# EQUATION OF THE LINES AT A PARTICULAR DISTANCE FROM THE GIVEN POINT

## About "Equation of the Lines at a Particular Distance From the Given Point"

Equation of the Lines at a Particular Distance From the Given Point :

Here we are going to see how to find the equation of the lines at a particular distance from the given point.

## Equation of the Lines at a Particular Distance From the Given Point - Examples

Question 1 :

Find the equations of two straight lines which are parallel to the line 12x + 5y +2 = 0 and at a unit distance from the point (1, − 1).

Solution :

Since the required line is parallel to the line 12x + 5y + 2  = 0.

Equation of the required line will be in the form

12x + 5y + k = 0

Distance between the point (1, -1) and 12x + 5y + k = 0 is 1.

Distance between the point and a line

=  |Ax + By + C|/√A2 + B2

= |12(1) + 5(-1) + k|/√122 + 52

1  = |7 + k|/√169

1  = |7 + k|/13

 13  =  7 + kk  =  13 - 7  =  6 -13  =  7 + kk  =  -13 - 7  =  -20

Hence the required lines are 12x + 5y + 6 = 0 and 12x + 5y - 20 = 0.

Question 2 :

Find the equations of straight lines which are perpendicular to the line 3x+ 4y −6 = 0 and are at a distance of 4 units from (2, 1).

Solution :

Since the required line is perpendicular to the line 3x+ 4y −6 = 0

Equation of the required line will be in the form

4x - 3y + k = 0

Distance between the point (2, 1) and 4x - 3y + k = 0 is 4.

Distance between the point and a line

=  |Ax + By + C|/√A2 + B2

4  = |4(2) - 3(1) + k|/√42 + (-3)2

4  = |8 - 3 + k|/√16 + 9

4  = |5 + k|/5

20  =  |5 + k|

 20  =  5 + kk  =  20 - 5  =  15 -20  =  5 + kk  =  -20 - 5  =  -25

After having gone through the stuff given above, we hope that the students would have understood "Equation of the Lines at a Particular Distance From the Given Point".

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