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Perpendicular distance of a point from a line

Given a point P (x1, y1), and a line with the equation ax + by + c = 0, the perpendicular distance of a point from a line is given by the equation –

d  =  {~delim{|}{ax_1~+~by_1~+~c}{|}~}/{sqrt{a^2~+~b^2}}

Remember the shortest distance of a point from a line is always the perpendicular distance.

Note : We put an absolute sign in the above formula, since the distance cannot be negative.

 

Example 1: Find the shortest distance of the point (1, -3) from 3x – 4y + 5 = 0.

The shortest distance of a point from a line is the perpendicular distance. And the perpendicular distance of a point from a line is given by the equation –

d  =  {~delim{|}{ax_1~+~by_1~+~c}{|}~}/{sqrt{a^2~+~b^2}}

Here x1 = 1, y1 = -3,   a = 3, b = -4 and c = 5

d  =  {~delim{|}{3~*~1~+~-4~*~3~+~5}{|}~}/{sqrt{3^2~+~(-4)^2}}

=  {~delim{|}{3~+~-12~+~5}{|}~}/{sqrt{9~+~16}}

=  20/5

=  4 units

 

Example 2: Show that the line 3x – 4y + 5 = 0 is the tangent to the circle x2 + y2 = 1

The tangent and radius of a circle are perpendicular to each other. Hence if we show that the perpendicular distance of the centre of the circle to the given line is equal to the radius, then the line will be tangent to the circle.

Equation of the line is 3x – 4y + 5 = 0

Center of the circle x2 + y2 = 1 is (0, 0) and radius = 1 unit.

Perpendicular distance of (0, 0) to the line is

d  =  {~delim{|}{ax_1~+~by_1~+~c}{|}~}/{sqrt{a^2~+~b^2}}

d  =  {~delim{|}{3~*~0~+~-4~*~0~+~5}{|}~}/{sqrt{3^2~+~(-4)^2}}

=  5/{sqrt{9~+~16}}

=  5/5

=  1 unit   =  radius of the circle

Hence the line is a tangent to the given circle.