Q74 / 180 · KCET 2022 · Mathematics
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
- A-5
- Correct answerB5
- C29
- D-29
Solution
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, $m_1 \cdot m_2=-1$
$$
\begin{aligned}
& \left(\frac{\beta-17}{8}\right) \times \frac{2}{3}=-1 \\
\Rightarrow \quad & \beta-17=-12 \Rightarrow \beta=5
\end{aligned}
$$
Hence, the correct answer is option: 'B'.