PascalTheorem::usage=" In any hexagon inscribed in a conical section, the continuations of opposite sides meet on a straight line, called the Pascal line. This property is repeated in Geometry, there is also the Monge's theorem and the Desargues' theorem. References: Personal, based in the Pascal's theorem. ";
Development:
Obtaining the vertices of the hexagon in numerical form:
In[]:=
r=2;k=4;d=-2;n=6;ρ=
2Sin
π
n

1-Sin
π
n

;x1=4;y1=6;k1=
3
5
;​​TablePoint{4-d,y1}+
2
k
2
((r+ρ)Cos[t1+2πk1]+x1-(4-d))
+
2
((r+ρ)Sin[t1+2πk1]+y1-6)
-
2
ρ
{(r+ρ)Cos[t1+2πk1]+x1-(4-d),(r+ρ)Sin[t1+2πk1]+y1-6},t1,0,2π,
2π
n
//FullSimplify
Point4-
2
5
,6-4
1-
2
5
,Point2-
2
-1+Sin
π
30

,6+
4Cos
π
30

-1+Sin
π
30

,Point2+
2
Csc
π
15

,6-4Cot
π
15
,Point24+
5
,6+4
5+2
5
,Point2+
2
1+Sin
π
30

,2+
8
1+Tan
π
60

,Point2+
2
Sec
π
15

,6+4Tan
π
15
,Point4-
2
5
,6-4
1-
2
5
;
In[]:=
​​​​​​Showr=2;k=4;d=-2;n=6;ρ=
2Sin
π
n

1-Sin
π
n

;x1=4;y1=6;k1=
3
5
;​​ContourPlot[
2
y
-16x-12y+84==0,{x,0,40},{y,-20,40},ContourStyle->{Black,Thick},PlotPoints->50],​​TableGraphicsBlue,Thick,Line{4-d,y1}+
2
k
2
((r+ρ)Cos[t1+2πk1]+x1-(4-d))
+
2
((r+ρ)Sin[t1+2πk1]+y1-6)
-
2
ρ
{(r+ρ)Cos[t1+2πk1]+x1-(4-d),(r+ρ)Sin[t1+2πk1]+y1-6},{4-d,y1}+
2
k

2
(r+ρ)Cost1+
2π
n
+2πk1+x1-(4-d)
+
2
(r+ρ)Sint1+
2π
n
+2πk1+y1-6
-
2
ρ
(r+ρ)Cost1+
2π
n
+2πk1+x1-(4-d),(r+ρ)Sint1+
2π
n
+2πk1+y1-6,t1,0.00001,2π,
2π
n
,TableGraphicsBlack,Thick,Dashed,InfiniteLine{4-d,y1}+
2
k
2
((r+ρ)Cos[t1+2πk1]+x1-(4-d))
+
2
((r+ρ)Sin[t1+2πk1]+y1-6)
-
2
ρ
{(r+ρ)Cos[t1+2πk1]+x1-(4-d),(r+ρ)Sin[t1+2πk1]+y1-6},{4-d,y1}+
2
k

2
(r+ρ)Cost1+
2π
n
+2πk1+x1-(4-d)
+
2
(r+ρ)Sint1+
2π
n
+2πk1+y1-6
-
2
ρ
(r+ρ)Cost1+
2π
n
+2πk1+x1-(4-d),(r+ρ)Sint1+
2π
n
+2πk1+y1-6,t1,0,2π,
2π
n
,GraphicsRed,PointSize[0.015],Point4-
2
5
,6-4
1-
2
5
,GraphicsBlue,PointSize[0.015],Point2-
2
-1+Sin
π
30

,6+
4Cos
π
30

-1+Sin
π
30

,GraphicsGreen,PointSize[0.015],Point2+
2
Csc
π
15

,6-4Cot
π
15
,GraphicsGray,PointSize[0.015],Point24+
5
,6+4
5+2
5
,GraphicsPink,PointSize[0.015],Point2+
2
1+Sin
π
30

,2+
8
1+Tan
π
60

,GraphicsPurple,PointSize[0.015],Point2+
2
Sec
π
15

,6+4Tan
π
15
,AxesTrue,BoxRatiosAutomatic,AxesOrigin{0,0},ImageSize700,PlotRange{{-2,40},{-20,44}}
Out[]=
Obtaining the equations of the opposite sides:
GreenBlue line and PinkPurple, form the point 1:
In[]:=
Solvey-6-4Cot
π
15
2-
2
-1+Sin
π
30

-2+
2
Csc
π
15

==6+
4Cos
π
30

-1+Sin
π
30

-6-4Cot
π
15
x-2+
2
Csc
π
15

,y-2+
8
1+Tan
π
60

2+
2
Sec
π
15

-2+
2
1+Sin
π
30

==6+4Tan
π
15
-2+
8
1+Tan
π
60

x-2+
2
1+Sin
π
30

,{x,y}//FullSimplify
x2,y2+
8
1+Cot
π
20

;
BlueRed line and GrayPink, form the point 2:
In[]:=
Solvey-6+
4Cos
π
30

-1+Sin
π
30

4-
2
5
-2-
2
-1+Sin
π
30

==6-4
1-
2
5
-6+
4Cos
π
30

-1+Sin
π
30

x-2-
2
-1+Sin
π
30

,y-6+4
5+2
5
2+
2
1+Sin
π
30

-24+
5
==2+
8
1+Tan
π
60

-6+4
5+2
5
x-24+
5
,{x,y}//FullSimplify
x2,y6+4
8
0.445
…
;
RedPurple line and GreenGray, form the point 3:
In[]:=
Solvey-6-4
1-
2
5
2+
2
Sec
π
15

-4-
2
5
==6+4Tan
π
15
-6-4
1-
2
5
x-4-
2
5
,y-6-4Cot
π
15
24+
5
-2+
2
Csc
π
15

==6+4
5+2
5
-6-4Cot
π
15
x-2+
2
Csc
π
15

,{x,y}//FullSimplify
x2,y4Cos
π
15
3
55-2
5

+15
5+2
5
+210-3
55+2
5

Cos
2π
15
+2-15-6
5
+2
55+2
5

Sin
2π
15
+5+3
5
2+3Cos
7π
30
-2Sin
7π
30
+
105+
5

-3+2Cos
7π
30
+3Sin
7π
30

165-30
15+6
5
+
55+2
5

Cos
π
15
+10
5-2
5
2
Sin
π
30

-25+2
5
Sin
π
15
;
Complete graph:
The graph shows Pascal’s line.