Using determinants show that the following points are collinear:

Solution:

(i) If the points $(5,5),(-5,1)$ and $(10,7)$ are collinear, then

$\Delta=\left|\begin{array}{ccc}5 & 5 & 1 \\ -5 & 1 & 1 \\ 10 & 7 & 1\end{array}\right|=0$

$=\left|\begin{array}{ccc}5 & 5 & 1 \\ -10 & -4 & 0 \\ 10 & 7 & 1\end{array}\right| \quad\left[\right.$ Applying $\left.R_{2} \rightarrow R_{2}-R_{1}\right]$

$=\left|\begin{array}{ccc}5 & 5 & 1 \\ -10 & -4 & 0 \\ 5 & 2 & 0\end{array}\right| \quad\left[\right.$ Applyin $\left.g R_{3} \rightarrow R_{3}-R_{1}\right]$

$=\left|\begin{array}{cc}-10 & -4 \\ 5 & 2\end{array}\right|=-20+20=0$

Thus, these points are colinear.

(ii) If the points $(1,-1),(2,1)$ and $(4,5)$ are collinear, then

$\Delta=\left|\begin{array}{rrr}1 & -1 & 1 \\ 2 & 1 & 1 \\ 4 & 5 & 1\end{array}\right|=0$

$=\left|\begin{array}{rrr}1 & -1 & 1 \\ 1 & 2 & 0 \\ 4 & 5 & 1\end{array}\right| \quad \quad\left[\right.$ Applying $\left.R_{2} \rightarrow R_{2}-R_{1}\right]$

$=\left|\begin{array}{rrr}1 & -1 & 1 \\ 1 & 2 & 0 \\ 3 & 6 & 0\end{array}\right| \quad \quad\left[\right.$ Applying $\left.R_{3} \rightarrow R_{3}-R_{1}\right]$

 

$=\left|\begin{array}{ll}1 & 2 \\ 3 & 6\end{array}\right|=6-6=0$

Thus, these points are collinear.

(iii) If the points $(3,-2),(8,8)$ and $(5,2)$ are collinear, then

$\Delta=\left|\begin{array}{ccc}3 & -2 & 1 \\ 8 & 8 & 1 \\ 5 & 2 & 1\end{array}\right|=0$

$=\left|\begin{array}{ccc}3 & -2 & 1 \\ 5 & 10 & 0 \\ 5 & 2 & 1\end{array}\right| \quad \quad\left[\right.$ Applying $\left.R_{2} \rightarrow R_{2}-R_{1}\right]$

$=\left|\begin{array}{ccc}3 & -2 & 1 \\ 5 & 10 & 0 \\ 2 & 4 & 0\end{array}\right| \quad \quad\left[\right.$ Applying $\left.R_{3} \rightarrow R_{3}-R_{1}\right]$

 

$=\left|\begin{array}{cc}5 & 10 \\ 2 & 4\end{array}\right|=20-20=0$

Thus the points are colinear.

(iv) If the points $(2,3),(-1,-2)$ and $(5,8)$ are collinear, then

$\Delta=\left|\begin{array}{ccc}2 & 3 & 1 \\ -1 & -2 & 1 \\ 5 & 8 & 1\end{array}\right|=0$

$=\left|\begin{array}{ccc}2 & 3 & 1 \\ -3 & -5 & 0 \\ 5 & 8 & 1\end{array}\right| \quad$ [Applying $R_{2} \rightarrow R_{2}-R_{1}$ ]

$=\left|\begin{array}{ccc}2 & 3 & 1 \\ -3 & -5 & 0 \\ 3 & 5 & 0\end{array}\right| \quad\left[\right.$ Applying $\left.R_{3} \rightarrow R_{3}-R_{1}\right]$

$=\left|\begin{array}{cc}-3 & -5 \\ 3 & 5\end{array}\right|=-15+15=0$

Thus the points are colinear.