Question

A value of $k$ such that the straight lines $y-3x+4=0$ and $(2k-1)x-(8k-1)y-6=0$ are perpendicular is

• A. $\frac{2}{7}$
• B. $-\frac{2}{7}$
• C. $1$
• D. $0$

Answer 1

Answer: (A)

$\frac{2}{7}$

Let $L_1:y-3x+4=0$ and $L_2:(2k-1)x-(8k-1)y-6=0$.

From the above equations, the slopes of the lines are:

$m_1=3$ and $m_2=\frac{2k-1}{8k-1}$

For two lines to be perpendicular, the necessary condition is $m_1{m}_2=-1$

$\therefore 3\times \frac{2k-1}{8k-1}=-1$

$\Rightarrow 6k-3=1-8k$

$\Rightarrow 14k=4$

$\Rightarrow k=\frac{2}{7}$