## mathmath333 one year ago Prove

1. 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}}

2. ParthKohli

Almost the same as the previous one - try to figure out the four lines.

3. 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}}

4. ParthKohli

Yeah. Now graph them and see what you get.

5. mathmath333

square with area $$k^2$$

6. ParthKohli

Same process for the second case.

7. mathmath333

8. ParthKohli

don't worry about them. figure out the equations first.

9. Here_to_Help15

! :D

10. ParthKohli

looks like the area is not $$k^2$$ though.

11. 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}}

12. ParthKohli

the diagonal of the square in the first case is $$2k$$, so the side length is $$k\sqrt 2$$

13. ParthKohli

yeah, can you manage the rest now?

14. mathmath333

area of the first is $$2k^2$$ ?

15. ParthKohli

that's correct.

16. mathmath333

and how to find 2nd one

17. ParthKohli

just find out the points where the lines meet and those would be the vertices of the square.

18. 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

19. mathmath333

would this be enough for proof

20. 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

21. mathmath333

ok thnx both

22. Michele_Laino

:)

23. 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.

24. 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