# 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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Updated on: 10-Oct-2022

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