Question

If the lines $x+y=|a|$ and $ax-y=1$ intersect each other in the first quadrant, then the range of $a$ is

• A. $(1,\infty )$
• B. $(-1,\infty )$
• C. $(-1,1)$
• D. $(0,\infty )$

Answer 1

The correct option is A $(1,\infty )$

Given lines are $x+y=|a|$ and $ax-y=1$
Let the point of intersection be $P$
$y=|a|-x\Rightarrow ax+x-|a|=1\Rightarrow x=\frac{1+|a|}{1+a}\Rightarrow y=\frac{a|a|-1}{1+a}$
Now, we get
$P=(\frac{1+|a|}{1+a},\frac{a|a|-1}{1+a})$

This point will lie in first quadrant iff
$\frac{1+|a|}{1+a}>0,\frac{a|a|-1}{1+a}>0\Rightarrow 1+a>0(\because 1+|a|>0)\Rightarrow a>-1$

Now,
$a|a|-1>0(\because 1+a>0)\Rightarrow a|a|>1$
Since, $|a|\ge 0$ so $a$ should be positive, we get
$a^2>1\therefore a\in (1,\infty )$