# Thread: Confusion on how to graph this eqn

1. ## Confusion on how to graph this eqn

Hi

I'm confused as to how I'm supposed to graph an eqn.

I was asked to find the area of the region for |x-3| + |y-3| = 3. I thought that I could solve it by sketching the graph. But I got stuck at the |y-3| portion.

Whoever who can help me, please don't solve the whole thing as it takes the whole fun out of it. If you could just point me in the right direction with some tips/ hints, that would be good enough.

2. Originally Posted by dd86
Hi

I'm confused as to how I'm supposed to graph an eqn.

I was asked to find the area of the region for |x-3| + |y-3| = 3. I thought that I could solve it by sketching the graph. But I got stuck at the |y-3| portion.

Whoever who can help me, please don't solve the whole thing as it takes the whole fun out of it. If you could just point me in the right direction with some tips/ hints, that would be good enough.
Happy New Year!

1. Re-arrange the equation:

$\displaystyle |y-3| = 3-|x-3|$

2. Use the definition of the absolute value:

$\displaystyle |y-3|=\left\{\begin{array}{rcl}y-3&if&y-3\geq 0~\implies~y\geq3 \\ -(y-3)&if&y-3< 0~\implies~y<3\end{array}\right.$

3. You'll get two equations:

$\displaystyle y = 6-|x-3|~\wedge~y\geq 3$
and
$\displaystyle y = |x-3|~\wedge~y< 3$

4. The graph of the complete (original) equation is a rhombus.

3. ARRRRGGGGHHH!

How could I miss that?! Sorry for being so blur!

Thanks!

4. Alternatively

$\displaystyle |x-3|+|y-3|=3$

$\displaystyle (1)\;\;\;x,\;y\ \ge\ 3\Rightarrow\ x-3+y-3=3\Rightarrow\ x+y=9$

$\displaystyle (2)\;\;\;x\ \ge\ 3,\;y<3\Rightarrow\ x-3-(y-3)=3\Rightarrow\ x-y=3$

Continue for

$\displaystyle (3)\;\;\;x<3,\;y\ \ge\ 3$

$\displaystyle (4)\;\;\;x<3,\;y<3$

5. Your approach seems more familiar to me, Archie.

Was just confused how to adapt it since I had 2 different variables.