If the straight line \(2 x-3 y+17=0\) is perpendicular to the line passing through the points \((7,17)\) and \((15, \beta)\), then \(\beta\) equals

Equation of line passing through \((7,17)\) and \((15, \beta)\) is
\[
\begin{aligned}
& (y-17)=\left(\frac{\beta-17}{15-7}\right)(x-7) \\
& \Rightarrow \quad y-17=\frac{(\beta-17)}{8}(x-7) \text {. } \\
& \Rightarrow \quad y=\left(\frac{\beta-17}{8}\right) x-\frac{7(\beta-17)}{8}+17 \\
&
\end{aligned}
\]
On comparing to \(y=m x+c\), we get
Slope, \(m_1=\frac{\beta-17}{8}\)
It is given that above line is perpendicular to
\[
2 x-3 y+17=0
\]

\[
\begin{aligned}
& \Rightarrow \quad y=\frac{2}{3} x+\frac{17}{3} \\
& \therefore \text { Slope, } m_2=\frac{2}{3}
\end{aligned}
\]
Condition for perpendicular lines, \(m_1 \cdot m_2=-1\)
\[
\begin{aligned}
& \left(\frac{\beta-17}{8}\right) \times \frac{2}{3}=-1 \\
\Rightarrow \quad & \beta-17=-12 \Rightarrow \beta=5
\end{aligned}
\]