KCET · Maths · Straight Lines
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 \( \frac{3}{2} \)
- C \( -\frac{2}{3} \)
- D \( -\frac{3}{2} \)
Answer & Solution
Correct Answer
(C) \( -\frac{2}{3} \)
Step-by-step Solution
Detailed explanation
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
\[
\begin{array}{l}
m_{1} m_{2}=-1 \\
\Rightarrow \frac{-2}{3} \times\left(-\frac{1}{k}\right)=-1 \\
\Rightarrow \frac{2}{3 k}=-1 \\
\Rightarrow k=\frac{-2}{3}
\end{array}
\]
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