Find the values of k, if the points A(k + 1, 2k),

Solution:

We know that, if three points are collinear, then the area of triangle formed by these points is zero.

Since, the points A(k + 1,2k), B(3k, 2k + 3) and C(5k -1, 5k) are collinear.

Then, area of ΔABC = 0

$\Rightarrow$ $\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right]=0\right.$

Here, $\quad x_{1}=k+1, x_{2}=3 k, x_{3}=5 k-1$ and $y_{1}=2 k_{1}, y_{2}=2 k+3, y_{3}=5 k$

$\Rightarrow \quad \frac{1}{2}[(k+1)(2 k+3-5 k)+3 k(5 k-2 k)+(5 k-1)(2 k-(2 k+3))]=0$

$\Rightarrow \quad \frac{1}{2}[(k+1)(-3 k+3)+3 k(3 k)+(5 k-1)(2 k-2 k-3)]=0$

$\Rightarrow \quad \frac{1}{2}\left[-3 k^{2}+3 k-3 k+3+9 k^{2}-15 k+3\right]=0$

$\Rightarrow$ $\frac{1}{2}\left(6 k^{2}-15 k+6\right)=0 \quad$ [multiply by 2$]$

$\Rightarrow \quad 6 k^{2}-15 k+6=0$ [by factorisation metbbd]

$\Rightarrow \quad 2 k^{2}-5 k+2=0$ [divide by 3]

$\Rightarrow \quad 2 k^{2}-4 k-k+2=0$

$\Rightarrow \quad 2 k(k-2)-1(k-2)=0$

$\Rightarrow \quad(k-2)(2 k-1)=0$

If $k-2=0$, then $k=2$

If $2 k-1=0$, then $k=\frac{1}{2}$

$\therefore$ $k=2, \frac{1}{2}$

Hence, the required values of $k$ are 2 and $\frac{1}{2}$