Question

Prove

Answer 1

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

Answer 2

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

Answer 3

\(\large \color{black}{\begin{align} x+y=k\hspace{1.5em}\\~\\
x-y=k\hspace{1.5em}\\~\\
-x+y=k\hspace{1.5em}\\~\\
-x-y=k\hspace{1.5em}\\~\\
\end{align}}\)

Answer 4

Yeah. Now graph them and see what you get.

Answer 5

square with area \(k^2\)

Answer 6

Same process for the second case.

Answer 7

what about 'a' and 'b'

Answer 8

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

Answer 9

! :D

Answer 10

looks like the area is not \(k^2\) though.

Answer 11

\(\large \color{black}{\begin{align} x+a-y-b=k\hspace{1.5em}\\~\\
x+a+y+b=k\hspace{1.5em}\\~\\
-x-a+y+b=k\hspace{1.5em}\\~\\
-x-a-y-b=k\hspace{1.5em}\\~\\
\end{align}}\)

Answer 12

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

Answer 13

yeah, can you manage the rest now?

Answer 14

area of the first is \(2k^2\) ?

Answer 15

that's correct.

Answer 16

and how to find 2nd one

Answer 17

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

Answer 18

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

Answer 19

would this be enough for proof

Answer 20

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

Answer 21

ok thnx both

Answer 22

:)

Answer 23

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.

Answer 24

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