(B) 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: $\Delta = \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)

(B) 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: $\Delta = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$ Given vertices are $(-3, 0)$,$(3, 0)$,and $(0, k)$,and the area $\Delta = 9$. Substituting the values: $9 = \frac{1}{2} |-3(0 - k) + 3(k - 0) + 0(0 - 0)|$ $9 = \frac{1}{2} |3k + 3k|$ $9 = \frac{1}{2} |6k|$ $9 = |3k|$ This implies $3k = 9$ or $3k = -9$. Therefore,$k = 3$ or $k = -3$.

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 $(-3, 0)$,$(3, 0)$,and $(0, k)$ is $9$ sq units. Find the value of $k$.

KCET 2023 All KCET

A
$6$
B
$3$
C
$-6$
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

Previous Year

KCET 2023 Paper All KCET years →

Consider three points $P(\cos \alpha, \sin \beta)$, $Q(\sin \alpha, \cos \beta)$, and $R(0,0)$, where $0 < \alpha, \beta < \frac{\pi}{4}$. Then:

MediumWBJEE 2025

View Solution

The area of the triangle with vertices $(3, 8)$, $(-4, 2)$ and $(5, 1)$ is $\frac{P}{4}$. The value of $P$ is:

DifficultKCET 2026

View Solution

In order that the matrix $\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 3 & \lambda & 5 \end{bmatrix}$ be non-singular,$\lambda$ should not be equal to:

Easy

View Solution

Let $\Delta = \left| \begin{array}{ccc} 1 & \omega & 2\omega^2 \\ 2 & 2\omega^2 & 4\omega^3 \\ 3 & 3\omega^3 & 6\omega^4 \end{array} \right|$ where $\omega$ is the cube root of unity,then

Easy

View Solution

If $D = \left|\begin{array}{ccc}1 & -\cos \theta & -1 \\ \cos \theta & 1 & -\cos \theta \\ 1 & \cos \theta & 1\end{array}\right|$,and $p$ and $q$ are the maximum and minimum values of $D$ respectively,then the value of $2p + 3q$ is . . . . . . .

EasyGUJCET 2021

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 $(-3, 0)$,$(3, 0)$,and $(0, k)$ is $9$ sq units. Find the value of $k$.

Similar Questions

Consider three points $P(\cos \alpha, \sin \beta)$, $Q(\sin \alpha, \cos \beta)$, and $R(0,0)$, where $0 < \alpha, \beta < \frac{\pi}{4}$. Then:

The area of the triangle with vertices $(3, 8)$, $(-4, 2)$ and $(5, 1)$ is $\frac{P}{4}$. The value of $P$ is:

In order that the matrix $\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 3 & \lambda & 5 \end{bmatrix}$ be non-singular,$\lambda$ should not be equal to:

Let $\Delta = \left| \begin{array}{ccc} 1 & \omega & 2\omega^2 \\ 2 & 2\omega^2 & 4\omega^3 \\ 3 & 3\omega^3 & 6\omega^4 \end{array} \right|$ where $\omega$ is the cube root of unity,then

If $D = \left|\begin{array}{ccc}1 & -\cos \theta & -1 \\ \cos \theta & 1 & -\cos \theta \\ 1 & \cos \theta & 1\end{array}\right|$,and $p$ and $q$ are the maximum and minimum values of $D$ respectively,then the value of $2p + 3q$ is . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module