Equation Of A Line

Equation of a Line DE12 = (E2-E1) P2:(N2,E2)

N2

N

Az Azimuth of the Line P:(N,E)

N1

P1 E1*m

DN12 = (N2-N1) Any point on Line DN=(length of the line)*Cos(Azimuth of the line) DE=(length of the line)*Sin(Azimuth of the line)

NE=0 m = slope = DN12/DE12 =Cot(Az)

E1

Tan(Az)= DE12 DN12

0 E1

E

E2

General Equation: N − N1 E − E1 = N 2 − N 1 E 2 − E1

(eq01)

Rearranged:

(eq02)

a

b

0

0

a * E + b * N + c = = ( N1 − N 2 ) * E + ( E 2 − E1 ) * N + ( E1 * ( N 2 − N1 ) − N1 * ( E2 − E1 )) =

c

(eq3) N axis intercept = -c/b = ( N1 * ( E 2 − E1 ) − E1 * ( N 2 − N1 )) /( E 2 − E1 ) = NE=0 = N Value when E=0 E axis intercept = -c/ a = ( N1 * ( E 2 − E1 ) − E1 * ( N 2 − N1 )) /( N1 − N 2 ) = E value when N=0 (eq4) = EN=0 Write equation of line passing through P1:(500.000,300.000) and P2:(800.000,900.000) -300.000*E+600.000*N-210,000.000= 0 = -3.00000*E+6.00000*N-2,100.000 = 0 -E+2.00000*N-700.000 = 0

(eq05)

N=(0.500000)*E+350.000

(eq06)

1/4

Equation of 2 Lines and their Intersection DE12 = (E2-E1) P2:(N2,E2)

N2 N3

P3 DN12 = (N2-N1)

N

P:(N,E)

N1

Let intersection be Pi:(Ni,Ei)

P1

DN34=( N4-N3)

P4

N4 DE34 = (E4-E3) 0 E3

E1

E

E2

E4

On the N,E Plane any two non-Parallel lines will intersect ( have a common point). This situation will have one solution. On the Line defined by P1 and P2, or P1 and Azimuth from P1 to P2 : N = ( N2-N1) * E + N1-E1*( N2-N1) = ( N2-N1) *(E-E1) + N1 = DN12*(E-E1)+N1 (eq07) ( E2-E1 ) ( E2-E1 ) ( E2-E1 ) DE12 N=(1/Tan(Az12))*E+N1-E1*(1/Tan(Az12)) = Cot(Az12)*E+(N1-E1*Cot(Az12))

(eq08)

On the Line defined by P3 and P4, or P3 and Azimuth from P3 to P4 : N = ( N4-N3) * E + N3-E3*( N4-N3) = ( N4-N3) *(E-E3) + N3 = DN34*(E-E3)+N3 (eq09) ( E4-E3 ) ( E4-E3 ) ( E4-E3 ) DE34 N=(1/Tan(Az34))*E+N3-E3*(1/Tan(Az34)) = Cot(Az34)*E+(N3-E3*Cot(Az34))

(eq10)

At the intersection (N, E) is the same (equal) on both lines. We will label that Pi:(Ni, Ei). ( N2-N1) * Ei + N1-E1*( N2-N1) = ( N4-N3) * Ei + N3-E3*( N4-N3) ( E2-E1 ) ( E4-E3 ) ( E4-E3 ) ( E2-E1 )

(eq11)

Cot(Az12)*Ei+(N1-E1*Cot(Az12)) = Cot(Az34)*Ei+(N3-E3*Cot(Az34))

(eq12)

2/4

⎡ ⎤ ⎡ N − N1 N 4 − N 3 ⎤ N − N1 N − N3 Ei = ⎢ N 3 − N1 + 2 * E1 − 4 * E3 ⎥ / ⎢ 2 − ⎥ E2 − E1 E4 − E3 ⎣ ⎦ ⎣ E2 − E1 E4 − E3 ⎦

(eq13)

Place this E in either line equation (eq07, eq09) to get N ( or do both for a check, both should be the same). Ei = [N 3 − N1 + Cot ( Az12 ) * E1 − Cot ( Az34 ) * E3 ]/[Cot ( Az12 ) − Cot ( Az34 )]

(eq14)

Place this E in either line equation (eq08, eq10) to get N ( or do both for a check, both should be the same). Note: if line is due N-S division by zero is encountered above.

Example: P#:( N# , E# ) P1:(500.000,300.000); P2:(800.000,900.000); P3:(850.000,200.000); P4:(150.000,950.000) Ei=479.070(from eq13), Ni=589.535(from line 12, eq07), Ni=589.535(from line 34, eq09) P1:(500.000,300.000); Az12:63±26’05.8”;

P3:(850.000,200.000); Az34:133±01’30.2”

Ei=479.070(from eq14), Ni=589.535(from P1, Az12,eq08), Ni=589.535(from P3,Az34,eq10) 5 Or approach the problem with triangles and trigonometry:

list Azimuths that you have for each line or computed from coordinate inverses for each line. Lines Az:(P1ØP2)= Az:(P1ØPi), Az:(P3ØP4)= Az:(P3ØPi), and Az:(P1ØP3)(suggested). Derive distance P1ØP3 from coordinate inverse Determine interior angles in triangle Pi, P1, P3 by the differences in Azimuth. Determine another distance (PiØP1 or PiØP3) in the triangle ( or both two provide access to a check) using the Sin Law. Use the distance(s) along with the Azimuth(s) to get the deltas ((DN1i, DE1i) or (DN3i, DE3i) or both to continue with the checking process).

3/4

P3

Now use the known coordinates of P1 or P3 (or both to carry on with access to a check) and the respective D’s to arrive at the coordinates for Pi, the point of intersection.

P2

Pi

N1+DN1i=Ni, E1+DE1i=Ei N3+DN3i=Ni, E3+DE3i=Ei P1 P4

If you have calculated from both points you should get the same Pi solution.

A good confirmation of the solution can also be devised by: Inverse coordinates P3ØPi. The Azimuth should be the same as for P3ØP4. Inverse coordinates P1ØPi. The Azimuth should be the same as for P1ØP2. Both should be true!

Example: P1:(500.000,300.000); P2:(800.000,900.000); P3:(850.000,200.000); P4:(150.000,950.000) Inverses: Az12:63±26’05.8”, Az34:133±01’30.2”, Az13:344±03’16.6”, Dist13:364.005 Angle @ P1=((63±26’05.8” + 360±00’00”)- 344±03’16.6”) = 79±22’49.2” Angle @ P3=((344±03’16.6”-180±00’00”) - 133±01’30.2”) = 31±01’46.4” Angle @ Pi=(133±01’30.2” - 63±26’05.8”) transversal = 69±35’24.4” SUM = 180±00’00” 5 Distance P1ØPi= Dist1i = Sin(31±01’46.4”)*364.005/Sin(69±35’24.4”)= 200.206 Distance P3ØPi= Dist3i = Sin(79±22’49.2”)*364.005/Sin(69±35’24.4”)= 381.735 Ni=500.000+200.206*Cos(63±26’05.8”) = 589.535 Ei=300.000+200.206*Sin(63±26’05.8”) = 479.070 Ni=850.000+381.735*Cos(133±01’30.2”) = 589.535 Ei=200.000+381.735*Sin(133±01’30.2”) = 479.070

5

Inverse Az1i = 63±26’05.8” compares to Az12:63±26’05.8” :yes Inverse Az3i = 133±01’30.1” compares to Az34:133±01’30.2” :yes within reason 5

4/4