Question

find the area of triangle whose vertices are (2,1) (10,1) and (6,9)​

Answer 1

let the three points be A(2,1) ,B(10,1) , C(6,9)

Area of triangle ABC =½[x¹(y²-y³)+x²(y³-y¹)+x³(y¹-y²)]

Here

x¹=2, y¹=1

x²= 10, y²=1

x³=6, y³=9

Putting values

Area of triangle ABC=½[2(1-9)+10(9-2)+6(1-1)]

=½[2(-8)+10(7)+6(0)]

=½[-16+70+0]

=½[54]

= 27 square units