The distance from the point to a line is considered to be the shortest distance from a point to a line. In short, it is the length of a perpendicular line segment from the line to the point. The distance from a point to a line can be simply calculated by finding the difference between the co-ordinates. |

**When the line is Vertical:**

In vertical lines, the

**x**- coordinates remains the same and the

**y**- coordinate changes. This line goes up and down and is always parallel to

**y**- axis. All the points lie on the line having same

**x**- coordinate and no slope is seen in the vertical line.

As mentioned earlier, finding the difference between the co-ordinate of the line is the only possible and the simplest way to calculate the distance from the point to the line in case of vertical line. In the below shown figure, we can see a vertical line with equation,

**x = 20**

This represents that all the points of the line have

**x**-coordinate of 20. And, all the

**x**co-ordinate of the second line passing through point C is 37. From the figure, the given point

**C**has the co-ordinate of (37, 8).

$\therefore$ The distance from the point to line is

**37 - 20,**that is

**17**.

We also have a very simple formula which is defined as

$Distance$ = $\left | P_x-L_x \right |$

Here,

**P**

_{x}= the

**x**co-ordinate of line

**P**

**L**

_{x}=

**x**co-ordinate of any point on the given vertical line.

**I I**(mode) - represents the absolute value, where we can consider the negative points into positive one.

**When the line is Horizontal:**

In horizontal line, the x co-ordinate changes and y co-ordinate remains the same. Here, the line goes left to right and is parallel to the x - axis. All the points lie on the line having same y co-ordinate and the slope is always valued as zero.

The below figure shows the horizontal line having an equation of y = 30, which represent all the points on the line having a y coordinate of 30, and C point is given with co - ordinate (35, 6). The y-coordinate of point C is 6.

$\therefore$ The distance between the points and line is 30 - 6, that is 24.

The formula for the distance from a point to line is defined as

$Distance$ = $\left | P_y-L_y \right |$

Here, P

_{y}= the y co-ordinate of the given lineL

_{y}= y co-ordinate of any pointI I (mode) - represents the absolute value, where you can consider the negative points into positive one.

The distance between a line and a point which is not on the line can be found by using a application of lines. This distance is defined as the length of the perpendicular line segment joining the point and the line.

**The Distance between the point (x**

_{1}, y_{1}) and the line Ax + By + C = 0 is given as

The below figure will give us the better idea on the distance from a point to a line. Here, we have drawn a line segment that is perpendicular to the line and ends at the point.

**Given:**

Point (30, 5)

Line, y = 0.50 x + 10

Let us first convert the line into the general form

-0.50 x + y - 10 = 0

Now, by using the formula, we get

$d$ = $\frac{\left | -0.50(30)+1(5)+(-10) \right |}{\sqrt{(-0.50)^2+(1)^2}}$

$d$ = $\frac{\left | -15+5-10 \right |}{\sqrt{0.25+1}}$

$d$ = $\frac{\left |-20 \right |}{1.1180}$

**$d$ = 17.9**

The length of that line segment is the distance what we are looking for and the distance is from point to the line, that is 17.9

Using Trigonometry, we find, calculate and analyse the mathematical properties of the quantities like the height of the ladder, tower or any vertical distances etc. In the same way, we can use the concept of trigonometry to find the distances from a point to a line.

A distances from a point to a line can be found using the trigonometry, when we have a line with an equation and a point with known co-ordinate.

In the figure given below, we have a line with a equation and

**C**with known co-ordinates. Now, first we should look for the perpendicular distances from

**C**to the line at

**B**and to find the distance

**CB.**We need to follow the steps below:

- Draw a horizontal line segment from
**C**until it intersects the line at**A**, forming the right triangle**CBA**. - Now, we need to find the co-ordinate of
**A**. The y co-ordinate of**A**must be the same as**'C'**, and the**x**co-ordinate is given by substituting that**y**into the line equation and solving for**x**. - Next, subtract the
**x**co-ordinate of**C**and**A**and find the length of the line**CA**. - Find the angle A.

m = slope of the given line (The slope of the line in degrees.)

arctan = inverse of tan function

II = absolute value

Finally, find the distance of CB

sin A = $\frac{CB}{CA}$

This CB is nothing but the distance from a point to the line.

Now, using the figure above, let me explain you the calculation to find the distance from a point to line using Trigonometry.

First, let us find the co-ordinate of A.

**Given:**

y = 15

By substituting y in the line eqaution, y = 0.50x + 5.0, we get

15 = 0.50x + 5.0

x = $\frac{15 - 5}{0.50}$

x = $\frac{10}{0.50}$

x = 20

$\therefore$ The co-ordinate of A is (20, 15).

Now, by subtracting the x co-ordinate C and A, we will get the length of line segment CA as 30 (that is, 50 - 20)

Now, our next step is to find the Angle A. That is, the slope of the line in degree.

$Angle\ A $ = $\left | \arctan\ m \right |$

$Angle\ A $ = $\left | \arctan\ 0.50 \right |$

$Angle\ A$ = 27Â°

Sin 27Â° = $\frac{CB}{30}$

CB = 30 sin 27Â°

CB = (30)(0.4540)

CB =

**13.62**

CB is the the distance from a point to the line.