Question

the area of the quadrilateral having vertices (2,-1),(-5,-4),(-6,7),(3,2) is??​

Answer 1

Answer:

The area of the quadrilateral is 56sq.units

Step-by-step explanation:

Given that,

The vertices of the quadrilateral are (2,-1),(-5,-4),(-6,7),(3,2) and we are required to find the area of the quadrilateral formed by these vertices.

The area of quadrilateral with vertices $(x_{1},y_{1}),(x_{2},y_{2}),(x_{3},y_{3}),(x_{4},y_{4})$ is

$A=\frac{1}{2}\left|\begin{array}{ccccc}x_{1}&x_{2}&x_{3}&x_{4}&x_{1}&y_{1}&y_{2}&y_{3}&y_{4}&y_{1}\\\\\end{array}\right|$

We have to find the determinant of above matrix in an cyclic order as shown below.

The area of the quadrilateral is

$A=\frac{1}{2}\left|\begin{array}{ccccc}2&-5&-6&3&2&-1&-4&7&2&-1\\\\\end{array}\right|\\=\frac{1}{2}|(-8-5)+(-35-24)+(-12-21)+(-3-4)|\\=\frac{1}{2}|-13-59-33-7|=56sq.units$

Therefore,

The area of the quadrilateral is 56sq.units

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