#### Please solve RD Sharma class 12 chapter Determinants exercise 5.3 question 2 subquestion (i) maths textbook solution

Hints: First by using the values of vertices find determinant. If value of determinant is zero, then points are collinear.

$Given\! : \left ( 5,5 \right ),\left ( -5,1 \right )\: and \: \left ( 10,7 \right )\! .$

$Explanation\! : V\! ertices\; are \left ( 5,5 \right ),\left ( -5,1 \right )\: and \: \left ( 10,7 \right )\! .$

$Determinant\!= \begin{vmatrix} X_{1} &Y_{1} &1 \\ X_{2} &Y_{2} &1 \\ X_{3} &Y_{3} &1 \end{vmatrix}$

$= \begin{vmatrix} 5 &5 &1 \\ -5 &1 &1 \\ 10 &7 &1 \end{vmatrix}R_{1}\rightarrow R_{1}+R_{2}$

$= \begin{vmatrix} 0 &6 &2 \\ -5 &1 &1 \\ 10 &7 &1 \end{vmatrix}R_{2}\rightarrow R_{2}-R_{3}$

$= \begin{vmatrix} 0 &6 &2 \\ -15 &-6 &0 \\ 10 &7 &1 \end{vmatrix}$

$= 0\begin{vmatrix} -6 &0 \\ 7 &1 \end{vmatrix}-6\begin{vmatrix} -15 &0 \\ 10 &1 \end{vmatrix}+2\begin{vmatrix} -15 &-6 \\ 10 &7 \end{vmatrix}$

$= 0-6\left ( -15-0 \right )+2\left ( -105+60 \right )$

$= 90+2\left ( -45 \right )$

$= 90-90$

$= 0$

Hence, points are collinear.