What is the value of K to make this true

EDIT:

It's actually PERPENDICULAR, not PARALLEL.

... I've been working on this for so much time and was wondering why k= 11/9 wasn't my answer haha

k=11/9 is the answer provided in the solution of the book.

We have two  different straight lines: {nl} (k-2)x - (k-1)y -7 = 0 and

2x - 7y + 5 = 0 {nl} Find the value of K if those two straight lines are PERPENDICULAR.

Slope line 1 = m₁:

$\begin{array}{|rcll|} \hline (k-2)x - (k-1)y -7 &=& 0 \quad & | \quad + 7 \\ (k-2)x - (k-1)y &=& 7 \quad & | \quad \cdot (-1) \\ -(k-2)x + (k-1)y &=& -7 \quad & | \quad + (k-2)x \\ (k-1)y &=& -7 + (k-2)x \quad & | \quad :(k-1) \\ y &=& \frac{-7}{k-1} + \underbrace{(\frac{k-2}{k-1} )}_{=m_1}\cdot x \\ \hline \end{array}$

Slope line 2 = m₂:

$\begin{array}{|rcll|} \hline 2x - 7y + 5 &=& 0 \quad & | \quad -5 \\ 2x - 7y &=& -5 \quad & | \quad \cdot (-1) \\ -2x + 7y &=& 5 \quad & | \quad + 2x \\ 7y &=& 5 + 2x \quad & | \quad :7 \\ y &=& \frac57 + \underbrace{\frac27}_{=m_2}\cdot x \\ \hline \end{array}$

PERPENDICULAR: $m_2 = -\dfrac{1}{m_1}$

$\begin{array}{|rcll|} \hline m_2 &=& -\frac{1}{m_1} \quad & | \quad m_1=\frac{k-2}{k-1} \quad m_2=\frac27 \\ \frac27 &=& -\frac{1}{\frac{k-2}{k-1}} \\ \frac27 &=& -\frac{k-1}{ k-2 } \\ 2\cdot (k-2) &=& -7\cdot (k-1) \\ 2k-4 &=& -7k+7 \quad & | \quad + 7k \\ 9k-4 &=& +7 \quad & | \quad + 4 \\ 9k &=& 11 \quad & | \quad : 9 \\ \mathbf{ k } & \mathbf{=} & \mathbf{ \frac{11}{9} } \\ \hline \end{array}$