In the seating arrangement of desks in a classroom three students Rohini,

The distance d between two points  and  is given by the formula

$d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$

For three points to be collinear the sum of distances between any two pairs of points should be equal to the third pair of points.

The given points are A(3,1), B(6,4) and C(8,6).

Let us find the distances between the possible pairs of points.

$A B=\sqrt{(3-6)^{2}+(1-4)^{2}}$

$=\sqrt{(-3)^{2}+(-3)^{2}}$

$=\sqrt{9+9}$

$A B=3 \sqrt{2}$

$A C=\sqrt{(3-8)^{2}+(1-6)^{2}}$

$=\sqrt{(-5)^{2}+(-5)^{2}}$

$=\sqrt{25+25}$

$A C=5 \sqrt{2}$

$B C=\sqrt{(6-8)^{2}+(4-6)^{2}}$

$=\sqrt{(-2)^{2}+(-2)^{2}}$

$=\sqrt{4+4}$

$B C=2 \sqrt{2}$

We see that $A B+B C=A C$.

Since sum of distances between two pairs of points equals the distance between the third pair of points the three points must be collinear.

Hence, the three given points are.