 & 1\\ x_3 & y_3 & 1\end{vmatrix}\), which equals \(\frac{1}{2}\left|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\right|\).
(A) \(\frac{1}{2}\left|0 + 4(6 - 0) + 0\right| = 12\) → (IV)
(B) \(\frac{1}{2}\left|1(5 - 8) + 3(8 - 2) + 5(2 - 5)\right| = \frac{1}{2}\left|-3 + 18 - 15\right| = 0\) → (I); the points are collinear.
(C) \(\frac{1}{2}\left|-1(0 - 5) + 3(5 - 0) + 1(0 - 0)\right| = \frac{1}{2}(5 + 15) = 10\) → (III)
(D) \(\frac{1}{2}\left|2(3 - 5) + 4(5 - 1) + 1(1 - 3)\right| = \frac{1}{2}\left|-4 + 16 - 2\right| = 5\) → (II)