Distance from Point to Line

[MathDefinition title="Perpendicular Distance"] The perpendicular distance from a point (x1, y1) to a line ax + by + c = 0 is: d = |ax1 + by1 + c|/√(a^2 + b^2) [/MathDefinition]

[MathDefinition title="Distance Between Parallel Lines"] The distance between two parallel lines ax + by + c1 = 0 and ax + by + c2 = 0 is: d = |c1 - c2|/√(a^2 + b^2) [/MathDefinition]

[MathKeyFormula title="Key Distance Formulas"]

1. Point to line: d = |ax1 + by1 + c|/√(a^2 + b^2)
2. Parallel lines: d = |c1 - c2|/√(a^2 + b^2)
3. Distance between two points: d = √((x2-x1)^2 + (y2-y1)^2)
4. Distance from origin to line: d = |c|/√(a^2 + b^2) [/MathKeyFormula]

[MathNote title="Important Notes"]

1. The formula works for any form of the line equation
2. For parallel lines, the coefficients of x and y must be the same
3. The distance is always positive (use absolute value) [/MathNote]

[MathExample title="Example 1: Point to Line Distance"] Problem: Find the perpendicular distance from the point (3, -2) to the line 3x - 4y - 26 = 0.

Solution: Using d = |ax1 + by1 + c|/√(a^2 + b^2):

a = 3, b = -4, c = -26 x1 = 3, y1 = -2

d = |3(3) + (-4)(-2) - 26|/√(9 + 16) = |9 + 8 - 26|/√25 = |-9|/5 = 9/5 = 1.8

Therefore, the distance is 9/5 units. [/MathExample]

[MathExample title="Example 2: Distance Between Parallel Lines"] Problem: Find the distance between the parallel lines 6x + 8y - 5 = 0 and 6x + 8y + 10 = 0.

Solution: The lines are already in the form ax + by + c = 0 with same coefficients.

a = 6, b = 8, c1 = -5, c2 = 10

d = |c1 - c2|/√(a^2 + b^2) = |-5 - 10|/√(36 + 64) = |-15|/√100 = 15/10 = 3/2

Therefore, the distance between the lines is 3/2 units. [/MathExample]

[MathExample title="Example 3: Finding Line at Given Distance"] Problem: Find the equation of the line parallel to 3x + 4y - 7 = 0 and at a distance of 5 units from it.

Solution: The required line is 3x + 4y + k = 0

Distance between parallel lines: d = |k - (-7)|/√(9 + 16) = 5 |k + 7|/5 = 5 |k + 7| = 25

k + 7 = 25 or k + 7 = -25 k = 18 or k = -32

Therefore, the required lines are: 3x + 4y + 18 = 0 or 3x + 4y - 32 = 0 [/MathExample]

[MathExample title="Example 4: Distance from Origin"] Problem: Find the distance from the origin to the line 5x + 12y - 26 = 0.

Solution: Using d = |c|/√(a^2 + b^2):

a = 5, b = 12, c = -26

d = |-26|/√(25 + 144) = 26/√169 = 26/13 = 2

Therefore, the distance from the origin is 2 units. [/MathExample]