Question

lf the slope of the line $\left(\frac{1}{a}+\frac{k}{b}\right)x+(\frac{1}{b}+\frac{k}{a})y-(1+k)=0$ is $-1$, then the value of $k$ is

• A. $2$
• B. $-1$
• C. $1$
• D. $-2$

Answer 1

Answer: (C)

$1$

Slope of given line is $-1.$
So, $(\frac{1}{a}+\frac{k}{b})x+(\frac{1}{b}+\frac{k}{a})y-(1+k)=0$
$\Rightarrow -1=\frac{-(\frac{1}{a}+\frac{k}{b})}{(\frac{1}{b}+\frac{k}{a})}=m$
$\Rightarrow \frac{1}{b}+\frac{k}{a}=\frac{1}{a}+\frac{k}{b}$
$\Rightarrow k(\frac{1}{a}-\frac{1}{b})=(\frac{1}{a}-\frac{1}{b})$
$\Rightarrow k=1.$