**Question:**

In the seating arrangement of desks in a classroom three students Rohini, Sandhya and Bina are seated at A(3, 1), B(6, 4), and C(8, 6). Do you think they are seated in a line?

**Solution:**

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.