Q101 / 180 · KCET 2016 · Mathematics
If the straight lines \( 2 x+3 y-3=0 \) and \( x+k y+7=0 \) are perpendicular, then the value of \( k \) is
- A\( \frac{2}{3} \)
- B\( \frac{3}{2} \)
- Correct answerC\( -\frac{2}{3} \)
- D\( -\frac{3}{2} \)
Solution
Given lines,
\[ \begin{array}{l} 2 x+3 y-3=0 \rightarrow(1) \\ x+k y+7=0 \rightarrow(2) \end{array} \]
From Eqs. (1) and (2), we get
\[ m_{1}=\frac{-2}{3} \text { and } m_{2}=-\frac{1}{k} \]
Since, both lines are perpendicular so
\[ \begin{array}{l} m_{1} m_{2}=-1 \\ \Rightarrow \frac{-2}{3} \times\left(-\frac{1}{k}\right)=-1 \\ \Rightarrow \frac{2}{3 k}=-1 \\ \Rightarrow k=\frac{-2}{3} \end{array} \]
\[ \begin{array}{l} 2 x+3 y-3=0 \rightarrow(1) \\ x+k y+7=0 \rightarrow(2) \end{array} \]
From Eqs. (1) and (2), we get
\[ m_{1}=\frac{-2}{3} \text { and } m_{2}=-\frac{1}{k} \]
Since, both lines are perpendicular so
\[ \begin{array}{l} m_{1} m_{2}=-1 \\ \Rightarrow \frac{-2}{3} \times\left(-\frac{1}{k}\right)=-1 \\ \Rightarrow \frac{2}{3 k}=-1 \\ \Rightarrow k=\frac{-2}{3} \end{array} \]
Hence, the correct answer is option: 'C'.