Find the area of quadrilateral ABCD, whose vertices are: $A( -3,\ -1) ,\ B( -2,\ -4) ,\ C( 4,\ -1)$ and$\ D( 3,\ 4) .$

Given: Here given a quadrilateral ABCD, where $A( -3,\ -1) ,\ B( -2,\ -4) ,\ C( 4,\ -1)$ and $D( 3,\ 4)$

To do: To find out the area of the given quadrilateral ABCD.

As shown in the figure, ABCD is the given quadrilateral.

Area of quadrilateral $ABCD=area( ΔABC) +area( ΔADC)$

And we know that area of a triangle with vertices $( x_{1} ,\ y_{1})$, $( x_{2}, y_{2})$and $( x_{3}, y_{3})$

$\frac{1}{2}[ x_{1}( y_{2} -y_{3}) +x_{2}( y_{3} -y_{1}) +x_{3}( y_{1} -y_{2})]$

here as shown in the fig. we find the vertices $A( 1,\ 3) ,\ B( -1,\ 0)$ and $C( 4,\ 0)$

$\therefore$ Area of $\vartriangle ABC=\frac{1}{2}[ -3( -4+1) -2( -1+1) +4( -1+4)]$

$=\frac{1}{2}( 9+0+12)$

$=\frac{21}{2}$

$=10.5$ sq.unit

And similarly, Area of $\vartriangle ADC\ =\frac{1}{2}[ -3( 4+1) +3( -1+1) +4( -1-4)]$

$=\frac{1}{2}( -15+0-20)$

$=\frac{-35}{2}$

$\because$ area can’t be negative,

$\therefore$ area of $\vartriangle ADC$ is $\frac{35}{2}$ square units 

Area of quadrilateral ABCD$=area( \vartriangle ABC) +area( \vartriangle ADC) =\frac{21}{2} +\frac{35}{2} =\frac{66}{2} =33$ square unit