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

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

(A) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by the formula: $\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$. Given vertices are $(K, 0), (4, 0), (0, 2)$ and $\text{Area} = 4$. 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: $-2K + 8 = 8$ or $-2K + 8 = -8$ Case $1$: $-2K = 0 \Rightarrow K = 0$. Case $2$: $-2K = -16 \Rightarrow K = 8$. Thus,the value of $K$ is $0$ or $8$.

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

The area of a triangle with vertices $(K, 0), (4, 0), (0, 2)$ is $4$ square units. Find the value of $K$.

KCET 2017 All KCET

A
$0$ or $8$
B
$0$ or $-8$
C
$0$
D
$8$

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

KCET 2017 Paper All KCET years →

The value of the determinant $\left| \begin{array}{ccc} 1 & 2 & 3 \\ 3 & 5 & 7 \\ 8 & 14 & 20 \end{array} \right|$ is

Difficult

View Solution

If $\left|\begin{array}{lll}x & x^2 & 1+x^3 \\ y & y^2 & 1+y^3 \\ z & z^2 & 1+z^3\end{array}\right|=0$ and $x, y, z$ are all distinct,then $x y z=$

MediumAP EAMCET 2021

View Solution

The set of all values of $\lambda$ for which the system of linear equations $x - 2y - 2z = \lambda x$,$x + 2y + z = \lambda y$,and $-x - y = \lambda z$ has non-zero solutions.

DifficultJEE MAIN 2019

View Solution

If one of the roots of $\left|\begin{array}{lll}3 & 5 & x \\ 7 & x & 7 \\ x & 5 & 3\end{array}\right|=0$ is $-10$, then the other roots are

EasyTS EAMCET 2009

View Solution

The system of equations $\lambda x + y + z = 0, -x + \lambda y + z = 0, -x - y + \lambda z = 0$ will have a non-zero solution if real values of $\lambda$ are given by

MediumIIT 1984

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 with vertices $(K, 0), (4, 0), (0, 2)$ is $4$ square units. Find the value of $K$.

Similar Questions

The value of the determinant $\left| \begin{array}{ccc} 1 & 2 & 3 \\ 3 & 5 & 7 \\ 8 & 14 & 20 \end{array} \right|$ is

If $\left|\begin{array}{lll}x & x^2 & 1+x^3 \\ y & y^2 & 1+y^3 \\ z & z^2 & 1+z^3\end{array}\right|=0$ and $x, y, z$ are all distinct,then $x y z=$

The set of all values of $\lambda$ for which the system of linear equations $x - 2y - 2z = \lambda x$,$x + 2y + z = \lambda y$,and $-x - y = \lambda z$ has non-zero solutions.

If one of the roots of $\left|\begin{array}{lll}3 & 5 & x \\ 7 & x & 7 \\ x & 5 & 3\end{array}\right|=0$ is $-10$, then the other roots are

The system of equations $\lambda x + y + z = 0, -x + \lambda y + z = 0, -x - y + \lambda z = 0$ will have a non-zero solution if real values of $\lambda$ are given by

Vedclass Test Series

Exam Paper Generator

Online Exam Module