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

Related Topics