9.2 Analytical Geometry, Coordinate Systems, and Vectors

Coordinates, Departures, and Latitudes

Surveying analytical geometry lives in a plane coordinate system where N (north) is the y-axis and E (east) is the x-axis. Azimuths are measured clockwise from north, ranging 0° to 360°. This convention flips the usual math definitions, so memorize the survey forms:

• Departure (east component): ΔE = distance · sin(azimuth)
• Latitude (north component): ΔN = distance · cos(azimuth)

A new point is then E₂ = E₁ + ΔE and N₂ = N₁ + ΔN. The signs of departure and latitude automatically place the line in the correct quadrant: NE has +ΔE/+ΔN, SE has +ΔE/−ΔN, SW has −ΔE/−ΔN, and NW has −ΔE/+ΔN.

Worked forward computation. From point A (E = 5,000.00, N = 5,000.00), run a line of azimuth 123°45'00" and distance 640.20 ft.

• ΔE = 640.20 · sin(123.750°) = 640.20 · 0.83147 = +532.31 ft
• ΔN = 640.20 · cos(123.750°) = 640.20 · (−0.55557) = −355.66 ft
• Point B: E = 5,532.31, N = 4,644.34

The positive departure and negative latitude correctly place B southeast of A, matching an azimuth between 90° and 180°.

Inverse: Recovering Distance and Direction

The inverse computation finds the length and azimuth between two known points — the backbone of COGO. Given ΔE = E₂ − E₁ and ΔN = N₂ − N₁:

• Distance: D = √(ΔE² + ΔN²)
• Azimuth: Az = atan2(ΔE, ΔN), then add 360° if negative

Using point B above back to a point C (E = 5,140.00, N = 4,720.00): ΔE = 5,140.00 − 5,532.31 = −392.31, ΔN = 4,720.00 − 4,644.34 = +75.66.

• D = √((−392.31)² + 75.66²) = √(153,907 + 5,724) = √159,631 = 399.54 ft
• The raw atan(ΔE/ΔN) = atan(−392.31/75.66) = atan(−5.1851) = −79.07°. Because ΔE is negative and ΔN positive (NW quadrant), the azimuth is 360° − 79.07° = 280.93° = 280°55'48".

The quadrant rule is non-negotiable. A bare atan returns only −90° to +90°, so you must inspect the signs of ΔE and ΔN to land in the right quadrant. The convenience of atan2(ΔE, ΔN) is that it does this automatically — but you still confirm by sketch.

Vectors, Projections, and the Coordinate Area Formula

Thinking in vectors organizes offsets and projections. To project the perpendicular offset of a point P onto a line through A in direction û (a unit vector), compute the vector AP, then the along-line component is the dot product (AP · û) and the perpendicular offset is the cross-product magnitude |AP × û|. A right offset (point to the right of the line, looking forward) yields a negative cross-product under the standard convention; carry the sign so left/right is preserved.

For enclosed areas, the coordinate method (a.k.a. the shoelace formula) is exact for any closed polygon listed in order:

Area = ½ |Σ (Eᵢ · Nᵢ₊₁ − Eᵢ₊₁ · Nᵢ)|

Worked area. A four-sided parcel has corners (in order) (1000,1000), (1300,1050), (1280,1380), (980,1300):

Cross-sum Σ(Eᵢ·Nᵢ₊₁ − Eᵢ₊₁·Nᵢ): (1000·1050 − 1300·1000) + (1300·1380 − 1280·1050) + (1280·1300 − 980·1380) + (980·1000 − 1000·1300) = (−250,000) + (450,000) + (311,600) + (−320,000) = 191,600. Area = ½·|191,600| = 95,800 ft² ≈ 2.20 acres (43,560 ft²/acre). A reasonableness check: the parcel spans roughly 300 ft × 330 ft, so an area near 99,000 ft² is sensible.

Line Intersections and Reasonableness Checks

Locating where two lines cross is a staple COGO operation — for example, finding where a proposed right-of-way meets an existing property line. The cleanest setup expresses each line parametrically from a known point along its azimuth, then solves for the two scale factors where the coordinates match. A bearing-bearing intersection uses two points and two directions; a distance-distance intersection uses two points and two radii (two circles, generally two solutions); and a bearing-distance intersection mixes the two. The FS exam favors the bearing-bearing case because it reduces to solving two simultaneous linear equations.

For a bearing-bearing intersection from point P₁ along azimuth α₁ and point P₂ along azimuth α₂, write E₁ + t·sin α₁ = E₂ + u·sin α₂ and N₁ + t·cos α₁ = N₂ + u·cos α₂, then solve the 2×2 system for the unknown distances t and u. If sin α₁·cos α₂ − cos α₁·sin α₂ = 0, the directions are parallel and no unique intersection exists — a condition worth checking before grinding through algebra.

Reasonableness is mandatory. Several quick checks catch gross errors before they reach a plat:

• A computed distance should be positive; a negative t means the intersection lies behind the starting point, so re-examine the azimuths.
• An azimuth must land in 0°–360°; add 360° to a negative result.
• A departure and latitude must carry signs consistent with the quadrant of the azimuth (NE +/+, SE +/−, SW −/−, NW −/+).
• A parcel area should agree with a rough length-times-width estimate to within a sensible margin.

These sanity checks are not optional polish; on the FS exam they distinguish a transposed-coordinate blunder from a correct answer in seconds.