Question

if A (- 5,7) B ( -4,-5) C (-1,-6) D (1,2) vertices of a quadrilateral and find the area of quadrilateral ABCD

Answer 1

Answer:

$Area$ $of$ $ABCD$ $quadrilateral$$=93/2$ $unit$

Step-by-step explanation:

Given,A(-5,7) B(-4,-5) C(-1,-6) D(1,2)

$x_{1}=-5,x_{2}=-4,x_{3}=-1,x_{4}=1$

$y_{1}=7,y_{2}=-5,y_{3}=-6,y_{4}=2$

Area of quadrilateral $ABCD$ in coordinate form

$=1/2\left [ \left | \begin{pmatrix} x_{1} & x_{2} \\ y_{1} & y_{2} \end{pmatrix} \right | +\left | \begin{pmatrix} x_{2} & x_{3} \\ y_{2} & y_{3} \end{pmatrix}\right |+\left | \begin{pmatrix} x_{3} & x_{4} \\ y_{3} & y_{4} \end{pmatrix} \right |+\left | \begin{pmatrix} x_{4} & x_{1} \\ y_{4} & y_{1} \end{pmatrix} \right |\right ]$

$=1/2\left [ \left | \begin{pmatrix} -5 & -4 \\ 7 & -5 \end{pmatrix} \right | +\left | \begin{pmatrix} -4 & -1 \\ -5 & -6 \end{pmatrix}\right |+\left | \begin{pmatrix} -1 & 1 \\ -6 & 2 \end{pmatrix} \right |+\left | \begin{pmatrix} 1 & -5 \\ 2 & 7 \end{pmatrix} \right |\right ]$

$=1/2\left [\left (-5\times -5 -7\times -4)+\left ( -4\times -6--5\times -1 \right )+\left ( -1\times 2--6\times 1 \right )+\left ( 1\times 7 -2\times -5 \right ) \right ]$$=1/2\left [ \left ( 25+28 \right )+\left ( 24-5 \right )+\left ( -2+6 \right )+\left ( 7+10 \right ) \right ]$

$=1/2\left [ 25+28+24-5-2+6+7+10 \right ]$

$=93/2 unit$

     Area of $ABCD$  quadrilateral is      

          $=93/2 unit$