CPhill

                                                       Origin of the Knot

As GA points out, Mr. Gumpach's tome, while illustrative, does not quite cover all the bases. For instance....sailing an equatorial path is always tricky.......crossing the Dateline insures problems because one never knows whether he/she is measuring speed in current terms - or in yesterday's terms. What might appear to be great speed yesterday might only appear to be a cipher, today.

Furthermore, historically......."knots"   were measured somewhat erratically. The once-accepted method was accomplished  by the "chip log."  Of course, finding logs at sea was sometimes as difficult as finding logs in Mathematics. So....if the ship carried livestock for provision of meat and/or dairy, the "cow chip" was sometimes an adequate substitute.  In the "Horse Latitudes,"  the whole cow was sometimes thrown overboard.  Lacking that, Mr. Newton himself was often shanghaied in port and thrown overboard later. (Not for any particular measurement purposes, but for the fact that few personages - if any - can abide a windbag scientist.)

So....the present dilemma remains.....the "knot"  is an unspecified quantity. In fact, it is the inspiration for the term "naught"  meaning  "nothingness"  or "nonexistence" or "insignificant." Therefore, I urge that we forget about chasing the quixotic and focus on the more attainable. For instance, the search for the Yeti immediately comes to mind.

I apologize if some of the facts I have laid out are errrant. However, mostly I told the truth....mostly........

Your Humble Servant and WebCalc Historian,

Sir CPhill, Esq 

Rewrite as

x^2 + 3x + y^2 - 2y = 1

Complete the square on x and y

x^2 + 3x + 9/4 + y^2 - 2y + 1  =  1  + 9/4  + 1

( x + 3/2)^2  +  ( y - 1)^2  =  2 + 9/4

( x + 3/2)^2  +  ( y - 1)^2  =  8/4 + 9/4

( x + 3/2)^2  +  ( y - 1)^2  =  17/4 

Look at the following :

I think you want to find the height of the triangle containing the isosceles triangle

AB = 2sqrt (2)   

Let  AEB  be isosceles......with  angles ABE, BAE  equal....so  AE  = BE

Bisect  AEB.......by SAS......BH = AH = EH

And ABE  = 90......so  ABE, BAE  = 45

And BH =  sqrt (2).......so EH = sqrt (2)

Let E be the intersection of medians BG and AF

Draw GF through D.....drop  perpendiculars from F, G to interesect AB at I and J  and by AAS,  triangle FIB is congruent to  triangle GJA.  ...then FI  = GJ ...so.. GF  is parallel to AB

And triangle AGF  is similar to triangle ABC.....so  DF  = 1/2  of HB  =  sqrt(2)/2

So  CD  =   1/2 CH   

And EDF  is right  and  angle DFE  = BAE  =  45

So DEF  = 45...so triangle EDF  is 45 - 45 - 90  with DF = DE  = sqrt (2) / 2

So.....HD  = EH + DE  =  sqrt (2) +  sqrt (2) / 2

And..........By  similar triangles......  2 CD  = 2 HD  =  HC

So  .....HC  =    2 [  sqrt (2) +  sqrt (2) / 2 ]   =  3sqrt (2)  = height of ABC

Note that   264  /  495  =    8  / 15

So......the ratio of the legs  is   8  :  15

And one of the Pythagorean "Triples"  is     8  : 15  : 17

So.........

264 / 8  =  33

And the hypotenuse must be......17 * 33  =   561

So......the  sides are    264 -  495  - 561

2nd quadrant.......

Just replace "x" in the equation with "6"

y  =  -2 (6) + 16  =  4   ⇒   (6, 4 )  

y  =  ax + b 

y - b =  ax

[ y - b]  /  a  =  x        swap x and y and for y write f^-1 (x)

So 

f^-1 (x)  = [ x - b ] / a  =  0         when     b  =  x 

3td quadrant

Here's the last one

Since    OA / OB  = OD / OC

We must have that triangle  OAB   is similar to triangle ODC

And  angle  CDB  = angle CDO  = angle BAO = angle BAC = angle CAB

 (2,1) and (3,4)

Slope

[ Subtract the y coordinates in some order]            [ 4 - 1]          3

_____________________________________ =  ______  =   __   =    3

[Subtract the x coordinates in the same order ]      [ 3 - 2 ]        1