Show that the following sets of points are collinear.

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.