If A(-4,8),B(-3,-4),C(0,-5) and D(5,6) are the …

If A(- 4,8), B(- 3, - 4), C(0, - 5) and D(5,6) are the vertices of a quadrilateral ABCD,then we have to find  its area.

Here , x₁=-4, y₁=8, x₂=-3,y₂=-4,  x₃ =0, y₃ = -5, x₄ = 5, y₄ = 6

Area of quadrilateral
{tex}= \frac { 1 } { 2 } \left[ \left( x _ { 1 } y _ { 2 } - x _ { 2 } y _ { 1 } \right) + \left( x _ { 2 } y _ { 3 } - x _ { 3 } y _ { 2 } \right)\right.{/tex} {tex}\begin{array} { c } { + \left( x _ { 3 } y _ { 4 } - x _ { 4 } y _ { 3 } \right) } {+ \left( x _ { 4 } y _ { 1 } - x _ { 1 } y _ { 4 } \right) ] } \end{array}{/tex}
Here co-ordinates are:
A(-4, 8), B(- 3, - 4), C(0, - 5), D(5, 6)
Area = {tex}\frac { 1 } { 2 } [ \{ - 4 \times ( - 4 ){/tex}- (- 3) (8)} + {(- 3) (- 5) - 0
{tex}× ( - 4 ) \} + \{ 0 \times 6 - 5 ( - 5 ) \} + \{ 5 \times 8 - ( - 4 ) ( 6 ) \}{/tex}]
= {tex}\frac { 1 } { 2 }{/tex}[16 + 24 +15 - 0 + 0 + 25 + 40 + 24]
= {tex}\frac 12{/tex}[40 + 15 + 25 + 40 + 24]
= {tex}\frac { 1 } { 2 } \times 144{/tex} = 72 sq units.