Find the values of $ k $ if the points $ \mathrm{A}(k+1,2 k), \mathrm{B}(3 k, 2 k+3) $ and $ \mathrm{C}(5 k-1,5 k) $ are collinear.
Given:
$A( k+1,\ 2k),\ B( 3k,\ 2k+3)$ and $C( 5k+1,\ 5k)$ are collinear.
To do:
We have to find the values of $k$.
Solution:
Let us assume that the points $A(x_1,\ y_1)=(k+1,\ 2k),\ B(x_2,\ y_2)=(3k,\ 2k+3)$ & $C( x_3,\ y_3)=( 5k+1,\ 5k)$ form a triangle.
Now, $Area( \vartriangle ABC)$ with vertices $A(x_1,\ y_1),\ B(x_2,\ y_2)$ and $C(x_3,\ y_3)$ is given by
$Area( \vartriangle ABC)=\frac{1}{2}[x_1( y_2-y_3)+x_2( y_3-y_1)+x_3( y_1-y_2)]$
$\therefore Area( \vartriangle ABC)=\frac{1}{2}[(k+1)(2k+3-5k)+3k(5k-2k)+(5k+1)(2k-2k-3)]$
$\therefore Area( \vartriangle ABC)=2k^2-5k+2$
Since, $A,\ B,\ C$ are collinear
$\Rightarrow Area( \vartriangle ABC)=0$
$\Rightarrow 2k^2-5k+2=0$
$\Rightarrow ( k-2)( 2k-1)=0$
$\Rightarrow k=2,\ \frac{1}{2}$
$\therefore$ for $k=2,\ \frac{1}{2}$, given points are collinear.
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