Given: A (4,4), B(3,- 16) and C(3,-2) are the vertices of the triangle ABC Let x1=4,y1=4,x2=3,y2=−16,x3=3 and y3=−2 As we know that, if A(x1,y1),B(x2,y2) and C(x3,y3) are the vertices of a Δ ABC then area of ΔABC=
1
2
.|
x1
y1
1
x2
y2
1
x3
y3
1
| ⇒ ⇒A=
1
2
.|
4
4
1
3
−16
1
3
−2
1
| ⇒A=7squnits Hence, option B is the correct answer.