2017-II-23

Ans: A
I is true. Note that $m_{L_1}={\displaystyle \frac{-1}{m}}$ and $m_{L_2}={\displaystyle \frac{-1}{p}}$. According to the given figure, we have

$\begin{array}{rcl}{m}_{L_1}& >& 0\\ {\displaystyle \frac{-1}{m}}& >& 0\\ m& <& 0\end{array}$

and

$\begin{array}{rcl}{m}_{L_2}& >& 0\\ {\displaystyle \frac{-1}{p}}& >& 0\\ p& <& 0\end{array}$

Hence, we have

$\begin{array}{rcl}{m}_{L_2}& >& m_{L_1}\\ {\displaystyle \frac{-1}{m}}& >& {\displaystyle \frac{-1}{p}}\\ {\displaystyle \frac{1}{m}}& >& {\displaystyle \frac{1}{p}}\\ p& >& m\end{array}$

II is true. Note that the $x$-intercepts of $L_1$ and $L_2$ are $n$ and $q$ respectively. According to the given figure, we have

$\begin{array}{rcl}n& >& q\end{array}$

III is not true. Consider the intersection point of $L_1$ and $L_2$.

$\{\begin{array}{ll}x+my=n& \dots \text{①}\\ x+py=q& \dots \text{②}\end{array}$

$\text{①}–\text{②}$, we have

$\begin{array}{rcl}my–py& =& n-q\\ y& =& {\displaystyle \frac{n-q}{m-p}}\end{array}$

According to the given figure, we have

$\begin{array}{rcl}{\displaystyle \frac{n-q}{m-p}}& <& -1\\ n-q& >& -m+p\\ m+n& >& p+q\end{array}$