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I'm trying to graph $|x+y|+|x-y|=4$. I rewrote the expression as follows to get a function that resembles the direction of unit vectors at $\pi/4$ to the horizontal axis (take it to be $x$)$$\biggl|\dfrac{x+y}{\sqrt{2}}\biggr|+\biggl|\dfrac{x-y}{\sqrt{2}}\biggr|=2\sqrt{2}$$

However, I'm not able to proceed further. Any hints are appreciated. Notice this is an exam problem, so time-efficient methods are key. Please provide any hints accordingly.

Paras Khosla
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7 Answers7

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The lines $x+y=0$ and $x-y=0$ define a coordinate system rotated $45^\circ$ from the canonical axes and scaled by a factor of $\sqrt2$. Accordingly, $|x+y|+|x-y|=4$ defines a diamond (rotated square) in this new coordinate system, which corresponds to a (axis-aligned, centred) square in the canonical one.

To determine the size of this square, note that $(x,y)=(2,0)$ satisfies the equation. Thus the graph of $|x+y|+|x-y|=4$ is a square of side $4$ centred on the origin.

Parcly Taxel
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The given graph forms a square with side 4.Simple logic is that when (X+y)+(x-y)=c When (X+y) approaches c,(x-y) approaches 0. From the above thing the upper boundary will be 2x=4. You get X=2,y=2. Because there is mod if you can solve one boundary it gets mirrored about both X and y axis

ASTRONO
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It is a square with sides as $x=\pm 2$ and $y=\pm 2$,and diagonals as $y=\pm x$. for the explanation see my answer in MSE below.

How to draw graph of $|x+y-1| + |2x + y + 1|=1$?

Z Ahmed
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I think you dismiss the solution by cases too quickly. Actually you only need to do one case, and the rest is developed by symmetry.

The easiest case is when $x+y\geq 0$ and $x - y \geq 0,$ equivalently $x\geq -y$ and $x \geq y,$ or in other words when $(x,y)$ is to the right of both of the lines $x = y$ and $x= -y.$

In that case the formula simplifies to $$ (x+y) + (x - y) = 4, $$ that is, $2x = 4,$ so $x = 2.$ We have a vertical line segment from $(2,-2)$ to $(2,2).$

Now observe that the result of $\lvert x+y\rvert +\lvert x-y\rvert$ does not change if we swap $x$ and $y$. So we also have the mirror image through the line $y=x,$ which says we also have the horizontal line segment from $(-2,2)$ to $(2,2).$

Notice how this is two adjacent sides of the square with vertices $(-2,2),$ $(2,2),$ $(2,-2),$ and $(-2,-2).$

Also observe that the result of $\lvert x+y\rvert +\lvert x-y\rvert$ does not change if we replace $x$ and $y$ with $-x$ and $-y$. So we also have symmetry via a reflection through the origin, which is also a $180$-degree rotation around the origin. That gives us the other two sides of the square.

David K
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I would take cases. When $x > y$

$$|x+y| + |x-y| = 4 \implies x = 2$$

Try to figure out the rest of the cases on your own then graph everything together.

Ninad Munshi
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  • Thanks for answering. But as I said, I would like to know about time-efficient methods. – Paras Khosla Sep 20 '19 at 02:23
  • @ParasKhosla It is only time efficient with practice. Then you recognize patterns. Don't be lazy, there are only 4 cases and they have a nice pattern to it like this one. – Ninad Munshi Sep 20 '19 at 02:25
  • As I said even spending a minute is a lot for this question. Besides this isn't a hint, making cases was the first obvious thing that comes to mind. I'm looking for innovative solutions. Could you think about other methods? Thanks – Paras Khosla Sep 20 '19 at 02:31
  • The only faster way is to know the solution ahead of time. Actually the way I solved this initially was recognizing that this was a rotated (by $\frac{\pi}{4}$) version of the function $|x| + |y| = C$, which is common knowledge to be a diamond with corners on the axes. That's why I said it is only time efficient with pattern recognition and practice. Don't be lazy, the cases doesn't take nearly as long as you're making it out to be if you don't see that immediately. 35 seconds, tops. – Ninad Munshi Sep 20 '19 at 02:35
  • Yes I found that method efficient, the one involving rotation of the axes. This was what you should've answered in the first place. Thanks – Paras Khosla Sep 20 '19 at 03:23
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Once $x$ is fixed you can look for solutions in three distinct regions: $y>x$, $-x<y<x$, $y<-x$. By doing this you can simplify the absolute values and find solutions.

fGDu94
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  • Working all the cases would take a long time, do you know of any other time-efficient methods? – Paras Khosla Sep 20 '19 at 02:24
  • I think this would take a couple of minutes, do you need something faster? – fGDu94 Sep 20 '19 at 02:24
  • This exam literally allows $2$ minutes a question and this is something which would comprise just one-tenth or so of the solution, so spending a minute on just this is a lot to ask for. – Paras Khosla Sep 20 '19 at 02:26
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Looks like you have said

$u = \frac {\sqrt {2}}{2} (x + y)\\ v = \frac {\sqrt {2}}{2} (x - y)$

Which rotates coordinate system 45 degrees.

And your expression becomes

$|\sqrt 2 u| + |\sqrt 2 v| = 4$ or $|u| + |v| = \sqrt 2$

So what does this graph look like in an unrotated frame?

Do you know about the taxicab metric?

https://en.wikipedia.org/wiki/Taxicab_geometry

What we have here is the set of points that are $\sqrt 2$ units from the origin by the taxicab metric.

enter image description here

And rotate coordinates back to the original system.

Doug M
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