Question

Find the area of the triangle whose vertices are x(3,2) y(-1,0)and z(1,-12)

Answer 1

The triangle whose vertices are x(3,2) y(-1,0)and z(1,-12).

=> xy = sqrt(4^2 + 2^2) = 4.472

=> xz = sqrt(2^2 + 14^2) = 14.142

=> yz = sqrt(2^2 + 12^2) = 12.166

=> Perimeter/2 = P = (xy + yz + xz)/2 = (4.472 + 14.142 + 12.166)/2 = 15.39

=> Applying Heron's formula, the area of triangle would be:

A = sqrt(P*(P-xy)*(P-xz)*(P-yz))

= sqrt(15.39*(15.39 - 4.472)*(15.39 - 14.142 )*(15.39 - 12.166))

= 26.001

= ~26

Hope this helps!

:)