In view of the coronavirus pandemic, we are making LIVE CLASSES and VIDEO CLASSES completely FREE to prevent interruption in studies
Maths > Straight Lines > Distance of Point From a Line
Straight Lines

Distance of Point From a Line

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.

Suggested Videos

Basics of Straight Lines
Introduction to Lines
Pair of Straight Lines

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(x1, y1) and B(x2, y2) lying on the same line. The length or the distance between the two is ((x2 − x1)2 + (y2 − y1)2)1/2 .

Also, we know the area of a triangle formed by three points is given by  ½ |x1(y2 − y3) + x2(y3 − y1) + x3(y1 − y2)|. Let us study the way to find the distance of point from a line.

You can download Straight Lines Cheat Sheet by clicking on the download button below

straight lines cheat sheet

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(x1, y1) 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.

Distance of Point From a Line

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 |x1(y2 − y3) + x2(y3 − y1) + x3(y1 − y2)|
Or, ½ | x1 (0 + C/B) + (−C/A) (−C/B − y1) +0 (y1 − 0)| = ½ |x1 C/B + y1 C/A + C2/AB|
Or, ½ |C/ (AB) |.|Ax1 + By1 + C|… (II)

Distance of the line AB = ((0 + C/A)2 + (C/B − 0)2)½ = |C| × ((1/A2) + (1/B2))½
Or, Distance, AB = |C/AB| (A2 + B2)½ … (III)

Putting (II) & (III) in (I), we have
NM = d = |Ax1 + By1 + C| / (A2 + B2)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 l1 and l2 in XY-plane with equal slope = m. The equations of the parallel lines:

y = mx + c1 … (I)
y = mx + c2 … (II)

Distance of Point From a Line

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 l1 from M (− c2/m, 0). We know that the distance between two lines is:

d =|Ax1 + By1 + C| / (A2 + B2)½.
Here, A = m, B = 1 and C = c1 (comparing y = mx + cand Ax + By + C = 0)
And, x1 = − c2/m and y1 = 0
So, d = |m (− c2/m) + 0 + c1| / (m2 + 1)½ = |c1 − c2| / (m2 + 1)½

Generalizing the above, we have, d = |C1 − C2| / (A2 + B2)½

(If l1: Ax + By + C1 = 0 and l2: Ax + By + C2 = 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, C1 = 6 and C2 = −6. The required distance between the two lines is,
d = |C1 – C2| / (A2 + B2)½ = |6 − (−6)| / (52 + 32)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, x1 = 2, y1 = 3. The required distance of the point from the line is d =|Ax1 + By1 + C | / (A2 + B2)½ = |2.2 + 5.3 + 10| / (22 + 52)½ = |4 + 15 + 10| /√29 =  √29.

Share with friends

Customize your course in 30 seconds

Which class are you in?
Get ready for all-new Live Classes!
Now learn Live with India's best teachers. Join courses with the best schedule and enjoy fun and interactive classes.
Ashhar Firdausi
IIT Roorkee
Dr. Nazma Shaik
Gaurav Tiwari
Get Started

Leave a Reply

1 Comment threads
0 Thread replies
Most reacted comment
Hottest comment thread
1 Comment authors
Teju Recent comment authors
newest oldest most voted
Notify of




Stuck with a

Question Mark?

Have a doubt at 3 am? Our experts are available 24x7. Connect with a tutor instantly and get your concepts cleared in less than 3 steps.
toppr Code

chance to win a

study tour

Download the App

Watch lectures, practise questions and take tests on the go.

Get Question Papers of Last 10 Years

Which class are you in?
No thanks.