Find the area of the quadrilateral whose vertices are A(-3,1); B(-2,-2

1.	Find the area of the quadrilateral whose vertices are A(-3,1); B(-2,-2); C(4,1); D(2,3): *
Answer» Answer: 35/2 sq.units Step-by-step explanation: Given, A = (-3 , 1) B = (-2 , -2) C = (4 , 1) D = (2 , 3) To Find :- AREA of the quadrilateral. Formula REQUIRED :- $Area=\dfrac{1}{2}\left\|\begin{vmatrix}x_1&y_1\\x_2&y_2\end{vmatrix}+\begin{vmatrix}x_2&y_2\\x_3&y_3\end{vmatrix}+\begin{vmatrix}x_3&y_3\\x_4&y_4\end{vmatrix}+..\begin{vmatrix}x_n&y_n\\x_1&y_1\end{vmatrix}\right\|$ SOLUTION :- x_1 = -3 , y_1 = 1 x_2 = -2 , y_2 = -2 x_3 = 4 , y_3 = 1 x_4 = 2 , y_4 = 3 $Area=\dfrac{1}{2}\left\|\begin{vmatrix}-3&1\\-2&-2\end{vmatrix}+\begin{vmatrix}-2&-2\\4&1\end{vmatrix}+\begin{vmatrix}4&1\\2&3\end{vmatrix}+\begin{vmatrix}2&3\\-3&1\end{vmatrix}\right\|$ $=\dfrac{1}{2}\left\|(6+2)+(-2+8)+(12-2)+(2+9)\right\|$ $=\dfrac{1}{2}\left\|8+6+10+11\right\|$ $=\dfrac{1}{2}\|35\|$ $=\dfrac{35}{2}sq.units$

Find the area of the quadrilateral whose vertices are A(-3,1); B(-2,-2); C(4,1); D(2,3): *

Answer»

Answer:

35/2 sq.units

Step-by-step explanation:

Given,

A = (-3 , 1)

B = (-2 , -2)

C = (4 , 1)

D = (2 , 3)

To Find :-

AREA of the quadrilateral.

Formula REQUIRED :-

$Area=\dfrac{1}{2}\left|\begin{vmatrix}x_1&y_1\\x_2&y_2\end{vmatrix}+\begin{vmatrix}x_2&y_2\\x_3&y_3\end{vmatrix}+\begin{vmatrix}x_3&y_3\\x_4&y_4\end{vmatrix}+..\begin{vmatrix}x_n&y_n\\x_1&y_1\end{vmatrix}\right|$

SOLUTION :-

x_1 = -3 , y_1 = 1
x_2 = -2 , y_2 = -2
x_3 = 4 , y_3 = 1
x_4 = 2 , y_4 = 3

$Area=\dfrac{1}{2}\left|\begin{vmatrix}-3&1\\-2&-2\end{vmatrix}+\begin{vmatrix}-2&-2\\4&1\end{vmatrix}+\begin{vmatrix}4&1\\2&3\end{vmatrix}+\begin{vmatrix}2&3\\-3&1\end{vmatrix}\right|$