Question

Find the area of the triangle whose vertices are A(-3/2 , 3) B(6, 2) and C(-3, 4)

Answer 1

Answer:

3 sq. units..

Step-by-step explanation:

Here A (-3/2,3)

X1=-3/2,y1=3

B(6,2)

X2=6,y2=2

C(-3,4)

X3=-3,y3=4

we know, Area of a triangle

= 1/2 {X1(y2-y3)+X2(y3-y1)+(y1-y2)}

putting the values

=1/2{-3/2(2-4)+6(4-3)+ -3(3-1}

=1/2{-3/2*-2 + 6*1 + -3*1}

=1/2{3+6-3}

=1/2*6

=3 sq. units