Prove

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\(\large \color{black}{\begin{align} &\text{Prove that area of the figure}\hspace{.33em}\\~\\ &|x|+|y|=k ,\ k>0,\ k\in \mathbb{R}\hspace{.33em}\\~\\ &\text{is same as area of figure }\hspace{.33em}\\~\\ &|x+a|+|y+b|=k,\ \{a,b\}\in \mathbb{R}\hspace{.33em}\\~\\ \end{align}}\)
Almost the same as the previous one - try to figure out the four lines.
\(\large \color{black}{\begin{align} x+y=k\hspace{1.5em}\\~\\ x-y=k\hspace{1.5em}\\~\\ -x+y=k\hspace{1.5em}\\~\\ -x-y=k\hspace{1.5em}\\~\\ \end{align}}\)

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Other answers:

Yeah. Now graph them and see what you get.
square with area \(k^2\)
Same process for the second case.
what about 'a' and 'b'
don't worry about them. figure out the equations first.
! :D
looks like the area is not \(k^2\) though.
\(\large \color{black}{\begin{align} x+a-y-b=k\hspace{1.5em}\\~\\ x+a+y+b=k\hspace{1.5em}\\~\\ -x-a+y+b=k\hspace{1.5em}\\~\\ -x-a-y-b=k\hspace{1.5em}\\~\\ \end{align}}\)
the diagonal of the square in the first case is \(2k\), so the side length is \(k\sqrt 2\)
yeah, can you manage the rest now?
area of the first is \(2k^2\) ?
that's correct.
and how to find 2nd one
just find out the points where the lines meet and those would be the vertices of the square.
hint: if we make this traslation: \[\large \left\{ \begin{gathered} x + a = X \hfill \\ y + b = Y \hfill \\ \end{gathered} \right.\] where X,Y are the new coordinates, then the second equation can be rewritten as follows: \[\large \left| X \right| + \left| Y \right| = k\] and such equation has the same shape of the first one
would this be enough for proof
Yes I think so, since we have applied a traslation, and not a dilation, namely when we traslate a geometrical shape its area will be unchanged
ok thnx both
:)
That is actually a nice way to see it, but it seems that he actually has to *prove* that translation has no effect on the shape, and in turn, has no effect on area.
I think that a proof of my statement above, can be this: when we traslate a rigid body into the euclidean space, its volume is unchanged, now we can imagine to build the geometrical shape involved into our exercise using rigid rods, so we can conclude that the area enclosed by our geometrical shape will be unchanged after a traslation

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