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

To find out the foot of perpendicular(N) drawn from the point P, we follow the steps given below.

**Step 1 :**

Since the lines are perpendicular, the product of their slopes will be equal to -1.

**Step 2 :**

Slope of perpendicular line = -1/Slope of given line

**Step 3 :**

Find the equation using the given point and slope what we have derived from step 2.

**Step 4 :**

Solve the system of equations to find point of intersection.

**Step 5 :**

Find the distance between two points.

Distance formula = √(x_{2}–x_{1})^{2} + (y_{2}–y_{1})^{2}

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)---->(x_{1}, y_{1})

m = 1/2

(y – y_{1}) = m(x
– x_{1})

(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)---->(x_{1}, y_{1})

m = 1/4

(y – y_{1}) = m(x
– x_{1})

(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 – y_{1}) = m(x
– x_{1})

(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 – y_{1}) = m(x
– x_{1})

(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.

3(3x–5) = -x-5 9x-15 = -x-5 x = 1 |
3y = -1-5 3y = -6 y = -2 |

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

2(3–2x) = x+6 6-4x = x+6 -5x = 0 x = 0 |
2y = 0+6 2y = 6 y = 3 |

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.

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