## Perpendicular Distance

In this page perpendicular distance from a point to a straight line we are going to see how to find the length  of the perpendicular from a given point.

Formula:

The length of the perpendicular from the point (x₁,y₁) to the line ax + by + c = 0 is

| (ax₁ + by₁ + c)/ va² + b² |

Example 1:

Find the length of the perpendicular from (2,-3) to the straight line

2x - y + 9 = 0

Solution:

The length of the perpendicular from the point (x₁,y₁) to the line ax + by + c = 0 is

| (ax₁ + by₁ + c)/ va² + b² |

Here the given point is (2,-3). We have to apply this point in the given equation.Here x₁ = 2 and y₁ = -3

= |(2(2) - (-3) + 9)/v(2² + (-1)²)|

= | (4 + 3 + 9)/v(4+1) |

= | 16/v5|

= 16/v5 units           perpendicular distance  perpendicular distance

Example 2:

Find the length of the perpendicular from (5,2) to the straight line

3x+2y-1 = 0

Solution:

The length of the perpendicular from the point (x₁,y₁) to the line ax + by + c = 0 is

| (ax₁ + by₁ + c)/ va² + b² |

Here the given point is (5,2). We have to apply this point in the given equation.Here x₁ = 5 and y₁ = 2

= |(3(5) + 2(2) - 1)/v(3² + (2)²)|

= | (15 + 4 - 1)/v(9+4) |

= | 18/v13|

= 18/v13 units

Example 3:

Find the length of the perpendicular from (0,1) to the straight line

x-3y+2 = 0

Solution:

The length of the perpendicular from the point (x₁,y₁) to the line ax + by + c = 0 is

| (ax₁ + by₁ + c)/ va² + b² |

Here the given point is (0,1). We have to apply this point in the given equation.Here x₁ = 0 and y₁ = 1

= |(1(0) - 3(1) + 2)/v(1² + (-3)²)|

= | (0 - 3 + 2)/v(1+9) |

= | -1/v10|

= 1/v10 units

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