(A) The area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by the formula: $\text{Area} = \frac{1}{2} |x_1(y_2 - y_3

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(A) The area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by the formula: $\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$. Substituting the given vertices $(1, -1)$,$(-1, 1)$,and $(-1, -1)$: $\text{Area} = \frac{1}{2} |1(1 - (-1)) + (-1)(-1 - (-1)) + (-1)(-1 - 1)|$ $= \frac{1}{2} |1(2) + (-1)(0) + (-1)(-2)|$ $= \frac{1}{2} |2 + 0 + 2|$ $= \frac{1}{2} |4| = 2$. Thus,the area is $2$ square units.

Class 12 Mathematics 3 and 4 .Determinants and Matrices Expansion of determinants, Solution of equation in the form of determinants and area of triangle and Equation of Line

Easy

The area of a triangle whose vertices are $(1, -1)$,$(-1, 1)$,and $(-1, -1)$ is given by

A
$2$
B
$0.5$
C
$1$
D
$3$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Mathematics Questions

Chapter

3 and 4 .Determinants and Matrices

Subtopic

Expansion of determinants, Solution of equation in the form of determinants and area of triangle and Equation of Line

$\left| \begin{array}{ccc} 0 & p-q & p-r \\ q-p & 0 & q-r \\ r-p & r-q & 0 \end{array} \right| = $

Easy

View Solution

The greatest value of $c \in R$ for which the system of linear equations $x - cy - cz = 0$,$cx - y + cz = 0$,$cx + cy - z = 0$ has a non-trivial solution,is

DifficultJEE MAIN 2019

View Solution

If $A = \begin{bmatrix} \alpha & 2 \\ 2 & \alpha \end{bmatrix}$ and $|A^3| = 125$,then $\alpha = $ . . . . . .

EasyGSEB 2016

View Solution

$A$ value of $\theta$ for which the following system of equations has a non-trivial solution is:
$(4 \sin \theta) x - 3y + z = 0$
$x - (6 \cos 2\theta) y + z = 0$
$3x - 12y + 4z = 0$

MediumTS EAMCET 2019

View Solution

The value of $a$ for which the system of equations $a^3x + (a + 1)^3y + (a + 2)^3z = 0$,$ax + (a + 1)y + (a + 2)z = 0$,and $x + y + z = 0$ has a non-zero solution is:

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

The area of a triangle whose vertices are $(1, -1)$,$(-1, 1)$,and $(-1, -1)$ is given by

Similar Questions

$\left| \begin{array}{ccc} 0 & p-q & p-r \\ q-p & 0 & q-r \\ r-p & r-q & 0 \end{array} \right| = $

The greatest value of $c \in R$ for which the system of linear equations $x - cy - cz = 0$,$cx - y + cz = 0$,$cx + cy - z = 0$ has a non-trivial solution,is

If $A = \begin{bmatrix} \alpha & 2 \\ 2 & \alpha \end{bmatrix}$ and $|A^3| = 125$,then $\alpha = $ . . . . . .

$A$ value of $\theta$ for which the following system of equations has a non-trivial solution is: $(4 \sin \theta) x - 3y + z = 0$ $x - (6 \cos 2\theta) y + z = 0$ $3x - 12y + 4z = 0$

The value of $a$ for which the system of equations $a^3x + (a + 1)^3y + (a + 2)^3z = 0$,$ax + (a + 1)y + (a + 2)z = 0$,and $x + y + z = 0$ has a non-zero solution is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

$A$ value of $\theta$ for which the following system of equations has a non-trivial solution is:
$(4 \sin \theta) x - 3y + z = 0$
$x - (6 \cos 2\theta) y + z = 0$
$3x - 12y + 4z = 0$