Area & partitioning
Partitioning a parcel by a line through a fixed point
A cut line must start at an existing monument on the frontage and sever 2.400 acres. Swing a trial line to a corner, then rotate it off that corner by the triangle the deficiency demands.
Given
- Parcel: LOT 14, five corners, US survey feet, northing first.
- A N 3000.00 E 4000.00
- B N 3480.00 E 4160.00
- C N 3380.00 E 4720.00
- D N 2960.00 E 4820.00
- E N 2860.00 E 4300.00
- Boundary courses: A–B N 18°26′06″ E 505.96 ft; B–C S 79°52′31″ E 568.86 ft; C–D S 13°23′33″ E 431.74 ft; D–E S 79°06′52″ W 529.53 ft; E–A N 64°58′59″ W 331.06 ft.
- An existing monument P stands on the A–B frontage at N 3189.74, E 4063.25 — 200.00 ft from A and 305.96 ft from B.
- A division line is to be run from P, and must be run from P: the monument cannot be moved and no other point on the frontage may be used.
- The division must sever exactly 2.400 acres on the B side of the line.
Required
- The point X on the boundary at which the line from P must terminate.
- The length and bearing of the division line P–X.
- The areas of both resulting parcels.
Work it through yourself before reading on — the solution below shows every step, so there is no way to skim it without giving the answer away.
Worked solution
Whole parcel, and the target for each piece
Shoelace the five corners: 2A = −728,000.00 ft², so LOT 14 is 364,000 ft², or 8.356 acres. The severance is 2.400 acres = 104,544 ft², leaving 259,456 ft² or 5.956 acres.
Because the line must pass through a fixed point, there is one free parameter left — where the far end lands — so the problem has a unique solution as long as that far end falls on a boundary the line can actually reach. Establishing which boundary is the first real decision.
2A = −728,000.00 ft² → A = 364,000 ft² = 8.356 ac
severance required = 2.400 × 43,560 = 104,544 ft²
remainder to expect = 364,000 − 104,544 = 259,456 ft² = 5.956 acSwing a trial line from P to a corner
The technique is to run the line first to a convenient corner, compute the area that produces, and then rotate the far end along the adjacent boundary by exactly enough to make up the difference. Corner C is the natural trial: the line P–C severs the triangle P–B–C, which is plainly less than the acreage wanted, so the far end will have to move from C toward D.
Compute the trial area by shoelace on the three points, using the same cross-product sum as for a full parcel. A triangle is where the method is easiest to check, since the area can also be had from ½ · base · height.
| Course | Eᵢ · Nᵢ₊₁ | Eᵢ₊₁ · Nᵢ | Difference |
|---|---|---|---|
| P–B | 4063.25 × 3480.00 = 14,140,110.00 | 4160.00 × 3189.74 = 13,269,318.40 | +870,791.60 |
| B–C | 4160.00 × 3380.00 = 14,060,800.00 | 4720.00 × 3480.00 = 16,425,600.00 | −2,364,800.00 |
| C–P | 4720.00 × 3189.74 = 15,055,572.80 | 4063.25 × 3380.00 = 13,733,785.00 | +1,321,787.80 |
| Σ = 2A | −172,220.60 |
The deficiency, and the triangle that must make it up
The trial line severs 86,110.30 ft². The requirement is 104,544 ft², so 18,433.70 ft² is short. That deficiency has to come from a triangle P–C–X with X somewhere on side C–D, because rotating the far end of the line from C toward D sweeps exactly that triangle into the severed piece.
The triangle P–C–X has base C–X lying along the straight side C–D and apex at P. Its height is therefore the perpendicular distance from P to the line containing C–D — a fixed number that does not depend on where X ends up. That is what makes the method exact rather than iterative: the area added is strictly proportional to the distance C–X.
Get the perpendicular distance from the cross product of the side vector with the vector to P, divided by the length of the side.
deficiency ΔA = 104,544 − 86,110.30 = 18,433.70 ft²
C→D vector = ( −420.00, +100.00 ), | C–D | = 431.74 ft
C→P vector = ( 3189.74 − 3380.00, 4063.25 − 4720.00 ) = ( −190.26, −656.75 )
cross = (−420.00)(−656.75) − (+100.00)(−190.26) = 275,835.00 + 19,026.00 = 294,861.00
h = | cross | / | C–D | = 294,861.00 / 431.74 = 682.96 ftSolve for the distance C–X and set the point
With the height fixed, the base follows directly from the area of a triangle. Check that the result lies within the side before using it: C–D is 431.74 ft long and C–X comes to 53.98 ft, so X falls well inside side C–D and the trial corner C was the right one to swing from. Had C–X exceeded 431.74 ft, the far end would have had to pass corner D onto side D–E, and a second trial from D would be needed.
Set X by running 53.98 ft from C on the C–D bearing, then inverse P to X to get the division line as it will be described.
ΔA = ½ · (C–X) · h
C–X = 2 ΔA / h = 2 × 18,433.70 / 682.96 = 36,867.40 / 682.96 = 53.98 ft
53.98 < 431.74, so X lies on side C–D ✓
X from C on S 13°23′33″ E, 53.98 ft: N 3327.49, E 4732.50
X–D = 431.74 − 53.98 = 377.76 ft
division line P–X: 683.28 ft on N 78°22′10″ EReport both pieces
The severed parcel is P–B–C–X, bounded by 305.96 ft of the A–B frontage, the whole of B–C, 53.98 ft of C–D, and the new line back to P. The remainder is P–X–D–E–A.
Shoelacing the severed piece from the staked coordinates returns 104,542.4 ft² against the 104,544 ft² designed. The 1.6 ft² difference is rounding X to 0.01 ft — setting the base at 53.98 ft rather than the exact 53.9819 ft costs about 0.6 ft², and rounding the two coordinates of X costs the rest. At 0.00004 acre it is far below the reporting precision, and the deed should call 2.400 acres.
Do not be tempted to publish 104,542 ft² because that is what the coordinates return. The area is the controlling term of the conveyance here; the coordinates are a means of setting a stake.
| Parcel | Corners in order | Area (ft²) | Area (acres) |
|---|---|---|---|
| Severed | P – B – C – X | 104,544 | 2.400 |
| Remainder | P – X – D – E – A | 259,456 | 5.956 |
| Whole | A – B – C – D – E | 364,000 | 8.356 |
Answer
- The division line runs from the monument P (N 3189.74, E 4063.25) to a new point X on side C–D at N 3327.49, E 4732.50.
- X is 53.98 ft from corner C and 377.76 ft from corner D, measured along the C–D boundary.
- The division line P–X is 683.28 ft long on a bearing of N 78°22′10″ E.
- Severed parcel P–B–C–X = 104,544 ft² = 2.400 acres.
- Remainder P–X–D–E–A = 259,456 ft² = 5.956 acres.
Check
Shoelace both pieces independently from their corner coordinates: P–B–C–X returns 104,542.4 ft² and P–X–D–E–A returns 259,456.3 ft². They sum to 363,998.7 ft² against the whole parcel's 364,000 ft², a discrepancy of 1.3 ft², or 4 parts per million, from rounding X to the nearest 0.01 ft.
Check the added triangle on its own: shoelacing P, C and X gives 18,432.1 ft², against the 18,433.70 ft² the deficiency demanded. Consistent to 1.6 ft², again the rounding of X.
Check the perpendicular distance a second way. The division line P–X is 683.28 ft and the perpendicular from P to C–D is 682.96 ft; since X is close to the foot of that perpendicular, the two must be nearly equal, and 683.28 against 682.96 is the expected relationship. A perpendicular distance wildly different from the length of the short cut line would mean the cross product was formed with the wrong pair of vectors.
Check the trial triangle by base and height instead of by coordinates: base B–C = 568.86 ft, and the perpendicular from P to line B–C computes to 302.75 ft, giving ½ × 568.86 × 302.75 = 86,110 ft² — the shoelace value for triangle P–B–C.
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