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If the points $(x+1,\ 2),\ (1,\ x+2)$ and $( \frac{1}{x+1},\ \frac{2}{x+1})$ are collinear, then find $x$.
Given: Points $(x+1,\ 2),\ (1,\ x+2)$ and $( \frac{1}{x+1},\ \frac{2}{x+1})$ are collinear.
To do: To find the value of $x$.
Solution:
As given, points $(x+1,\ 2),\ (1,\ x+2)$ and $( \frac{1}{x+1},\ \frac{2}{x+1})$ are collinear.
$\therefore$ Slope of $BA=$ Slope of $CB$
$\Rightarrow \frac{x+2-2}{1-x-1}=\frac{\frac{2}{x+1}-( x+2)}{\frac{1}{x+1}-1}$
$\Rightarrow \frac{x}{-x}=\frac{\frac{2-( x+1)( x+2)}{x+1}}{\frac{1-x-1}{x+1}}$
$\Rightarrow -1=\frac{2-( x+1)( x+2)}{-x}$
$\Rightarrow 2-( x+1)( x+2)=x$
$\Rightarrow 2-x^2-2x-x-2=x$
$\Rightarrow -x^2-3x=x$
$\Rightarrow x+x^2+3x=0$
$\Rightarrow x^2+4x=0$
$\Rightarrow x( x+4)=0$
$\Rightarrow x=0,\ x=-4$
Thus, for $x=0,\ -4$, given points are collinear.
Updated on: 10-Oct-2022
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