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# Perpendicular Distance of a Point From a Line

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Last updated date: 12th Aug 2024
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## What's the Perpendicular Distance of a Point from a Line?

The shortest distance between a point and a line in geometry is the distance of a point from a line. There are an endless number of lines that can be drawn in a plane from one point to another. Drawing a perpendicular line segment on the line passing through the specified location is the only way to get the shortest distance. This is so that we can create a triangle by connecting the point and line with more than one line. Knowing the distance from a point to a line can be useful in various real-life situations-for example, to find the distance between two objects like two trees. In this article, we will study how to find the length of a perpendicular from a point to a line and solved examples.

## Perpendicular Distance

The distance between two objects measured along a line perpendicular to one or both is referred to as the perpendicular distance in geometry. In two- dimensional space, the distance between two points and two lines is the perpendicular distance. For the perpendicular distance from the origin to a plane in three dimensions, the point on the plane closest to the origin.

Distance between a point and a plane in three dimensions, for the perpendicular distance between any two points.

## Perpendicular Distance of a Point from a Line

The length of the perpendicular drawn from a point to a line represents the shortest separation between the two. The stages to get at the formula for calculating the perpendicular distance of a point from a line are listed below.

Step 1: In the first step, think about a line $L: A x+B y+C=0$ that is $d$ distance from the point $P\left(x_{1}, y_{1}\right)$

Step 2: As illustrated in the diagram below, draw a perpendicular PM from point $P$ to line L.

Step 3: Assume that $Q$ and $R$ are the points where the line intersects the $x, y$ - axis respectively.

Step 4: The point coordinates can be expressed as

$Q\left(\dfrac{-C}{A}, 0\right), R\left(0, \dfrac{-C}{B}\right)$

Point Coordinates

## Perpendicular Distance Formula

We have the line Ax+By-C-0 i.e. DE and has a slope $\dfrac{-A}{B}$

Perpendicular Distance

Now, we have the point $\mathrm{P}$ as $\left(H^{\prime} H\right)$. So, we have to find a perpendicular distance from $\mathrm{P}$ to DE.

So, construct a line that is parallel to $D E$ and has a slope $\dfrac{-A}{B}$ as it is parallel to DE and then name it FG.

Now, we have to construct line $\mathrm{PQ}$ which passes through the origin and this line is parallel to slope $\dfrac{B}{A}$ as it is perpendicular to DE named as RS then extend it at $(0,0)$.

Slope

As FG has slope $\dfrac{-A}{B}$ and equation we have

$y-n=-\dfrac{A}{B}(x-m)$

$y=\dfrac{-A x+A m+B n}{B}$

We have

$y=\dfrac{B}{A} x$

$\dfrac{B}{A} x=\dfrac{-A x+A m+B n}{B}$

Prove that point $\mathrm{R}$ is $\left(\dfrac{A(A m+B n)}{A^{2}+B^{2}}, \dfrac{B(A m t+B n t)}{A^{2}+B^{2}}\right)$

$y=\dfrac{B}{A} x, A x+B y+C=0$

Implies $y=\dfrac{A x+C}{B}$

$=\dfrac{A x+C}{B}-\dfrac{B}{s} x$

$x=\dfrac{-A C}{A^{2}+B י}$

$y=\dfrac{B}{A} x$

$y=\dfrac{B}{A}\left(\dfrac{A C}{A^{2}+B^{2}}\right)=\dfrac{B C}{A^{2}+B^{2}}$

So, $\mathrm{S}$ is $\left(\dfrac{-A C}{A^{2}+B^{2}}, \dfrac{-B C}{A^{2}+B^{2}}\right)$

$d=\sqrt{\left(\dfrac{-A C}{A^{2}+B^{2}}-\dfrac{A(A m+B n)}{A^{2}+B^{2}}\right)^{2}+\left(\dfrac{-B C}{A^{2}+B^{2}}-\dfrac{B(A m+B n)}{A^{2}+B^{2}}\right)}$

Thus, distance RS is

$=\sqrt{\dfrac{\{-A(A m+B n+C)\}^{2}+\{-B(A m+B n+C)\}^{2}}{\left(A^{2}+B^{2}\right)^{2}}}$

$=\sqrt{\dfrac{\left(A^{2}+B^{2}\right)(A m-B n+C)^{2}}{\left(A^{2}+B^{2}\right)^{2}}}$

$=\sqrt{\dfrac{(A m+B n+C)^{2}}{A^{2}+B^{2}}}$

$=\dfrac{|A m+B n+C|}{\sqrt{A^{2}+B^{2}}}$

Thus, perpendicular distance formula is $d=\dfrac{|A m+B n+C|}{\sqrt{A^{2}+B^{2}}}$

If the perpendicular distance of point P is 5 units away from the axis perpendicularly and the foot of the perpendicular is in the opposite direction of the x-axis then point P has an ordinate which is 5 or -5. As y- coordinate of a point is equal to its perpendicular distance from the x-axis. The perpendicular distance of point P lies on the X-negative axis direction.

The point P therefore has a y-coordinate of 5 or -5.

## Length of the Perpendicular from a Point to a Line

Step 1: We have a line $L: A x+B y+C=0$ that is ' $\mathrm{d}$ ' distance from the point $P\left(x_{1}, y_{1}\right)$

Step 2: As illustrated in the diagram below, draw a perpendicular PM from point $P$ to line L.

Step 3: Assume that Q and R are the points where the line intersects the x-axis and y-axis respectively i.e. the length of the perpendicular from a point to a line.

Length of the Perpendicular from a Point to a Line

## Solved Examples

Q 1. Locate the distance between the point (-3, 5) and the line 4x - 3y - 26 = 0.

Ans: We have 4x-3y-26 = 0 at (-3, 5)

Implies A = 4, B = -3, C = -26, x = -3, y = 5

$d=\dfrac{|A x+B y+C|}{\sqrt{A^{2}+B^{2}}}$

$d=\dfrac{|4(-3)+(-3) 5+(-26)|}{\sqrt{4^{2}+(-3)^{2}}}=\dfrac{|-53|}{\sqrt{25}}=\dfrac{53}{5}$

Q 2. A line y = 3x + 1 and a point (5,1) must be separated by this distance.

Ans: We have y = 3x+1 at (5, 1)

Implies A = 3, B = -1, C = 1, x = 5, y = 1

$d=\dfrac{|A x+B y+C|}{\sqrt{A^{2}+B^{2}}}$

$d=\dfrac{|3.5+(-1) 1+1|}{\sqrt{3^{2}+(-1)^{2}}}=\dfrac{15}{\sqrt{10}}$

## Practice Questions

Q 1. Determine the length of the perpendicular that connects the point (3,2,1) to the line.

Ans: $\dfrac{x-7}{-2}=\dfrac{y-7}{2}=\dfrac{z-6}{3}$

Q 2. Using the formula for the distance of a point from a line, determine the distance between the point K(3,7) and the line PQ.

Ans: $y=\dfrac{6}{5} x+2$

Q 3. Find the perpendicular distance of the point (−1,1) from the line 12(x+6)=5(y−2).

Ans: 5 units

## Summary

As we know, the perpendicular distance between a point and a line is the shortest distance. Let that distance be ' $\mathrm{d}$ ' between $P\left(x_{1}, y_{1}\right)$ and line $L: a x+b y+c=0$ as $D=\dfrac{\left|a x_{1}+b y_{1}+c\right|}{\sqrt{a^{2}+b^{2}}}$

By calculating the perpendicular distance between any point on the line and the other, we may determine how far apart two parallel lines are from one another.

Competitive Exams after 12th Science

## FAQs on Perpendicular Distance of a Point From a Line

1. What type of lines is perpendicular?

Perpendicular lines are those that cross at a straight angle to one another. The symbol used to indicate two parallel lines is | |.

If a figure has an "L" form, the vertex's angle corresponds to a right angle. Perpendicular lines have two key characteristics. In every case, perpendicular lines cross or intersect.

2. What does the graphing term "perpendicular" mean?

Lines that cross each other at a straight angle are said to be perpendicular. The result of multiplying the slopes of two perpendicular lines in plane lines is minus one. In other words, perpendicular lines' slopes are their opposing reciprocals. When two lines cross at a straight angle or 90 degrees, perpendicular lines are created. Perpendicularity is the name given to this characteristic of lines.

3. List the Properties of Perpendicular Lines.

The Properties of Perpendicular Lines are:

• They cross each other at right angles.

• All crossing lines are not necessarily perpendicular to one another, but perpendicular lines always cross each other.

• In every case, perpendicular lines cross or intersect.

• The result of multiplying the slopes of two perpendicular lines in the plane is minus one.