Distance of a Point From a Line: Formula, Derivation & Examples

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Distance of a Point From a Line is the shortest distance between that point and the line in Geometry.

The length of a line segment drawn from the point to the closest point on the line is the shortest distance from that point.
It is the perpendicular distance of a point to the given line.
Drawing a vertical line segment on the line passing through the specified point is used to get the shortest distance.
An infinite number of lines can be drawn in a plane from one point to another.
A triangle can be created by connecting the point and line with more than one line.
Distance Formula can be used in real-life situations like calculating the distance between two nails, buildings, roads, etc.

Distance of a Point from a Line (d) is given by the formula:

d = [\|Ax₁ + By₁ + C\|]/ √(A² + B²)

Where, A, B, and C are the coefficients of the line equation and x₁ and y₁ are the coordinates of the point.

Table of Content

What is Distance of a Point From a Line?
Perpendicular Distance of a Point From a Line
Derivation of Distance of a Point From a Line
Distance Formula Solved Example
Things to Remember
Previous Year Questions
Sample Questions

Key Terms: Distance Formula, Perpendicular Distance, Triangle, Line Segment, Point, Geometry, Hypotenuse, Perpendicular Line

What is Distance of a Point From a Line?

[Click Here for Sample Questions]

Distance of a Point From a Line is the shortest possible distance between a point and the line. The length of a line segment joining point and the nearest point on the line is the minimum distance between them. It calculates the smallest length or distance needed to move a point along a line.

The distance of minimum length can be considered as a line perpendicular to that line. Let line ‘L’ and a point x that does not lie on line L be shown below:

Distance of a Point From a Line

Distance of a Point From a Line

Here, we need to measure the distance between a point ‘x’ and line L. Thus, follow the given steps:

First, draw a perpendicular so we can get a right-angle triangle.
The hypotenuse is a right triangle's longest side as well.
Drawing the foot of the perpendicular from the point to the line and any other segment connecting the point to the line yields a right triangle as a result.

perpendicular from the point to the line

Perpendicular from the point to the line

The hypotenuse of the right-angled triangle that results from this second line segment will always be drawn.

Using the Distance Formula, the distance between the given point and line can be calculated.

Read More:

Relevant Concepts
Properties of Parallel Lines	Collinear Points	Straight Lines
Distance Between Two Lines	Plane	Perpendicular Line Through a Point

Perpendicular Distance of a Point from a Line

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The shortest distance between two objects is represented by the length of the perpendicular drawn between them. The steps to get at the formula for calculating how far a point is from a line perpendicularly are listed below.

Step 1: Consider a line L: Ax + By + C = 0 whose distance from the point N (x₁, y₁) is d.
Step 2: Draw a perpendicular NM from point N to line L as shown in the figure below.
Step 3: On the x-and-y-axis, let's assume N and B be the points where the line segment meets.
Step 4: Coordinates of the points can be written as B(-C/A, 0) and A(0, -C/B).

Perpendicular Distance of a Point from a Line

Perpendicular Distance of a Point from a Line

Derivation of Distance of a Point From a Line

[Click Here for Sample Questions]

Distance of a Point From a Line can be derived using the area of a triangle in various aspects:

In triangle ANB,

Derivation of Distance of a Point From a Line

Derivation of Distance of a Point From a Line

ar (ΔANB) = (1/2) × Base × Height = (1/2) × NM × BA

⇒ NM = [2 ar(ΔANB)]/BA….(i)

In coordinate geometry, the area of a triangle with vertices (x₁, y₁), (x₂, y₂), and (x₃, y₃) is

Δ = (1/2) |x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)|

Here,

(x₁, y₁) = N(x₁, y₁)
(x₂, y₂) = B(-C/A, 0)
(x₃, y₃) = A(0, -C/B)

Now,

ar(ΔANB) = (1/2) |x₁(0 + C/B) + (-C/A)(-C/B – y₁) + 0(y₁ – 0)|

ar(ΔANB) = (1/2) |x₁(C/B) + (C²/AB) + y₁(C/A)|

⇒ 2 ar(ΔANB) = |C/AB|. |Ax₁ + By₁ + C|….(ii)

Distance BA = √[(0 + C/A)² + (C/B – 0)²]

= √[(C²/A²) + (C²/B²)]

Let’s find BA using the distance formula.

QR = |C/AB| √(A² + B²)….(iii)

Substituting (ii) and (iii) in (i),

PM = [|C/AB|. |Ax₁ + By₁ + C|] / [|C/AB| √(A² + B²)]

d = |Ax₁ + By₁ + C|]/ √(A² + B²)

Therefore, the perpendicular distance (d) of a line Ax + By+ C = 0 from a point (x₁, y₁) is given by

d = [\|Ax₁ + By₁ + C\|]/ √(A² + B²)

Distance Formula Solved Example

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Given below is a solved example on Distance Formula for the better understanding of the concept:

Example: Find the distance between the point (0,0) and the line 3x + 4y + 10 = 0.

Solution: From 3x + 4y + 10 = 0, we have,

a = 3
b = 4
c = 10

The distance between the point (0,0) and the line 3x + 4y + 10 = 0 can now be determined by substituting the values into the Distance Formula:

d = (ax₀ + by₀ +c )/ √a² +b²

d = [3(0)+4(0)+10]/ √3²+4²

d = (0+0+10)/ √9+16

d = 10/√25

d = 10/5

d = 2

∴ The distance between the point (0, 0) and the line 3x + 4y +10 = 0 is 2 units.

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Related Topics
Cartesian Coordinate System	Properties of Parallel Lines	Slope-Intercept Form Formula
Point of Intersection Formula	Constructing of Copy of Line Segments	Coplanarity of Two Lines
Angle Between a Line and a Plane	Introduction to Three-Dimensional Geometry	Angle Between Two Lines

Things to Remember

Distance of a Point from a Line is the perpendicular distance between the point and the line.
It is the measure of the smallest distance or length to move a point along a line.
Distance of a point to the line is given by the formula d = [|Ax₁ + By₁ + C|]/ √(A² + B²).
The distance between two lines is given by d = |C₁ – C₂| / (A² + B²)½ .
Distance Formula is used in a wide range of applications such as construction, buildings, etc.

Previous Years’ Questions

The distance of the point (1, 2, -4) from the line… (KCET - 2020)
Find perpendicular distance of the line joining the points…
The perpendicular distance from the point (1,−1) to the… (KEAM)
The shortest distance between the lines…
If the sum of the distance of a point from two perpendicular lines… (JEE Advanced - 1992)
The shortest distance between the straight lines through the points… (VITEEE - 2008)
The distance of the point (3, 4, 5) from X-axis is…
What is the perpendicular distance of the point P(6, 7, 8)...
The shortest distance from the point (3,0) to the parabola… (COMEDK UGET - 2005)
Equation of the line of the shortest distance between the lines… (JEE Main - 2014)

Sample Questions

Ques. Find the distance of the point (-3, 5) from the line 4x – 3y – 26 = 0. (3 Marks)

Ans. Comparing these with the standard forms,

A = 4
B = -3
C = -26

Given,

x₁ = -3, y₁ = 5

We know that the perpendicular distance (d) of a line Ax + By+ C = 0 from a point (x₁, y₁) is given by

d = |Ax₁ + By₁ + C|]/ √(A² + B²)

Substituting the values,

d = |4(-3) + (-3)(5) + (-26)|/√[(4)² + (-3)²]

= |-12 – 15 – 26|/√(16 + 9)

= |-53|/√25

= 53/5

Thus, the distance is 53/5 units.

Ques. Find the distance between two lines 5x + 3y + 6 = 0 and 5x + 3y – 6 = 0. (3 Marks)

Ans. Here,

A = 5
B = 3
C₁ = 6
C₂ = −6

The required distance between the two lines is,

d = |C₁ – C₂| / (A² + B²)½ = |6 − (−6)| / (5² + 3²)½ = 12/√34 Units

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

Ans. 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

Thus, the distance is √29 Units.

Ques. Find the distance between the point (5,1) and the line y = 3x + 1. (3 Marks)

Ans. Given point (5,1) and line y = 3x + 1.

We need to find the distance between them.

y = 3x + 1

First write the line equation in standard form as

3x + 1 – y = 0

Comparing these with the standard forms,

A = 3
B = -1
C = 1

Given,

x₁ = 5, y₁ = 1

We know that the perpendicular distance (d) of a line Ax + By+ C = 0 from a point (x₁, y₁) is given by

d = |Ax₁ + By₁ + C|]/ √(A² + B²)

Substituting the values,

d = [|3x(5) + (-1)(1) + 1|]/√(3² + (-1)²)

d = 15/√10 Units

Ques. Find the perpendicular distance from the point P (5, 6) to the line AB, – 2x + 3y + 4 = 0, using the distance of the point from a line formula. (3 Marks)

Ans. Point P (x₁,y₁) = (5, 6).

Given line equation is -2x + 3y + 4 = 0

Comparing with the standard form of the equation, we get,

A = -2
B = 3
C = 4

d = |((-2)(5) + (3)(6) + 4)/ √((-2)²+(3)²)

= |-10 + 18 + 4|/ √(4 + 9)|

= |12/√(13)|

d = 3.328 Units

Ques. Find the distance from the point K (−3,7) to the line PQ y = (6/5)x + 2 using the distance of the point from a line formula. (3 Marks)

Ans. The line PQ can be simplified as:

y = (6/5)x + 2

5y = 6x +10

Thus, 6x - 5y + 10 = 0

Here, the coordinates of the point K is K(x₁,y₁) = (-3, 7), and A = 6, B =-5 and C = 10

d = |(6)(-3) + (-5)(7) + 10|/ √((6)²+(-5)²)

= |-18 -35 + 10|/ √(36 + 25)

= |-43|/√(61)

d = |-5.506|

Ques. Point P is given (k, -4). Find the value of k such that distance between point P and the line 6x - 8y = 5 is 9/2. (3 Marks)

Ans. The given point is P(k, -4)

So,

x₀ = k, y₀ = -4

The line equation 6x - 8y = 5 is in standard form, so convert it into general form.

6x - 8y - 5 = 0

Thus,

a = 6
b = -8
c = -5

We know, the distance formula is given by

d = ax₀ + by₀ + c/ √a² +b²

Substituting the values, we get

9/2 = |6(k) + (-8)(-4) - 5|/ √(6)² + (8)²

9/2 = |6k - 27|/10

45 = |6k - 27|

6k - 27 = ±45

Solving above equation we get two values of k as

k = 3 and 12

To find which value is correct, substitute the value of k in the distance formula.

Thus from cross-checking, we get

k = 3.

Ques. Find the distance between the point (10,5) and the line y= 5/3x+7. (3 Marks)

Ans. From the given point, we know that x₀ =10 and y₀ =5. However, the line is not written in the general form.

y = 5/3x+7

3[y = 5/3x+7]

3y = 5x+21

3y-5x=5x - 5x +21

3y - 5x = 21

3y - 5x - 21 = 21 - 21

-5x + 3y - 21 =0

Thus,

d = Ax₀ + By₀ + c/ √a² +b²

= -50 +15 -21/√25+9

= -56/√34

= 56/√34

d = 9.6 (approx)

Ques. If the perpendicular distance from the origin to the line x/a + y/b = 1 is m. Find the relation in a, b, and m. (3 Marks)

Ans. The equation of a line is x/a + y/b = 1

Expressing in general form, we get,

⇒ bx + ay - ab = 0

The point is the origin (0, 0)

x₀ = 0, y₀ = 0

Given, the distance of origin to the line is m. So, from the distance formula

m = |b(0) + a(0) - ab|/ √b² + a²

m = ab/√a² + b²

Squaring both sides of the equation

m²= a²b²/ (a² + b²)

⇒ 1/m² = a² + b²/ (a²b²)

Hence,

1/m² = 1/a² + 1/b²

Ques. Find the coordinates of the points on the x-axis, such that their distances from the line x/3 + y/4 = 1 are 4. (3 Marks)

Ans. The line equation is x/3 + y/4 = 1

In general form,

4x + 3y - 12 = 0 …(i)

On comparing with the general equation: ax + by + c = 0 we get,

a = 4, b = 3, c = -12

Any point on the x-axis is (p, 0) and its distance from the line is 4.

We know that the perpendicular distance (d) of a line Ax + By+ C = 0 from a point (x₁, y₁) is given by

d = |ax₁ + by₁ + c|]/ √(a² + b²)

Substituting the values we get

4 = |4(p) + 3(0) - 12|/ √(4² + 3²)

⇒ 4 = |4p - 12|/ 5

⇒ 20 = |4p - 12|

± (4p - 12) = 20

⇒ 4p -12 = 20 or 4p - 12 = -20

⇒ 4p = 32 or 4p = -8

⇒ p = 8 or -2

∴ The points on the x-axis are (8, 0) and (-2, 0).

Ques. If m and n are the perpendicular distance length from the origin to the lines xsecθ + y cosecθ = k and x cosθ - y sinθ = kcos²θ respectively. Then prove that n² + 4m² = k². (5 Marks)

Ans. The line equations are

xsecθ + y cosecθ = k … (1)

x cosθ - y sinθ = kcos²θ …(2)

We know that the perpendicular distance (d) of a line Ax + By+ C = 0 from a point (x₁, y₁) is given by

d = |ax₁ + by₁ + c|]/ √(a² + b²)

On comparing with the general form of the equation ax + by + c = 0, we get

a = secθ, b = cosecθ, c = -k

Given, the length of perpendicular distance from the origin (0, 0) is m

m = |secθ(0) + cosecθ(0) - k|/ (√sec²θ + cosec²θ)

⇒ m = |-k|/ (√sec²θ + cosec²θ)

⇒ m = k/ (√1/cos²θ + 1/sin²θ) = kcosθsinθ/(√cos²θ + sin²θ) = kcosθsinθ

⇒ m = kcosθsinθ

Multiply the equation by 2 on both sides

2m = k(2cosθsinθ)

⇒ 2m = ksin2θ

Square both sides of the equation,

4m² = k² sin²2θ …(3)

Now, for eq(2)

x cosθ - y sinθ = kcos2θ

X cos θ - ysinθ - kcos2θ

Comparing with the general equation ax + by + c = 0,

a = cosθ, b = -sinθ, c = – kcos2θ

Its distance from the origin is n

Thus,

n = |cosθ(0) - sinθ(0) - kcos2θ|/ (√cos²θ + sin²θ)

n = |-kcos2θ|/ (√cos²θ + sin²θ)

n = k cos2θ

Square both sides of the equation

n² = k² cos²2θ …(4)

Add eq (3) and (4), we get

n² + 4m² = k² cos²2θ + k² sin²2θ

n² + 4m² = k² (cos²2θ + sin²2θ)

n² + 4m² = k²

Hence Proved.

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Distance of a Point From a Line: Formula, Derivation & Examples

What is Distance of a Point From a Line?

Perpendicular Distance of a Point from a Line

Derivation of Distance of a Point From a Line

Distance Formula Solved Example

Things to Remember

Previous Years’ Questions

Sample Questions

CBSE CLASS XII Related Questions

1.
Find a point on the line \( \frac{x - 2}{3} = \frac{1 - y}{2} = \frac{z - 3}{2} \) at a distance of \( \sqrt{2} \) units from the point \( (1, 2, 3) \).

2.
Find:

The shortest distance between the lines: \[ \vec{r}=(4+\lambda)\hat{i}+(2\lambda-1)\hat{j}-3\lambda\hat{k} \] and \[ \vec{r}=(1+2\mu)\hat{i}+(4\mu-1)\hat{j}+(2-5\mu)\hat{k} \]

3.
Find:

The principal value of \[ \sec^{-1}(\sqrt{2})+2\csc^{-1}(-2) \] is:

6.
Find:

If the function \[ f(x)= \begin{cases} \frac{\sin x}{x}+\cos x, & x\neq0\\ k, & x=0 \end{cases} \] is continuous at \(x=0\), then find the value of \(k\).

Similar Mathematics Concepts

Comments

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Distance of a Point From a Line: Formula, Derivation & Examples

What is Distance of a Point From a Line?

Perpendicular Distance of a Point from a Line

Derivation of Distance of a Point From a Line

Distance Formula Solved Example

Things to Remember

Previous Years’ Questions

Sample Questions

CBSE CLASS XII Related Questions

1.Find a point on the line \( \frac{x - 2}{3} = \frac{1 - y}{2} = \frac{z - 3}{2} \) at a distance of \( \sqrt{2} \) units from the point \( (1, 2, 3) \).

2. Find: The shortest distance between the lines: \[ \vec{r}=(4+\lambda)\hat{i}+(2\lambda-1)\hat{j}-3\lambda\hat{k} \] and \[ \vec{r}=(1+2\mu)\hat{i}+(4\mu-1)\hat{j}+(2-5\mu)\hat{k} \]

3. Find: The principal value of \[ \sec^{-1}(\sqrt{2})+2\csc^{-1}(-2) \] is:

6. Find: If the function \[ f(x)= \begin{cases} \frac{\sin x}{x}+\cos x, & x\neq0\\ k, & x=0 \end{cases} \] is continuous at \(x=0\), then find the value of \(k\).

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Find a point on the line \( \frac{x - 2}{3} = \frac{1 - y}{2} = \frac{z - 3}{2} \) at a distance of \( \sqrt{2} \) units from the point \( (1, 2, 3) \).

2.
Find:

The shortest distance between the lines: \[ \vec{r}=(4+\lambda)\hat{i}+(2\lambda-1)\hat{j}-3\lambda\hat{k} \] and \[ \vec{r}=(1+2\mu)\hat{i}+(4\mu-1)\hat{j}+(2-5\mu)\hat{k} \]

3.
Find:

The principal value of \[ \sec^{-1}(\sqrt{2})+2\csc^{-1}(-2) \] is:

6.
Find:

If the function \[ f(x)= \begin{cases} \frac{\sin x}{x}+\cos x, & x\neq0\\ k, & x=0 \end{cases} \] is continuous at \(x=0\), then find the value of \(k\).