If $A(x_1, y_1), B(x_2, y_2),$ and $C(x_3, y_3)$ are the vertices of $\Delta ABC,$ then the area of $\Delta ABC$ is equal to:

  • A
    $\frac{1}{2}|x_1(x_2 - x_3) + y_1(y_2 - y_3)|$
  • B
    $\frac{1}{2}|x_1 y_2 - x_2 y_3 + x_3 y_1 - x_1 y_3|$
  • C
    $\frac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
  • D
    $|x_1(x_2 - y_3) + x_2(x_3 - y_1) + x_3(x_1 - y_2)|$

Explore More

Similar Questions

The distance between $A(\cos \theta, 0)$ and $B(0, \sin \theta)$ is ...........

The vertices of $\Delta ABC$ are $A(2, 4)$,$B(5, 0)$,and $C(5, 4)$. Then the area of $\Delta ABC = \ldots$

If the distance between points $A(a, 2)$ and $B(3, -5)$ is $\sqrt{53},$ find the possible values of $a$.

If the vertices of $\Delta ABC$ are $A(-4, -1)$,$B(1, 2)$,and $C(4, -3)$,then the area of the triangle is:

Which of the following groups is true for $\square ABCD?$
$1$. $\square ABCD$ is a rhombus. $a$. $\overline{AC}$ and $\overline{BD}$ bisect each other.
$2$. $\square ABCD$ is a parallelogram. $b$. $\overline{AC}$ and $\overline{BD}$ bisect each other at right angles.
$3$. $\square ABCD$ is a rectangle. $c$. $\overline{AC}$ and $\overline{BD}$ are congruent and bisect each other at right angles.
$4$. $\square ABCD$ is a square. $d$. $\overline{AC}$ and $\overline{BD}$ are congruent and bisect each other.

Difficult
View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo