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

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

$Given\! : \left ( 2,3 \right ),\left ( -1,-2 \right )\: and \: \left ( 5,8 \right )\! .$

$Explanation\! : V\! ertices\; are \left ( 2,3 \right ),\left ( -1,-2 \right )\: and \: \left ( 5,8 \right )\! .$

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

$= \begin{vmatrix} 2 &3 &1 \\ -1 &-2 &1 \\ 5 &8 &1 \end{vmatrix}$

$= 2\! \begin{vmatrix} -2 &1 \\ 8 &1 \end{vmatrix}-3\! \begin{vmatrix} -1 &1 \\ 5 &1 \end{vmatrix}+1\! \begin{vmatrix} -1 &-2 \\ 5 &8 \end{vmatrix}$

$= 2\! \left (-2-8 \right )-3\! \left ( -1-5 \right )+1\! \left ( -8-\left ( -10 \right ) \right )$

$= 2\! \left ( -10 \right )-3\! \left ( -6 \right )+1\! \left ( -8+10 \right )$

$= -20+18+2$

$= -20+20$

$= 0$

Hence, points are collinear.