If A (-3, 5), B (-2,-7), C (1,-8) and D (6, 3) are the vertices of a quadrilateral ABCD, find its area.

Given vertices of a quadrilateral ABCD are A(−3, 5), B(−2, −7), C(1, −8) and D(6, 3)

Area of the quadrilateral ABCD = Area of ∆ABC + Area of ∆ACD

Area of the triangle having vertices (x₁,y₁), (x₂,y₂) and (x₃,y₃)

= |x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)|

Area of ∆ABC = | − 3[ − 7 − ( − 8 )] + ( − 2) ( − 8 – 5 ) + 1 [ 5 − (− 7) ] |

= | − 3 + 26 + 12 |

=sq. units

Area of ∆ACD = |− 3( − 8 – 3 ) + 1( 3 – 5 ) + 6[ 5 − ( − 8 ) ] |

= | 33 – 2 + 78 |

= sq. units

Area of the quadrilateral ABCD = + = 72 sq. units

∴Hence, the area of the quadrilateral is 72 sq. units.