Let O(0, 0), P(6, 0) and Q(0, 4) be three points. Consider the following statements:
(A) The area of triangle OPQ is 12 square units.
(B) The triangle with vertices (1, 1), (7, 1) and (1, 5) has area 13 square units.
(C) The triangle with vertices (0, 0), (12, 0) and (0, 8) has area 48 square units.
(D) Taking the vertices in the order O, Q, P makes the area of the triangle \(-12\) square units.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A) and (C) only
The area of a triangle is the absolute value of
\(\displaystyle \Delta = \frac{1}{2}\begin{vmatrix}x_1 & y_1 & 1\\ x_2 & y_2 & 1\\ x_3 & y_3 & 1\end{vmatrix}\)
For O, P, Q:
\(\displaystyle \Delta = \frac{1}{2}\begin{vmatrix}0 & 0 & 1\\ 6 & 0 & 1\\ 0 & 4 & 1\end{vmatrix} = \frac{1}{2}(24 - 0) = 12\)
The points (1, 1), (7, 1), (1, 5) are O, P, Q each moved by (1, 1):
\(\displaystyle \frac{1}{2}\left|1(1 - 5) + 7(5 - 1) + 1(1 - 1)\right| = \frac{1}{2}(24) = 12\)
The points (0, 0), (12, 0), (0, 8) are O, P, Q with both coordinates doubled:
\(\displaystyle \frac{1}{2}\left|0 + 12(8 - 0) + 0\right| = 48\)
Interchanging two vertices interchanges two rows, so the determinant changes sign, but the area is its absolute value.
So the area of OPQ is 12, a shift keeps it 12, doubling the coordinates makes it 48, and reordering the vertices keeps it 12.
Checking the statements
(A) Area of OPQ is 12 — Correct. \(\frac{1}{2}(6 \times 4) = 12\).
(B) Area 13 for (1, 1), (7, 1), (1, 5) — Incorrect. Its area is \(\frac{1}{2}(24) = 12\); moving all vertices does not change the area.
(C) Area 48 for (0, 0), (12, 0), (0, 8) — Correct. \(\frac{1}{2}(12 \times 8) = 48\), four times 12.
(D) Area \(-12\) for the order O, Q, P — Incorrect. Only the determinant becomes \(-12\); the area is \(|-12| = 12\).