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

Question

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

My attempt

I can operate modulus in equations like $|x-1| + |2x + 1|=1$ but I don't know how to proceed in this question since it contains $2$ variable, $x$ and $y$.

Answer 1

$|x+y-1| + |2x + y + 1|=1$

As usual, you can do this in cases depending on whether $x+y-1$ and $2x+y+1$ are great or equal to zero, or less than zero.

The trick is to realize that $x + y -1 = 0$ and $2x+y+1$ equal zero are two intersecting lines that cut the plane into four regions and each region will represent one of the four cases.

For instance. The region I) above the two lines and to the right of both lines will be the region where $x + y - 1>0$ and $2x + y +1 > 0$

In this region graph $(x+y-1)+(2x+ y + 1) = 1\implies 3x + 2y = 1$. This will be a line but restrict it to the wedge area.

The region II) above and to the right of $2x+y+1 = 0$ and below and to the left of $x+y-1 = 0$ will be the region where $x + y - 1 < 0$ and $2x + y+1 > 0$.

In this region graph $-(x+y-1) + (2x + y + 1) = 1\implies x=-1$. This is just a vertical line. Graph it but restrict it to the wedge area.

And so on. The region III) below and to the left of both lines is wer $x + y -1 < 0$ and $2x +y + 1> 0$.

Graph $-(x+y-1) - (2x + y+1) = 1\implies -3x - 2y = 1 \implies 3x+2y = -1$. Graph this but restriction. It's Worth noting that region I and III graph parallel lines. Try to Think about why that is actually to be expected.

And region IV) is The region II) above and to the right of $x+y-1 = 0$and below and to the left of $2x+y+1 = 0$ will be the region where $x + y - 1 > 0$ and $2x + y+1 <0$.

In this region graph $(x+y-1)- (2x + y + 1) = 1\implies x=-3$. This is just a vertical line. Graph it but restrict it to the wedge area.

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Here is an image of what I am talking about:

Answer 2

The equation: $$ A |L_1| +B |L_2| =C $$, if $L_1,L_2$ are non-paralal lines, and $A,B,C >0$, represents a parallelogram with $L_1$ and $L_2$ as diagonals and their point of intersection is the center of the 'gram.

If $L_1$ and $L_2$ are perpendicular it is square or rhombus. If $L_1,L_2$ are non-perpendicular it is rectangle or a parallelogram. Where

$$S_1= A L_1 + B L_2-C=0, S_2=A L_1-B L_2-C=0, S_3=-A L_1 +B L_2-C=0, S_4=-A L_1 -B L_2-C=0$$ are the equations of the sides. If $S_1=0, S_2=0$ are perpendicular then we have square or rectangle.

In your case the figure is a parallelogram (which is not square, rectangle or rhombus).

Answer 3

If you know already a bit of linear algebra you may proceed as follows:

You surely can draw $|u|+|v|=1$.

Now, consider the mapping

$$\begin{pmatrix}u \\ v\end{pmatrix} = \begin{pmatrix}1 & 1 \\ 2 & 1\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix}+\begin{pmatrix}-1 \\ 1\end{pmatrix}$$

The inverse mapping (just solve for $\begin{pmatrix}x \\ y\end{pmatrix}$) is

$$\begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}-1 & 1 \\ 2 & -1\end{pmatrix}\begin{pmatrix}u\\ v\end{pmatrix}+\begin{pmatrix}-2 \\ 3\end{pmatrix}$$

Since we are dealing with a linear mapping plus a shift (a so-called affine mapping), you only need to find the images of the corners of $|u|+|v|=1$:

• $\begin{pmatrix}1 \\ 0\end{pmatrix},\begin{pmatrix}-1 \\ 0\end{pmatrix},\begin{pmatrix}0 \\ 1\end{pmatrix},\begin{pmatrix}0 \\ -1\end{pmatrix}$

Plugging these into the inverse mapping you get as graph the sides of the parallelogram spanned by $$\boxed{\begin{pmatrix}-3 \\ 5\end{pmatrix},\begin{pmatrix}-1 \\ 1\end{pmatrix},\begin{pmatrix}-1 \\ 2\end{pmatrix},\begin{pmatrix}-3 \\ 4\end{pmatrix}}$$