Question

find area of triangles whose vertices at (1,0),(2,2),(4,3)​

Answer 1

Answer:

let A(1,0) B(2,2) and C(4,3) are vertices of ∆ABC

Area of triangle=1/2[X1(y2-y3)+X2(y3-y1)+x3(y1-y2)]

Here, (X1,y1)=(1,0)

(X2,y2)=(2,2)

(x3,y3)=(4,3)

= 1/2[1(2-3)+2(3-0)+4(1-2)]

= 1/2 [2-3+6+4-2]

=1/2[9]

= 9/2 sq, unit