Let A be the area of a triangle having vertices (x₁, y₁), (x₂, y₂) and (x₃, y₃). Which of the…

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
$x_1$ $y_1$ 1
$x_2$ $y_2$ 1
$x_3$ $y_3$ 1
= $+A$
b
$x_1$ $y_1$ 1
$x_2$ $y_2$ 1
$x_3$ $y_3$ 1
= $\pm 2A$
c
$x_1$ $y_1$ 1
$x_2$ $y_2$ 1
$x_3$ $y_3$ 1
= $\pm \frac{A}{2}$
d
$x_1$ $y_1$ 1
$x_2$ $y_2$ 1
| $x_3$ | $y_3$ | 1 |$^2 = A^2$

Answer

Official answer key

Correct option: b

Answer: (b) the determinant $= \pm 2A$ The area is $A = \frac{1}{2}\left|\Delta\right|$ where $\Delta$ is the determinant, which may be positive or negative, so $\Delta = \pm 2A$.

$x_1$	$y_1$	1
$x_2$	$y_2$	1
$x_3$	$y_3$	1

$x_1$	$y_1$	1
$x_2$	$y_2$	1
$x_3$	$y_3$	1

$x_1$	$y_1$	1
$x_2$	$y_2$	1
$x_3$	$y_3$	1

$x_1$	$y_1$	1
$x_2$	$y_2$	1