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

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 . . . . . . .

The value of $a$ for which the matrix $A = \begin{bmatrix} a & 2 \\ 2 & 4 \end{bmatrix}$ is singular is:

If the points with position vectors $60 \hat{i}+3 \hat{j}$,$40 \hat{i}-8 \hat{j}$,and $a \hat{i}-52 \hat{j}$ are collinear,then $a$ is equal to

The number of real values of $t$ such that the system of homogeneous equations
$\begin{aligned}
t x+(t+1) y+(t-1) z &=0 \\
(t+1) x+t y+(t+2) z &=0 \\
(t-1) x+(t+2) y+t z &=0
\end{aligned}$
has non-trivial solutions is

Let $0 \neq a \in \mathbb{Z}$ and $A = \begin{bmatrix} a & a & a-y \\ a & a+x & a \\ a & a & a \end{bmatrix}$ be a matrix. Then,the equation $\det(A) = 16$ represents:

Vedclass Test Series

Exam Paper Generator

Online Exam Module