Image of parallelogram $f(A)$

Question

Let $f:\mathbb{R}^2\to \mathbb{R}^2$ be function $f(x,y)=(\frac{1}{2}x+y,x-2y)$. Find a image of set $A\subset\mathbb{R}^2$ bounded with lines $x-2y=0, x-2y+2=0, x+2y-2=0, x+2y-3=0.$

Set $A$ is parallelogram with vertices $(1,\frac{1}{2}), (\frac{3}{2},\frac{3}{4}), (\frac{1}{2},\frac{5}{4}), (0,1)$.

What is $f(A)$?

Any help is welcome. Thanks in advance.

Answer 1

Your function $f$ is clearly a linear function. If you check the determinant of the matrix ($\frac 12\cdot -2 - 1\cdot 1$) you find that it is non-zero so the linear function is one-to-one. That means that lines segments map to line segments, parallel lines map to parallel lines, etc.

Thus the image of the parallelogram is another parallelogram. You can find the image parallelogram by finding the images of the vertices of the source parallelogram--these images will be the vertices of the image parallelogram.

Answer 2

Just plug in the coordinates. So $$(1,{1\over 2}) \longmapsto (1,0)$$ $$({3\over 2},{1\over 4}) \longmapsto ({3\over 2},0)$$

$$({1\over 2},{5\over 4}) \longmapsto ({3\over 2},-1)$$

$$(0,1) \longmapsto (1,-2)$$

Now calculate the line betwen $(1,0),$ and $({3\over 2},0)$ and so on.