modulus graphs

This question really confused me.

Usually with mod graphs, I believe we just consider 2 cases, the positive and negative and then join them together.

I can follow the ms until the "cross graph", but then I just cant follow.

so take the simplified case of |x-2| + |y-2| <= 4, to get a foothold
then simplify it more X = x-2, Y= y-2
so |X| + |Y| <= 4
take +ve case of X,Y so you get y = 4-x,
then +ve Y, -ve X; y = 4+x
and just keep going until you get a full diagonal square.
(this part required some brain power and time, but I think if you just sketch each case, you can see its just a diagonal square)

the shifting part I can understand, but where it becomes a cross, I just dont follow.
If anyone could clarify, help would be appreciated.

(edited 1 month ago)

Reply 1

mqb2766

Original post by mosaurlodon

This question really confused me.
Usually with mod graphs, I believe we just consider 2 cases, the positive and negative and then join them together.

I can follow the ms until the "cross graph", but then I just cant follow.
so take the simplified case of |x-2| + |y-2| <= 4, to get a foothold
then simplify it more X = x-2, Y= y-2
so |X| + |Y| <= 4
take +ve case of X,Y so you get y = 4-x,
then +ve Y, -ve X; y = 4+x
and just keep going until you get a full diagonal square.
(this part required some brain power and time, but I think if you just sketch each case, you can see its just a diagonal square)
the shifting part I can understand, but where it becomes a cross, I just dont follow.
If anyone could clarify, help would be appreciated.

In b, the light shade area is the part you want (quadrant 1), then reflect in the x-y axes to get the 4 quadrants (or not bother and just get the area of the light shaded area and multiply by 4)

Personally, Id just consider quadrant 1 so positive x and y and mulitiply the area by 4 so consider the outside boundary of
|x-2| + |y-2| <= 4
the axis crossing points are at (4,0), (0,4) and as |x-2| or |y-2| decrease to 0 then (4,0) -> (6,2) and (0,4) -> (2,6) and those two points are then joined by a line as x-2 and y-2 are both positive. So area 4*4/2 + 4sqrt(2)*2sqrt(2) = 24 and multiply by 4.

Youd have to justify theres no hole at (0,0) but thats simple enough (but just).

Edit. Its fairly obvious that (4,4) is on the boundary of this shape and its roughly a circle equation with radius 4sqrt(2). So guestimating pi*r^2 would roughly give the answer.

(edited 1 month ago)

Reply 2

mosaurlodon

Ok I think I managed to get both ways now:

that circle trick is so cool, might be useful in a time crunch - thank you!

(edited 1 month ago)

Reply 3

mqb2766

Original post by mosaurlodon

Ok I think I managed to get both ways now:

that circle trick is so cool, might be useful in a time crunch - thank you!

NP. One trick that slighly simplifies their area calc is to note that the square areas are the product of the diagonals / 2, which is slighlty simpler as the diagonals are along the coordinate axes. So 8^2/2 - 4^2/2 and obv multiply by 4. More generally, for the area of a quadrilateral its the product of the diagonals multiplied by sin of the angle between them / 2.

(edited 1 month ago)

Reply 4

mosaurlodon

Oh yeah because any quadrilateral is just 4 triangles and then the sin gives you the height.
Triangles always seem to be the way to make everything easier.