#### Provide solution for RD Sharma maths class 12 chapter Determinants exercise 5.3 question 11

Answer: $x$=5

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: (3,-2), ($x$,2) and (8,8).

Explanation: Vertices are (3,-2), ($x$,2) and (8,8).

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

$\Rightarrow \left|\begin{array}{lll} 3 & -2 & 1 \\ x & 2 & 1 \\ 8 & 8 & 1 \end{array}\right|=0$

$\Rightarrow 3\left|\begin{array}{ll} 2 & 1 \\ 8 & 1 \end{array}\right|-(-2)\left|\begin{array}{cc} x & 1 \\ 8 & 1 \end{array}\right|+1\left|\begin{array}{cc} x & 2 \\ 8 & 8 \end{array}\right|=0$

$\Rightarrow 3(2-8)+2(x-8)+1(8x-16)=0$

$\Rightarrow 3(-6)+2(x-8)+8x-16=0$

$\Rightarrow -18+2x-16+8x-16=0$

$\Rightarrow 10x-50=0$

$\Rightarrow x=\frac{50}{10}$

$\Rightarrow x=5$

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