How can you find the distance between two objects say two street lights? By measuring the length between them. By the length between the two street lights, we can find the distance. How can you measure the distance between your house and your friend’s? We have to measure the length between the two. Similarly, for finding the distance between two points, we measure the length between them. Here, we will study how to find the distance of point from a line.

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## Distance Between Two Points

We are familiar with the representation of points on a graph sheet. In coordinate geometry, we learned to find the distance between two points, say A and B. Suppose the coordinates of two points are A(x_{1}, y_{1}) and B(x_{2}, y_{2}) lying on the same line**. **The length or the distance between the two is** **((x_{2} − x_{1})^{2} + (y_{2} − y_{1})^{2})^{1/2}** **.

Also, we know the area of a triangle formed by three points is given by _{ ½} |x_{1}(y_{2} − y_{3}) + x_{2}(y_{3} − y_{1}) + x_{3}(y_{1} − y_{2})|. Let us study the way to find the distance of point from a line.

## Perpendicular Distance of Point from a Line

Have you ever thought of measuring the distance of the point from a line? How can we do this? The point is not lying on the line. How to do it? Let’s find out.

Before we start, learn the formula of distance here.

Suppose a line *l* in XY−plane and N(x_{1}, y_{1}) is any point at a distance d from the line *l*. This line is represented by Ax + By + C = 0. The distance of point from a line, ‘d’ is the length of the perpendicular drawn from N to *l*. The x and y-intercepts are −C/A and −C/B respectively.

The line meets the y and the x axes at points A and B respectively. The coordinates of the points are A (0, −C/B) and B (−C/A, 0). The area of the triangle is given by

area (Δ NAB) = ½ base** × **height = ½ AB × NM**,
**⇒

**NM = 2 area (Δ NAB) / AB … (I)**

Also, area (Δ NAB) = ½ |x_{1}(y_{2} − y_{3}) + x_{2}(y_{3} − y_{1}) + x_{3}(y_{1} − y_{2})|

Or, ½ | x_{1} (0 + C/B) + (−C/A) (−C/B − y_{1}) +0 (y_{1} − 0)| = ½ |x_{1} C/B + y_{1} C/A + C^{2}/AB|

Or, ½ |C/ (AB) |.|Ax_{1} + By_{1} + C|… (II)

Distance of the line AB = ((0 + C/A)^{2} + (C/B − 0)^{2})^{½} = |C| × ((1/A^{2}) + (1/B^{2}))^{½
}Or, Distance, AB = |C/AB| (A^{2} + B^{2})^{½} … (III)

Putting (II) & (III) in (I), we have

NM = d = |Ax_{1} + By_{1} + C| / (A^{2} + B^{2})^{½}.

It is interesting to find out the distance between two parallel lines.

**Browse more Topics Under Straight Lines**

## Distance between Two Parallel Lines

Two lines are parallel to each other if the distance between them at any point remains the same. In other words, if the slopes of both the lines are the same, they will be parallel to each other. Suppose there are two parallel lines *l _{1 }*and

*l*in XY-plane with equal slope = m. The equations of the parallel lines:

_{2}y = mx + c_{1} … (I)

y = mx + c_{2} … (II)

The distance between two parallel lines is calculated by the distance of point from a line. It is equal to the length of the perpendicular distance from any point to one of the lines. Let N be the point through which the perpendicular or normal is drawn to *l _{1}* from M (− c

_{2}/m, 0). We know that the distance between two lines is:

d =|Ax_{1} + By_{1} + C| / (A^{2} + B^{2})^{½}.

Here, A = m, B = 1 and C = c_{1}** **(comparing y = mx + c_{1 }and Ax + By + C = 0)

And, x_{1}** **= − c_{2}/m and y_{1} = 0

So, d = |m (− c_{2}/m) + 0 + c_{1}| / (m^{2} + 1)^{½} = |c_{1} − c_{2}| / (m^{2} + 1)^{½}

Generalizing the above, we have, d = |C_{1} − C_{2}| / (A^{2} + B^{2})^{½}

(If *l _{1}*: Ax + By + C

_{1}= 0 and

*l*: Ax + By + C

_{2}_{2}= 0)

Understand the concept of coordinates here in detail.

## Solved Example for You

Problem: Find the distance between two lines 5x + 3y + 6 = 0 and 5x + 3y – 6 = 0.

Solution: Here, A = 5, B = 3, C_{1} = 6 and C_{2} = −6. The required distance between the two lines is,

d = |C_{1} – C_{2}| / (A^{2} + B^{2})^{½} = |6 − (−6)| / (5^{2} + 3^{2})^{½} = 12/√34.

Problem: Find the distance between the line x/5 + y/2 + 1 = 0 and a point (2, 3).

Solution: The equation of the line can be written as 2x + 5y + 10 = 0. Here, A = 2, B = 5, C = 10, x_{1} = 2, y_{1} = 3. The required distance of the point from the line is d =|Ax_{1} + By_{1} + C | / (A^{2} + B^{2})^{½} = |2.2 + 5.3 + 10| / (2^{2} + 5^{2})^{½} = |4 + 15 + 10| /√29 = √29.

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