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!

:)

1.3

step-by-step explanation:

3x2=6   9x+6

bring down the 6 on both sides

6+6=12

12= 9x

divide by 9

12/9 =

1.3 =x

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step-by-step explanation: