FIND THE DISTANCE BETWEEN A GIVEN POINT AND LINE

The distance from the point to the line is the distance from the point to the foot of the perpendicular.

Find the distance from :

Example 1  :

(8, - 5) to y  =  - 2x – 4

Solution  :

Step 1 :

y  =  - 2x – 4 -----(1)

By comparing the given equation with slope intercept form (y  =  mx + b)

slope (m)  =  - 2

Step 2 :

slope of perpendicular line  =  1/2

Step 3 :

Find the equation of the perpendicular line passes through (8, - 5)

(8, - 5)---->(x₁, y₁)

m  =  1/2

 (y – y₁)  =  m(x – x₁)

(y + 5)  =  1/2(x - 8)

2(y + 5)  =  x – 8

2y  =  x – 8 -10

2y  =  x – 18 -----(2)

Step 4 :

Find the point of intersection of two lines,

By applying (1) in (2), we get

2(- 2x – 4)  =  x – 18

- 4x – 8  =  x – 18

x  =  2

2y  =  2 – 18

2y  =  - 16

y  =  - 8

Foot of perpendicular is (2, -8)

Step 5 :

Distance between two points :

=  √(2 - 8)² + (- 8 + 5)²

=  3√5 units.

So, the required distance is 3√5 units.

Example 2  :

(- 10, 9) to y  =  - 4x + 3

Solution  :

y  =  - 4x + 3 -----(1)

slope (m)  =  - 4

slope of perpendicular line  =  1/4

Equation of perpendicular line :

(- 10, 9)---->(x₁, y₁)

m  =  1/4

 (y – y₁)  =  m(x – x₁)

(y - 9)  =  1/4(x + 10)

4y – 36  =  x + 10

4y  =  x + 46 -----(2)

By applying (1) in (2), we get

4(- 4x + 3)  =  x + 46

-16x + 12  =  x + 46

x  =  - 2

4y  =  x + 46

4y  =  - 2 + 46

y  =  44/4

y  =  11

Foot of perpendicular is (- 2, 11)

Distance between two points :

=  √(- 2 + 10)² + (11 - 9)²

=  2√17 units.

So, the required distance is 2√17 units.

Example 3  :

(- 2, 8) to 3x – y  =  6

Solution  :

- y  =  - 3x + 6

y  =  3x – 6 -----(1)

slope (m)  =  3

slope of perpendicular line  =  - 1/3

Equation of perpendicular line :

 (y – y₁)  =  m(x – x₁)

(y - 8)  =  - 1/3(x + 2)

3y – 24  =  - x – 2

3y  =  - x + 22 -----(2)

By applying (1) in (2), we get

3(3x – 6)  =  - x + 22

9x – 18  =  - x + 22

x  =  4

3y  =  - x + 22

3y  =  - 4 + 22

y  =  6

Foot of perpendicular is (4, 6)

Distance between two points :

=  √(4 + 2)² + (6 - 8)²

=  2√10 units.

So, the required distance is 2√10 units.

Example 4  :

(1, 7) to 4x – 3y  =  8

Solution  :

4x – 3y  =  8

- 3y  =  - 4x + 8

y  =  4/3x – 8/3 -----(1)

slope (m)  =  4/3

slope of perpendicular line  =  - 1/(4/3)

=  - 3/4

Equation of perpendicular line :

 (y – y₁)  =  m(x – x₁)

(y - 7)  =  - 3/4(x - 1)

4y – 28  =  - 3x + 3

4y  =  - 3x + 31 -----(2)

By applying (1) in (2), we get

4(4/3x – 8/3)  =  - 3x + 31

16/3x – 32/3  =  - 3x + 31

x  =  5

4y  =  - 3x + 31

4y  =  - 3(5) + 31

y  =  4

Foot of perpendicular is (5, 4)

Distance between two points :

=  √(5 - 1)² + (4 - 7)²

=  5 units

So, the required distance is 5 units.

Foot of perpendicular is (1, -2).

Step 5 :

Distance between two points :

=  √(1–7)² + (-2+4)²

=  √40

=  2√10 units.

So, the required distance is 2√10 units.

Example 2 :

(-6, 0) to y  =  3–2x

Solution :

y  =  3–2x  ----(1)

By comparing the given equation with y = mx+b, we come so know slope (m)

Slope (m)  =  - 2

Slope of perpendicular line  =  1/2

Equation of perpendicular line :

(y - 0)  =  1/2(x + 6)

2y  =  x+6  ----(2)

Applying (1) in (2), we get

So, foot of perpendicular is (0, 3).

Distance between (-6, 0) and (0, 3).

=  √(0+6)² + (3-0)²

=  √(36+9)

=  √45

=  3√5 units.

So, the required distance is 3√5 units.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.