Points \(A(3,1),\ B(12,-2),\ C(0,2)\) cannot be vertices of a triangle.
True.
Step 1: Three points form a triangle only if they are not collinear. If the area of the triangle formed by them is 0, then they are collinear and cannot form a triangle.
Step 2: Use the area of triangle formula for points \(A(x_1,y_1)\), \(B(x_2,y_2)\), \(C(x_3,y_3)\):
\(\text{Area} = \dfrac{1}{2}\left|x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)\right|\).
Step 3: Substitute \(A(3,1)\), \(B(12,-2)\), \(C(0,2)\).
Step 4: Compute the expression inside the absolute value:
\(x_1(y_2-y_3) = 3\big((-2)-2\big) = 3(-4) = -12\)
\(x_2(y_3-y_1) = 12\big(2-1\big) = 12(1) = 12\)
\(x_3(y_1-y_2) = 0\big(1-(-2)\big) = 0\times 3 = 0\)
Step 5: Add them:
\(-12 + 12 + 0 = 0\)
Step 6: Area of \(\triangle ABC\):
\(\text{Area} = \dfrac{1}{2}|0| = 0\)
Final Step: Since the area is 0, points A, B, and C are collinear. Therefore, they cannot be vertices of a triangle, so the statement is True.