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Perpendicular Distance from a Point to a Line Simplified Revision Notes

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Perpendicular Distance from a Point to a Line

What is Perpendicular Distance?

The perpendicular distance from a point P(x1,y1)P(x_1, y_1) to a line ax+by+c=0ax + by + c = 0 is the shortest distance between the point and the line. This distance is always measured along the line perpendicular to the given line.

Formula for Perpendicular Distance

The perpendicular distance dd from a point P(x1,y1)P(x_1, y_1) to the line ax+by+c=0ax + by + c = 0 is:

d=ax1+by1+ca2+b2d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}

Components of the Formula

  • a,b,ca, b, c: Coefficients from the equation of the line.
  • x1,y1x_1, y_1: Coordinates of the given point.
  • | \cdot |: Absolute value ensures the distance is non-negative.
  • a2+b2\sqrt{a^2 + b^2}: Normalizes the direction of the line.

Derivation of the Formula

  1. Start with the equation of the line: ax+by+c=0ax + by + c = 0
  2. Identify the perpendicular line passing through P(x1,y1)P(x_1, y_1), which has slope ab-\frac{a}{b} (negative reciprocal of ba\frac{b}{a}).
  3. Find the intersection of the two lines.
  4. Use the distance formula to compute the shortest distance between P(x1,y1)P(x_1, y_1) and the line.

Worked Examples

infoNote

Example 1: Find the Perpendicular Distance

Problem: Find the perpendicular distance from P(3,2)P(3, -2) to the line 4x3y+5=04x - 3y + 5 = 0


Solution:

Step 1: Using the formula:

d=ax1+by1+ca2+b2d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}

Step 2: Substitute a=4,b=3,c=5,x1=3,y1=2a = 4, b = -3, c = 5, x_1 = 3, y_1 = -2

d=4(3)+(3)(2)+542+(3)2=12+6+516+9d = \frac{|4(3) + (-3)(-2) + 5|}{\sqrt{4^2 + (-3)^2}} = \frac{|12 + 6 + 5|}{\sqrt{16 + 9}} =2325=235= \frac{|23|}{\sqrt{25}} = \frac{23}{5}

Answer: The perpendicular distance is 235\frac{23}{5} or 4.64.6 units.


infoNote

Example 2: Check Perpendicularity

Problem: Determine if the point P(1,1)P(1, 1) lies on the line 2xy1=02x - y - 1 = 0


Solution:

The perpendicular distance will be zero if the point lies on the line.

Use the formula:

d=2(1)1(1)122+(1)2=2114+1=05=0d = \frac{|2(1) - 1(1) - 1|}{\sqrt{2^2 + (-1)^2}} = \frac{|2 - 1 - 1|}{\sqrt{4 + 1}} = \frac{|0|}{\sqrt{5}} = 0

Answer: The point P(1,1)P(1, 1) lies on the line.


Summary

  • Perpendicular distance formula:
d=ax1+by1+ca2+b2d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}
  • The distance is the shortest path between a point and a line.
  • If d=0d = 0, the point lies on the line.
  • Practice applying the formula to compute distances efficiently.
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