Draw a quadrilateral in the Cartesian plane, whose vertices are `(-4, 5)` , `(0, 7)`, `(5, -5)`,and `(-4, -2)`. Also, find its area.

Explanation

In the figure the vertices A = (-4,5), B = (0,7), C = (5,-5) and D = (-4,-2) of the quadrilateral are plotted on the graph.

From the given vertices,
`x_(1)=-4, y_(1)=5`
`x_(2)=0, y_(2)=7`
`x_(3)=5, y_(3)=-5`
`x_(4)=-4, y_(4)=-2`
Join the diagonal AC.
Area of `DeltaABC`,
`=(1)/(2)[(x_(1)y_(2)+x_(2)y_(3)+x_(3)y_(1))-(y_(1)x_(2)+y_(2)x_(3)+y_(3)x_(1))]`
`=(1)/(2)[{-4xx7+0xx(-5)+5xx5}-{5xx0+7xx5+(-5)xx(-4)}]`
`=(1)/(2)[(-28+0+25)-(0+35+20)]`
`=(1)/(2)[(-3)-(55)]`
`=(1)/(2)(-58)=29" (neglecting negative sign)"`
Area of `DeltaACD`
`=(1)/(2)[(x_(1)y_(3)+x_(3)y_(4)+x_(4)y_(1))-(y_(1)x_(3)+y_(3)x_(4)+y_(4)x_(1))]`
`=(1)/(2)[{-4xx-5+5xx(-2)+5xx(-4)}-{5xx5+(-5)xx(-4)+(-2)xx(-4)}]`
`=(1)/(2)[(20-10-20)-(25+20+8)]`
`=(1)/(2)[(-10)-(53)]`
`=(1)/(2)[-63] = (63)/(2)" (neglecting negative sign)"`
`=31.5`
Therefore, are of quadrilateral ABCD
= Area of `DeltaABC` + Area of `Delta ACD`
`=29+31.5=60.5` units
Therefore, are of quadrilateral ABCD = 60.5 units.