HP 35S: Intersection
Point of a Quadrilateral
Introduction
Let A, B, C, and D be four vertices of a quadrilateral, with
two lines:
Line AC connects points A and C.
Line BC connects points B and D.
Designate the following points as:
A: (ax, ay)
B: (bx, by)
C: (cx, cy)
D: (dx, dy)
The center point (px, py) can be found by the following
formulas:
px = (IBD – IAC)/(SAC – SBD)
py = SAC * px + IAC
Where:
Slope:
SAC = (cy – ay) / (cx – ax)
SBD = (dy – by) / (dx – bx)
Intercept:
IAC = ay – SAC * ax
IBD = by – SBD * bx
You can see the derivation of these formulas here: https://edspi31415.blogspot.com/2017/08/geometry-intersection-point-of.html
Program (Pedro Daniel Leiva)
The following program calculates the point – developed by Pedro
Daniel Leiva. I used the recall arithmetic available on the HP 35S to shorten
the program. Program listed here with
permission.
Variables Used:
R_A = ax
R_B = ay
R_C = bx
R_D = by
R_E = cx
R_F = cy
R_G = dx
R_H = dy
R_I = SAC
R_J = SBD
R_K = IAC
R_L = IBD
R_P = px
R_Y = py
Program:
I001 LBL I
I002 SF 10
I003 ENTER
XA^YA \\ EQN
I004 STO B
I005 x<>y
I006 STO A
I007 ENTER
XB^YB \\ EQN
I008 STO D
I009 x<>y
I010 STO C
I011 ENTER XC^YC \\
EQN
I012 STO F
I013 x<>y
I014 STO E
I015 ENTER
XD^YD \\ EQN
I016 STO H
I017 x<>y
I018 STO G
I019 CF 10
I020 RCL F
I021 RCL - B \\ [RCL] [ - ] ( B )
I022 RCL E
I023 RCL - A \\ [RCL] [ - ] ( A )
I024 ÷
I025 STO I
I026 STO P \\ advanced storage to calculate px to save
steps
I027 RCL× A
I028 +/-
I029 RCL+ B
I030 STO K
I031 RCL H
I032 RCL - D
I033 RCL G
I034 RCL – C
I035 ÷
I036 STO J
I037 STO – P
I038 RCL × C
I039 +/-
I040 RCL + D
I041 STO L
I042 RCL – K
I043 RCL ÷ P
I044 STO P
I045 ENTER
I046 RCL × I
I047 RCL + K
I048 STO Y
I049 x<>y
I050 RTN
50 steps
Instructions:
At each prompt, enter the x point, press [ENTER], enter the
y point, press [ R/S ]. The result shows
py on the Y stack, and px on the X stack.
Example:
A: (0, 8)
B: (11, 12)
C: (10, 4)
D: (3, 5)
Results:
py = 6.2353
px = 4.4118
Eddie
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2011-2018. Edward Shore. Unauthorized use and/or unauthorized
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