**Question:**

Show that the following sets of points are collinear.

(a) $(2,5),(4,6)$ and $(8,8)$

(b) $(1,-1),(2,1)$ and $(4,5)$

**Solution:**

The formula for the area ' $A$ ' encompassed by three points $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$ is given by the formula,

We know area of triangle formed by three points $\left(x_{1}, y_{1}\right),\left(x_{2} y_{2}\right)$ and $\left(x_{3} y_{3}\right)$ is given by $\Delta=\frac{1}{2}\left|x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right|$

If three points are collinear the area encompassed by them is equal to 0.

The three given points are *A*(2*,* 5)*, B*(4*,* 6) and *C*(8*,* 8). Substituting these values in the earlier mentioned formula we have,

$A=\frac{1}{2}|2(6-8)+4(8-5)+8(5-6)|$

$=\frac{1}{2}|2(-2)+4(3)+8(-1)|$

$=\frac{1}{2}|-4+12-8|$

$=\frac{1}{2}|-12+12|$

$=0$

Since the area enclosed by the three points is equal to 0 , the three points need to be collinear.

The three given points are $A(1,-1), B(2,1)$ and $C(4,5)$. Substituting these values in the earlier mentioned formula we have,

$A=\frac{1}{9}|1(1-5)+2(5+1)+4(-1-1)|$

$=\frac{1}{2}|1(-4)+2(6)+4(-2)|$

$=\frac{1}{2}|-4+12-8|$

$=\frac{1}{2}|-12+12|$

$=0$

Since the area enclosed by the three points is equal to 0 , the three points need to be collinear.