# 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 =| ax _{1}
+by_{1} +c|/[(a² +b²)]**

### Rule to find the distance between parallel lines

- Choose a point on one of the given parallel lines.
- 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

- 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 - Find the distance of the point (0, - 1) from the line joining the points (1, 3) and (-2, 6).
- 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).
- 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.
- 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.
- 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.
- 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.
- 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