Distance of a Point from a Line

The perpendicular distance d of a point P (x 1, y 1) from the line ax +by +c = 0 is given by
           d =| ax₁ +by₁ +c|/[(a² +b²)]

                        

Rule to find the distance between parallel lines

1. Choose a point on one of the given parallel lines.
2. Find the perpendicular distance from this point to the other line.

Illustrative Examples

Example

Find the equation of a straight line, with a positive gradient, which passes through the point (-5, 0) and is at a perpendicular distance of 3 units from the origin.

Solution

Let m (> 0) be the gradient of the line, then any line through (-5, 0) and with gradient m is
      y -0 = m(x +5) i.e. mx -y +5m = 0 ...(i)
It will be the required line if its perpendicular distance from origin (0, 0) is 3 units
=> |m.0 -0 +5 m| /[(m² +(-1)²)] = 3      => | 5 m | = 3 [m² +1]
=> 25 m² = 9 (m² +1)      => 16 m² = 9     => m = 3/4        (m > 0)
Substituting this value of m in (i), the equation of the required line is
      (3/4) x -y + 5.3/4 = 0 or 3 x -4 y +15 = 0

Example

Find the distance between the lines 3 x -4 y +7 = 0 and 6 x -8 y = 18.

Solution

The given lines are
       3 x -4 y +7 = 0       ...(i)
and 6 x -8 y -21 = 0      ...(ii)
We note that the slope of (ii) = - 6/(-8) = 3/4 = the slope of (i)
=> the given lines are parallel.
To find distance between these lines, we choose a point on (i).
On putting x = 0 in (i), we get -4y +7 = 0 => y = 7/4
Thus (0, 7/4) is a point on (i).
Required distance between given parallel lines
   = perpendicular distance from (0, 7/4) to the line (ii)
    = = |-35 | = units

Exercise

1. Find the distance of the point P from the line AB in the following cases:
   (i) P (2, -3), line AB is 2 x -3 y -25 = 0
   (ii) P (4, 1), line AB is 3 x -4 y -9 = 0
   (iii) P (0, 0), line AB is h (x +h) +k (y +k) = 0
2. Find the distance of the point (0, - 1) from the line joining the points (1, 3) and (-2, 6).
3. Calculate the length of the perpendicular from (7, 0) to the straight line 5 x +12 y -9 = 0 and show that it is twice the length of perpendicular from (2, 1).
4. Find the value (s) of k, given that the distance of the point (4, 1) from the line 3 x -4y +k = 0 is 4 units.
5. The points A (0, 0), B (1, 7), C (5, 1) are the vertices of a triangle. Find the length of perpendicular from A to BC and hence the area of ABC.
6. Find the lengths of altitudes of the triangle whose sides are given by x -4 y = 5, 4 x +3 y = 5 and x +y = 1.
7. Find the length of perpendicular from the point (4, -7) to the line joining the origin and the point of intersection of the lines 2 x -3 y +14 = 0 and 5 x +4 y -7 = 0.
8. A vertex of a square is at the origin and its one side lies along the line x -4 y -10 = 0. Find the area of the square.

Answers

1. (i) 12/13 units         (ii) 1/5 units           (iii) [h² +k²] units
2. 5/2 units                   3. 2 units               4. 12, -28
5. 17/13 units, 17 sq. units
6. 1 unit, 1/7 units, 1/[52] units
7. 1 unit                              8. 4 sq. units