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

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!

:)

all the angles opposite to the angle 150 would be 150 and all the angles that are next to 150 are 30 because they equal 180 it you add it up.

step-by-step explanation: