Let be the area of a triangle having vertices , and . Which of the following is correct?
Let $A$ be the area of a triangle having vertices $(x_1, y_1)$, $(x_2, y_2)$ and $(x_3, y_3)$. Which of the following is correct?
- a
= $+A$$x_1$ $y_1$ 1 $x_2$ $y_2$ 1 $x_3$ $y_3$ 1 - b
= $\pm 2A$$x_1$ $y_1$ 1 $x_2$ $y_2$ 1 $x_3$ $y_3$ 1 - c
= $\pm \frac{A}{2}$$x_1$ $y_1$ 1 $x_2$ $y_2$ 1 $x_3$ $y_3$ 1 - d
| $x_3$ | $y_3$ | 1 |$^2 = A^2$$x_1$ $y_1$ 1 $x_2$ $y_2$ 1
Answer
Answer
AIWritten by AI - it can contain mistakes.
Correct option: b
(b) | $x_1$ $y_1$ 1 | ; | $x_2$ $y_2$ 1 | ; | $x_3$ $y_3$ 1 | = $\pm 2A$
The area of the triangle formed by three vertices is given by $A = \frac{1}{2}|\Delta|$, where $\Delta = \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}$.
Since the determinant value may be positive or negative, $\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = \pm 2A$.
From ISC 2027 Specimen Mathematics Paper 1, question 1(iii).