mathmath333
  • mathmath333
Prove
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} &\text{Prove that area of the figure}\hspace{.33em}\\~\\ &|x|+|y|=k ,\ k>0,\ k\in \mathbb{R}\hspace{.33em}\\~\\ &\text{is same as area of figure }\hspace{.33em}\\~\\ &|x+a|+|y+b|=k,\ \{a,b\}\in \mathbb{R}\hspace{.33em}\\~\\ \end{align}}\)
ParthKohli
  • ParthKohli
Almost the same as the previous one - try to figure out the four lines.
mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} x+y=k\hspace{1.5em}\\~\\ x-y=k\hspace{1.5em}\\~\\ -x+y=k\hspace{1.5em}\\~\\ -x-y=k\hspace{1.5em}\\~\\ \end{align}}\)

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ParthKohli
  • ParthKohli
Yeah. Now graph them and see what you get.
mathmath333
  • mathmath333
square with area \(k^2\)
ParthKohli
  • ParthKohli
Same process for the second case.
mathmath333
  • mathmath333
what about 'a' and 'b'
ParthKohli
  • ParthKohli
don't worry about them. figure out the equations first.
Here_to_Help15
  • Here_to_Help15
! :D
ParthKohli
  • ParthKohli
looks like the area is not \(k^2\) though.
mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} x+a-y-b=k\hspace{1.5em}\\~\\ x+a+y+b=k\hspace{1.5em}\\~\\ -x-a+y+b=k\hspace{1.5em}\\~\\ -x-a-y-b=k\hspace{1.5em}\\~\\ \end{align}}\)
ParthKohli
  • ParthKohli
the diagonal of the square in the first case is \(2k\), so the side length is \(k\sqrt 2\)
ParthKohli
  • ParthKohli
yeah, can you manage the rest now?
mathmath333
  • mathmath333
area of the first is \(2k^2\) ?
ParthKohli
  • ParthKohli
that's correct.
mathmath333
  • mathmath333
and how to find 2nd one
ParthKohli
  • ParthKohli
just find out the points where the lines meet and those would be the vertices of the square.
Michele_Laino
  • Michele_Laino
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
mathmath333
  • mathmath333
would this be enough for proof
Michele_Laino
  • Michele_Laino
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
mathmath333
  • mathmath333
ok thnx both
Michele_Laino
  • Michele_Laino
:)
ParthKohli
  • ParthKohli
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.
Michele_Laino
  • Michele_Laino
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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