Distances

The distance from a point to a line always means the shortest distance, which is along the perpendicular.

Point to line. The distance of (x₁, y₁) from Ax + By + C = 0 is

• d = |Ax₁ + By₁ + C| / √(A² + B²): put the point into the left side, take the size (drop any minus sign), and divide by √(A² + B²).
• (2, 1) from 4x + 3y – 1 = 0: |8 + 3 – 1|/√(16 + 9) = 10/5 = 2.

Between two parallel lines Ax + By + C₁ = 0 and Ax + By + C₂ = 0: d = |C₁ – C₂| / √(A² + B²). First make A and B the same in both equations.

• 6x + 8y – 5 = 0 and 3x + 4y + 5 = 0: divide the first by 2 to get 3x + 4y – 2.5 = 0.
• d = |–2.5 – 5|/5 = 7.5/5 = 1.5.

Families of lines.

• Parallel to Ax + By + C = 0: keep A and B, change only the constant: Ax + By + k = 0.
• Perpendicular to it: swap A and B and change one sign: Bx – Ay + k = 0.
• Then find k from the extra condition.

Examples with the line 2x – y + 3 = 0:

• Parallel, through (1, 4): 2x – y + k = 0 with 2 – 4 + k = 0, so k = 2: 2x – y + 2 = 0.
• Perpendicular, through (1, 4): x + 2y + k = 0 with 1 + 8 + k = 0, so k = –9: x + 2y – 9 = 0.

A parallel line at a given distance. Lines parallel to 4x + 3y + 2 = 0 at distance 2 from it: 4x + 3y + k = 0 with |k – 2|/5 = 2, so k – 2 = ±10: 4x + 3y + 12 = 0 or 4x + 3y – 8 = 0. There are two, one on each side.

The line midway between 2x + y – 4 = 0 and 2x + y + 6 = 0 is equally far from both, so its constant is the average: 2x + y + 1 = 0.

Foot of the perpendicular. From P(5, 0) to x – y + 1 = 0 (slope 1):

• The perpendicular through P has slope –1: x + y = 5.
• Solve with y = x + 1: 2x + 1 = 5, so the foot is (2, 3).
• Check: the distance formula gives |5 – 0 + 1|/√2 = 3√2, and the distance from (5, 0) to (2, 3) is √18 = 3√2.

Altitude of a triangle. For B(0, 0), C(3, 4), A(5, 0): line BC is 4x – 3y = 0, so the altitude from A is |20|/5 = 4. With BC = 5, the area is ½ × 5 × 4 = 10.