Match the vertices of the triangles in List-I with their areas (in square units) in List-II
| List-I | List-II | ||
| (A) | (0, 0), (4, 0), (0, 3) | (I) | 4 |
| (B) | (1, 2), (4, 6), (7, 2) | (II) | 12 |
| (C) | (−1, 0), (3, 0), (1, 5) | (III) | 10 |
| (D) | (2, 1), (3, 4), (5, 2) | (IV) | 6 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(IV), (B)-(II), (C)-(III), (D)-(I)
The area of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), \((x_3, y_3)\) is the absolute value of
\( \displaystyle \frac{1}{2}\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \)
\( \displaystyle = \frac{1}{2}\left[x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\right] \)
(A) (0, 0), (4, 0), (0, 3)
This matches (IV) 6.
\( \displaystyle \frac{1}{2}\left[0(0 - 3) + 4(3 - 0) + 0(0 - 0)\right] = \frac{1}{2}(12) = 6 \)
(B) (1, 2), (4, 6), (7, 2)
This matches (II) 12.
\( \displaystyle \frac{1}{2}\left[1(6 - 2) + 4(2 - 2) + 7(2 - 6)\right] = \frac{1}{2}(4 + 0 - 28) = -12 \)
The area is the absolute value, 12.
(C) (−1, 0), (3, 0), (1, 5)
This matches (III) 10.
\( \displaystyle \frac{1}{2}\left[-1(0 - 5) + 3(5 - 0) + 1(0 - 0)\right] = \frac{1}{2}(5 + 15) = 10 \)
(D) (2, 1), (3, 4), (5, 2)
This matches (I) 4.
\( \displaystyle \frac{1}{2}\left[2(4 - 2) + 3(2 - 1) + 5(1 - 4)\right] = \frac{1}{2}(4 + 3 - 15) = -4 \)
The area is the absolute value, 4.