Question

Question 5
The points A(3,1), B(12,-2) and C(0,2) cannot be vertices of a triangle.

Answer 1

True
Let $A=(x_1,y_1)=(3,1),B=(x_2,y_2)=(12,-2)$
and $C=(x_3,y_3)=(0,2)$
$\because Areaof\Delta ABC=\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]=\frac{1}{2}\left[3\right(-2-2)+12(2-1)+0(1-(-2)\left)\right]=\frac{1}{3}\left[3\right(-4)+12(1)+0]=\frac{1}{3}(-12+12)=0\because Areaof\Delta ABC=0$
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) cannot be the vertices of a triangle.