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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.
Read More: NCERT Solutions For Class 11 Maths Straight Lines
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Key Terms: Distance Formula, Perpendicular Distance, Triangle, Line Segment, Point, Geometry, Hypotenuse, Perpendicular Line
What is Distance of a Point From a Line?
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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
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
- 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
Read More: Straight Lines Important Questions
Derivation of Distance of a Point From a Line
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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
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) |
Read Also: Different Forms of the Equation of Line
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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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.
Read More: Perpendicular Distance of a Point from a Plane Formula
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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