**TI-84 Plus: Intersection of two lines – program by Jack Kesler**

Special thanks
to Jack Kessler for providing the program CPXINT. CPXINT uses complex numbers to determine:

* The equation
of a line between two points. CPXINT
uses two lines:

y = Ax + B for coordinates N1, E1
and N2, E2.

y = Cx + D for coordinates N3, E3
and N4, E4

* The
intersection of the two lines, labeled by point N5, E5.

Keep in mind N
is for North (y axis) and E is for East (x axis).

**TI-84 Plus Program CPXINT – Jack Kesler**

Note: Initialize all the variables by choosing
option 1, SETUP. Choose 2 for input, 3
for equations of the lines, and 4 for intersection.

ClrHome

Disp "LINE
INT"

Disp "USNG
CPX"

Disp "VER.
1.0"

Pause

ClrHome

Lbl 50

Menu("CPXINT","SETUP",89,"INP
PTS",90,"COMP LINS",91,"COMP
INT",92,"EXIT",93)

End

Lbl 89

ClrHome

ClrAllLists

11→dim(L₁)

{2,2}→dim([A])

{2,1}→dim([B])

Goto 50

Lbl 90

ClrHome

Disp "INP
CRDS:"

Input "N1=
",T

T→L₁(1)

Input "E1=
",T

T→L₁(2)

Input "N2=
",T

T→L₁(3)

Input "E2=
",T

T→L₁(4)

Input "N3=
",T

T→L₁(5)

Input "E3=
",T

T→L₁(6)

Input "N4=
",T

T→L₁(7)

Input "E4=
",T

T→L₁(8)

ClrHome

Goto 50

Lbl 91

ClrHome

a+b

*i*
L₁(2)+L₁(1)

**→I***i*
L₁(4)+L₁(3)

**→J***i*
angle(J-I)→θ

If θ=90 or θ=90

Then

real(I)→B

0→[A](1,1)

1→[A](1,2)

B→[B](1,1)

0→A

Else

tan(θ)→A

imag(I)-A*real(I)→B

1→[A](1,1)

A→[A](1,2)

B→[B](1,1)

End

Disp
"Y=A*X+B"

Disp "A=
",A

Disp "B=
",B

Pause

ClrHome

L₁(6)+L₁(5)

**→I***i*
L₁(8)+L₁(7)

**→J***i*
angle(J-I)→θ

If θ=90 or θ=90

Then

real(I)→D

0→[A](2,1)

1→[A](2,2)

D→[B](2,1)

0→C

Else

tan(θ)→C

imag(I)-C*real(I)→D

1→[A](2,1)

C→[A](2,2)

D→[B](2,1)

End

Disp
"Y=C*X+D"

Disp "C=
",C

Disp "D=
",D

Pause

ClrHome

Real

Goto 50

Lbl 92

ClrHome

If A=C

Then

Disp "PARLLEL
LINES"

Disp "NO
SOLUTION"

Pause

ClrHome

Goto 50

End

[A]^-1*[B]→[D]

Disp "N5=
",[D](1,1)

[D](1,1)→L₁(9)

Disp "E5=
",[D](2,1)

[D](2,1)→L₁(10)

Pause

ClrHome

Goto 50

Lbl 93

ClrHome

Stop

**Example:**

Line 1: N1 = 0, E1 = 1, N2 = 7, E2 = 2.

Line 2: N3 = 4, E3 = 3, N4 = 4, E3 = -3

Results:

Y = AX +
B: A = 0, B = 4

Y = CX +
D: C = 7, D = -7

Intersection: N5 = 4,
E5 = 1.571428571

Thank you
Jack!

Eddie

This blog is
property of Edward Shore, 2016