(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: Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(

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: Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$. Given vertices are $(k, 0), (4, 0)$ and $(0, 2)$ and Area $= 4$. Substituting the values: $4 = \frac{1}{2} |k(0 - 2) + 4(2 - 0) + 0(0 - 0)|$. $4 = \frac{1}{2} |-2k + 8|$. $8 = |-2k + 8|$. This implies two cases: Case $1$: $-2k + 8 = 8 \implies -2k = 0 \implies k = 0$. Case $2$: $-2k + 8 = -8 \implies -2k = -16 \implies k = 8$. Thus,the values of $k$ are $0$ and $8$.

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

If the area of a triangle is $4$ sq. units whose vertices are $(k, 0), (4, 0)$ and $(0, 2)$,then the value of $k$ is . . . . . . .

GSEB 2021 All GSEB

A
$0, 8$
B
$0, -8$
C
$0, 16$
D
$0, -16$

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

Previous Year

GSEB 2021 Paper All GSEB years →

$A$ determinant is chosen at random from the set of all determinants of order $2 \times 2$ with elements $0$ or $1$ only. The probability that the determinant chosen is non-zero is .........

EasyAP EAMCET 2020

View Solution

The matrix $\begin{bmatrix} \lambda & -1 & 4 \\ -3 & 0 & 1 \\ -1 & 1 & 2 \end{bmatrix}$ is invertible,if

Easy

View Solution

The values of $\alpha$,for which $\left|\begin{array}{ccc}1 & \frac{3}{2} & \alpha+\frac{3}{2} \\ 1 & \frac{1}{3} & \alpha+\frac{1}{3} \\ 2 \alpha+3 & 3 \alpha+1 & 0\end{array}\right|=0$,lie in the interval

DifficultJEE MAIN 2024

View Solution

If $\left| {\begin{array}{*{20}{c}}{\cos (A + B)}&{ - \sin (A + B)}&{\cos 2B}\\{\sin A}&{\cos A}&{\sin B}\\{ - \cos A}&{\sin A}&{\cos B}\end{array}} \right| = 0$,then $B =$

Medium

View Solution

The roots of the equation $\left| \begin{array}{ccc} 1 & 4 & 20 \\ 1 & -2 & 5 \\ 1 & 2x & 5x^2 \end{array} \right| = 0$ are

MediumIIT 1987

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

If the area of a triangle is $4$ sq. units whose vertices are $(k, 0), (4, 0)$ and $(0, 2)$,then the value of $k$ is . . . . . . .

Similar Questions

$A$ determinant is chosen at random from the set of all determinants of order $2 \times 2$ with elements $0$ or $1$ only. The probability that the determinant chosen is non-zero is .........

The matrix $\begin{bmatrix} \lambda & -1 & 4 \\ -3 & 0 & 1 \\ -1 & 1 & 2 \end{bmatrix}$ is invertible,if

The values of $\alpha$,for which $\left|\begin{array}{ccc}1 & \frac{3}{2} & \alpha+\frac{3}{2} \\ 1 & \frac{1}{3} & \alpha+\frac{1}{3} \\ 2 \alpha+3 & 3 \alpha+1 & 0\end{array}\right|=0$,lie in the interval

If $\left| {\begin{array}{*{20}{c}}{\cos (A + B)}&{ - \sin (A + B)}&{\cos 2B}\\{\sin A}&{\cos A}&{\sin B}\\{ - \cos A}&{\sin A}&{\cos B}\end{array}} \right| = 0$,then $B =$

The roots of the equation $\left| \begin{array}{ccc} 1 & 4 & 20 \\ 1 & -2 & 5 \\ 1 & 2x & 5x^2 \end{array} \right| = 0$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module