If the point $(a,a)$ falls between the lines $|x+y|=2$ then prove that $|a|<1$

Question

The lines are $$x+y=2$$ $$x+y=-2$$

This question is very easy to solve using a graph, but is there a way to solve it mathematically, ie. without any drawning, because it’s not always possible to draw a graph.

Answer 1

Note that the solution to the inequality $|x+y|<2$ is the region bounded by the two given parallel lines $|x+y|=2$.

Since the point $(a,a)$ lies between the these two lines, we have $|a+a|<2$, hence $|a|<1$.

Answer 2

Hint:

Actually you want to know where the line $y=x$ cuts $|x+y|=2$. So solve this system and you will get an answer.

Answer 3

The point $(a,a)$ lies on the line $x=y$, which meets the line $x+y=-2$ at $x=-1$. So $(a,a)$ lies above the line $x+y=-2$ if $a>-1$. Similarly, $x=y$ meets the line $x+y=2$ at $x=1$, so $(a,a)$ lies below the line $x+y=2$ if $a<1$.

So the condition for $(a,a)$ to lie between the lines is $|a|<1$.