If left|begin{array}{ccc}2 a & x_{1} & y_{1} | vedclass

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

Question

(B) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\Delta = \frac{1}{2} \left| \begin{array}{ccc} x_1 & y_1 &

Vedclass · Accepted Answer

$1/8$

Vedclass · Answer

(B) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\Delta = \frac{1}{2} \left| \begin{array}{ccc} x_1 & y_1 & 1 \ x_2 & y_2 & 1 \ x_3 & y_3 & 1 \end{array} ight|$.Given the determinant $\left|\begin{array}{ccc}2 a & x_{1} & y_{1} \ 2 b & x_{2} & y_{2} \ 2 c & x_{3} & y_{3}\end{array}ight|=\frac{a b c}{2}$.Taking $2$ common from the first column,we get $2 \left|\begin{array}{ccc} a & x_{1} & y_{1} \ b & x_{2} & y_{2} \ c & x_{3} & y_{3}\end{array}ight|=\frac{a b c}{2}$.Dividing by $abc$,we get $2 \left|\begin{array}{ccc} 1 & x_{1}/a & y_{1}/a \ 1 & x_{2}/b & y_{2}/b \ 1 & x_{3}/c & y_{3}/c \end{array}ight|=\frac{1}{2}$.Thus,$\left|\begin{array}{ccc} 1 & x_{1}/a & y_{1}/a \ 1 & x_{2}/b & y_{2}/b \ 1 & x_{3}/c & y_{3}/c \end{array}ight|=\frac{1}{4}$.The area of the triangle is $\frac{1}{2} \left| \begin{array}{ccc} x_1/a & y_1/a & 1 \ x_2/b & y_2/b & 1 \ x_3/c & y_3/c & 1 \end{array} ight|$.Since the determinant value is $\frac{1}{4}$,the area is $\frac{1}{2} 	imes \frac{1}{4} = \frac{1}{8}$.

Similar Questions

For $0 < \theta < \frac{\pi}{2}$,if $A = \begin{bmatrix} 1 & -\cos \theta & -1 \\ \cos \theta & 1 & -\cos \theta \\ 1 & \cos \theta & 1 \end{bmatrix}$,then which of the following is true regarding $\operatorname{det}(A)$?

If $f(x)=\left|\begin{array}{ccc}x-3 & 2x^2-18 & 2x^3-81 \\ x-5 & 2x^2-50 & 4x^3-500 \\ 1 & 2 & 3\end{array}\right|$,then the value of $f(1) \cdot f(3)+f(3) \cdot f(5)+f(5) \cdot f(1)$ is:

If $\alpha, \beta, \gamma$ $(\alpha < \beta < \gamma)$ are the values of $x$ such that $\begin{vmatrix} x-2 & 0 & 1 \\ 1 & x+3 & 2 \\ 2 & 0 & 2x-1 \end{vmatrix} = 0$ is a singular matrix,then $2\alpha + 3\beta + 4\gamma = $

The set of all values of $t \in R$,for which the matrix $\left[\begin{array}{ccc}e^t & e^{-t}(\sin t-2 \cos t) & e^{-t}(-2 \sin t-\cos t) \\e^t & e^{-t}(2 \sin t+\cos t) & e^{-t}(\sin t-2 \cos t) \\e^t & e^{-t} \cos t & e^{-t} \sin t \end{array}\right]$ is invertible.

The area of the triangle whose vertices are $(3,5), (2,2)$ and $(k, 2)$ is $3$ sq. unit. Then,the value of $k$ is . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module