Question

find the area of triangle abc whose vertices are a(2,2) b(3,4) and c(-1,3)​

Answer 1

answer:

Area of ΔABC whose vertices are is (x1,y1),(x2,y2) and (x3,y3) are

Area of ΔABC=21[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]

In ΔABC, vertices are A(1,2),B(−2,3) and C(−3,−4)

Area of triangle =1/2(1(−2+3)−2(−4−2)−3(2−3))

=1/2(1+12+3)

=8 sq units.

here you go!! ;-)