Question

find the area of the triangle ABC whose vertices A(3,2) B (-2,8) C(6,-10) ​

Answer 1

Area of triangle=1/2[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]
After putting all the values as
(x1,y1)=A(3,2)
(x2,y2)=B(-2,8)
(x3,y3)=C(6,-10)
We get Area of(∆)=1/2(54+24+(-36))
=1/2(32)=16

Answer 2

Answer:

Step-by-step explanation:

Area of triangle=1/2[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]

After putting all the values as

(x1,y1)=A(3,2)

(x2,y2)=B(-2,8)

(x3,y3)=C(6,-10)

We get Area of(∆)=1/2(54+24+(-36))

=1/2(32)=16