Question

find co ordinate of vertex c . also find the area of quadrilateral ABCD

Answer 1

A (-5, 7), B (-4, -5) and C (4, 5) are the vertices of ΔABC
. Then

(x1 = -5, y1 = 7), (x2 = -4, y2 = -5), (x3 = 4, y3 = 5)

Area of triangle ABC

= 12{x1(y2−y3)+x2(y3−y1)+x3(y1−y2)}
= 12∗{−5(−5???5)+(−4)(5−7)+4(7−(−5))}
= 12∗{−5(−10)−4(−2)+4(12)}
= 12{50+8+48}
= 12{106}
= 53 sq.units