The points A (3, 1), B (12, - 2) and C (0, 2) cannot be vertices of a triangle.
True
Let the coordinates of A = (x1, y1) = (3, 1)
Coordinates of B = (x2, y2) = (12, - 2)
Coordinates of C = (x3, y3) = (0, 2)
Area of ∆ABC = ∆ = 1/2 [x1 (y2 - y3 ) + x2 (y3 - y1 ) + x3 (y1 - y2 )]
Δ= 1/2 [3 - (2 - 2) + 12(2 - 1) + 0{1 - ( - 2)}]
Δ =1/2 [3( - 4) + 12(1) + 0]
Δ =1/2 ( - 12 + 12)=0
Area of ΔABC = 0
Hence, the points A (3, 1), B (12, - 2) and C (0, 2) are collinear.
So, the points A (3, 1), B (12, - 2) and C (0, 2) can’t be the vertices of a triangle.
5