Please solve RD Sharma class 12 chapter Determinants exercise 5.3 question 10 maths textbook solution

Answer: $x$=3

Hints: Use vertices in the determinant formula and equate it with zero because if points are collinear then determinant will be zero and then find the value of x.

Given: ($x$, -2), (5,2) and (8,8) are collinear.

Explanation: Vertices are ($x$, -2), (5,2) and (8,8)

$\text { Determinant }=\left|\begin{array}{lll} X_{1} & Y_{1} & 1 \\ X_{2} & Y_{2} & 1 \\ X_{3} & Y_{3} & 1 \end{array}\right|=0$

$\Rightarrow \left|\begin{array}{lll} x & -2 & 1 \\ 5 & 2 & 1 \\ 8 & 8 & 1 \end{array}\right|=0$

$\Rightarrow x\left|\begin{array}{ll} 2 & 1 \\ 8 & 1 \end{array}\right|-(-2)\left|\begin{array}{cc} 5 & 1 \\ 8 & 1 \end{array}\right|+1\left|\begin{array}{cc} 5 & 2 \\ 8 & 8 \end{array}\right|=0$

$\Rightarrow x(2-8)+2(5-8)+1(40-16)=0$

$\Rightarrow x(-6)+2(-3)+1(24)=0$

$\Rightarrow -6x-6+24=0$

$\Rightarrow -6x+18=0$

$\Rightarrow 6x=18$

$\Rightarrow x=3$