Esercizio 
.
.
Per quale valore di 
le seguenti rette sono perpendicolari?
![\[r: \quad (3-k)x+y = -5 \qquad s: \quad (k-2)y-x=2.\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-db92527ef51c69abcb460252d8930ec8_l3.png "Rendered by QuickLaTeX.com")
Svolgimento.
Ricaviamo il coefficiente angolare da entrambe le rette
![\[(3-k)x+y = -5 \quad \Leftrightarrow \quad y = -(3-k)x-5 \quad \Rightarrow \quad m_r = k-3\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-d79c3e45abf32be36cde67c7ec78ba23_l3.png "Rendered by QuickLaTeX.com")
e
![\[(k-2)y-x=2\quad \Leftrightarrow \quad y = \dfrac{1}{k-2} x + 2 \quad \Rightarrow \quad m_s = \dfrac{1}{k-2}\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-9eae1fb807e5a2f91c86cf4fc7d19865_l3.png "Rendered by QuickLaTeX.com")
ed affinché siano parallele deve accadere
![\[m_r = - \dfrac{1}{m_s}\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-0f8da48a2cb6d38dfbf2f7a4846f44b6_l3.png "Rendered by QuickLaTeX.com")
cioè
![\[k-3 = - \dfrac{1}{1/(k-2)} \quad \Leftrightarrow \quad k-3 = -k+2 \quad \Leftrightarrow \quad 2k = 5 \quad \Leftrightarrow \quad \boxed{k = \dfrac{5}{2}}.\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-e46901df38576e6c165c3630fa6094da_l3.png "Rendered by QuickLaTeX.com")