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Points A (3, 1), B(12, –2) and C(0, 2) cannot be the vertices of a triangle.

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Answer.      [True]
Solution.         If they are not the vertices of a triangle then
Area of \triangleABC = 0
x1 =3    x2 =12     x3 =0
y1=1     y2=-2     y3=2
Let us find the area of \triangleABC
Area \, of \triangle ABC = \frac{1}{2}\left [ x_{1} \left ( y_{2} -y_{3}\right )+x_{2} \left ( y_{3} -y_{1}\right )+x_{3} \left ( y_{1} -y_{2}\right )\right ]
                                   = \frac{1}{2}\left [ 3\left ( -2-2 \right )+12\left ( 2-1 \right ) +0\left ( 1+2 \right )\right ]
                                   = \frac{1}{2}\left [ 3\left ( -4 \right ) +12\left ( 1 \right )+0\right ]
                                    = \frac{1}{2}\left [ -12+12 \right ]
                                     = \frac{0}{2}
                                      =   0

Area of \triangleABC = 0
Hence, they are collinear or not the vertices of a triangle.

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