Distance of a Point From a Line: Definition and Examples

In this article, we will cover the concept of Distance of a point from a line. This category falls under the broader category of Coordinate Geometry, which is a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination(JEE Main) and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. A total of eleven questions have been asked on JEE MAINS( 2013 to 2023) from this topic including one in 2014, one in 2015, three in 2019, two in 2020, one in 2021, and two in 2022.

Distance of a Point From a Line

The distance between a point and a line is the distance between them. It measures the minimum distance or length required to move a point on the line. The shortest distance of a point from a line is the length of the perpendicular drawn from the point to the line.

Formula to Calculate the Distance of a Point From a Line

Perpendicular length from a point $\left(x_1, y_1\right)$ to the line $L: A x+B y+C=0$ is
$
\frac{\left|\mathrm{Ax}_1+\mathrm{By}_1+\mathrm{C}\right|}{\sqrt{\mathrm{A}^2+\mathrm{B}^2}}
$

The steps to derive the formula for finding the shortest distance between a point and line.

Step 1: Consider a line $L: A x+B y+C=0$ whose distance from the point $P$ $\left(\mathrm{x}_1, \mathrm{y}_1\right)$ is d .

Step 2: Draw a perpendicular PM from the point $P$ to the line $L$ as shown in the figure below.

Step 3: Let $Q$ and R be the points where the line meets the $x$-and $y$-axes, respectively.

Step 4: The coordinates of the points can be written as $Q(-C / A, 0)$ and $R(0,-$ C/B).

Derivation of distance of a point to the line

Let $\mathrm{L}: \mathrm{Ax}+\mathrm{By}+\mathrm{C}=0$ be a line, whose distance from the point $\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1\right)$ is
d. Draw a perpendicular $P M$ from the point $P$ to the line $L$

The line meets the x -and y -axes at the points Q and R , respectively. Then, the coordinates of the points are $Q\left(-\frac{C}{A}, 0\right)$ and $R\left(0,-\frac{C}{B}\right)$. Thus, the area of the triangle PQR is given by
$
\begin{aligned}
\operatorname{area}(\triangle \mathrm{PQR})=\frac{1}{2} \mathrm{PM} & \cdot \mathrm{QR} \Rightarrow \mathrm{PM}=\frac{2(\text { area } \Delta \mathrm{PQR})}{\mathrm{QR}} \\
\text { also. area }(\triangle \mathrm{PQR}) & =\frac{1}{2}\left|\mathrm{x}_1\left(0+\frac{\mathrm{C}}{\mathrm{B}}\right)+\left(-\frac{\mathrm{C}}{\mathrm{A}}\right)\left(-\frac{\mathrm{C}}{\mathrm{B}}-\mathrm{y}_1\right)+0\left(\mathrm{y}_1-0\right)\right| \\
& =\frac{1}{2}\left|\mathrm{x}_1 \frac{\mathrm{C}}{\mathrm{B}}+\mathrm{y}_1 \frac{\mathrm{C}}{\mathrm{A}}+\frac{\mathrm{C}^2}{\mathrm{AB}}\right|
\end{aligned}
$
also. area $(\triangle \mathrm{PQR})=\frac{1}{2}\left|\mathrm{x}_1\left(0+\frac{\mathrm{C}}{\mathrm{B}}\right)+\left(-\frac{\mathrm{C}}{\mathrm{A}}\right)\left(-\frac{\mathrm{C}}{\mathrm{B}}-\mathrm{y}_1\right)+0\left(\mathrm{y}_1-0\right)\right|$
or $2 \times \operatorname{area}(\triangle P Q R)=\left|\frac{C}{A B}\right|\left|A x_1+B y_1+C_1\right|$
$
\mathrm{QR}=\sqrt{\left(0+\frac{C}{A}\right)^2+\left(\frac{C}{B}-0\right)^2}=\left|\frac{C}{A B}\right| \sqrt{A^2+B^2}
$

Substituting the values
$
\mathrm{PM}=\frac{\left|\mathrm{Ax}_1++\mathrm{By}_1+\mathrm{C}\right|}{\sqrt{\mathrm{A}^2+\mathrm{B}^2}}
$

What is the distance between two parallel lines?

The equation of two parallel lines is $a x+b y+c=0$ and $a x+b y+d=0$, then the distance between them is the perpendicular distance of any point on one line from the other line.

Formula to calculate the distance between two parallel lines

If $\left(\mathrm{x}_1, \mathrm{y}_1\right)$ is any point on the line $a x+b y+c=0$
Then, $a x_1+b y_1+c=0$
Now, the perpendicular distance of the point $\left(x_1, y_1\right)$ from the line $a x+b y+d=0$ is $\frac{\left|\mathrm{ax}_1+\mathbf{b y}_1+\mathrm{d}\right|}{\sqrt{\mathbf{a}^2+\mathrm{b}^2}}=\frac{|\mathbf{d}-\mathrm{c}|}{\sqrt{\mathbf{a}^2+\mathbf{b}^2}}$

Recommended Video Based on the Distance of a Point from a Line

Solved Examples Based on the Distance of a point from a line

Example 1: Let R be the point $(3,7)$ and let P and Q be two points on the line $x+y=5$ such that PQR is an equilateral triangle. Then the area of $\triangle \mathrm{PQR}$ is:
[JEE MAINS 2022]

Solution

$
\begin{aligned}
& \sin 60^{\circ}=\frac{5 / \sqrt{2}}{\mathrm{a}} \\
& \mathrm{a}=\frac{5 \sqrt{2}}{3}
\end{aligned}
$

Area of $\triangle \mathrm{PQR}=\frac{\sqrt{3}}{4} \mathrm{a}^2=\frac{25}{2 \sqrt{3}}$
Hence the correct answer is $\frac{25}{2 \sqrt{3}}$

Example 2: Let a circle C of radius 5 lie below the x -axis. The line $\mathrm{L}_1: 4 \mathrm{x}+3 \mathrm{y}+2=0$ passes through the center P of the circle C and intersects the line $\mathrm{L}_2: 3 \mathrm{x}-4 \mathrm{y}-11=0$ at $Q$. The line $\mathrm{L}_2$ touches C at the point $Q$. Then the distance $P$ from the line $5 x-12 y+51=0$ is $\qquad$ [JEE MAINS 2022]

Solution: The point of intersection of $\mathrm{L}_1: 4 \mathrm{x}+3 \mathrm{y}+2=0$ and $\mathrm{L}_2: 3 \mathrm{x}-4 \mathrm{y}-11=0$ is
$
\mathrm{Q}(1,-2)
$

The Centre P of the circle lies on $\mathrm{L}_1$
Slope of $\mathrm{L}_1: \tan \theta=-\frac{4}{3} \Rightarrow \cos \theta: \frac{-3}{5}, \sin \theta=\frac{4}{5}$
Coordinates of $\mathrm{P}:(1-5 \cos \theta,-2-5 \sin \theta)$
$
=(4,-6)
$
distance of $\mathrm{P}(4,-6)$ from $5 \mathrm{x}-12 \mathrm{y}+51$ is
$
\mathrm{d}=\left|\frac{5 \times 4-12 \times(-6)+51}{\sqrt{5^2+122}}\right|=\frac{20+72+51}{13}=\frac{143}{13}=11
$

Hence, the answer is 11

Example 3: If p and q are the lengths of the perpendiculars from the origin on the lines, $x \operatorname{cosec} \alpha-y \sec \alpha=k \cot 2 \alpha$ and $x \sin \alpha+y \cos \alpha=k \sin 2 \alpha$ respectively, then $k^2$ is equal to :
[JEE MAINS 2021]

Solution:

$
\begin{aligned}
& p=\frac{|k \cot 2 \alpha|}{\sqrt{\operatorname{cosec}^2 \alpha+\sec ^2 \alpha}} \\
& \Rightarrow p^2=\frac{k^2 \frac{\cos ^2 2 \alpha}{\sin ^2 2 \alpha}}{\frac{\sin ^2 \alpha+\cos ^2 \alpha}{\left(\sin ^2 \alpha \cos ^2 \alpha\right)^2}}=\frac{k^2 \cos ^2 2 \alpha}{4 \sin ^2 \alpha \cos ^2 \alpha} \times \sin ^2 \alpha \cos ^2 \alpha \\
& \Rightarrow p^2=\frac{k^2}{4} \cos ^2 2 \alpha \\
& \Rightarrow \cos ^2 2 \alpha=\frac{4 p^2}{k^2}---(1) \\
& q=\left|\frac{k \sin 2 \alpha}{\sin ^2 \alpha+\cos ^2 \alpha}\right| \\
& \Rightarrow q=k^2 \sin ^2 2 \alpha \Rightarrow \sin ^2 2 \alpha=\frac{q^2}{k^2}---(2) \\
& (1)+(2) \Rightarrow \frac{4 p^2}{k^2}+\frac{q^2}{k^2}=1 \Rightarrow k^2=4 p^2+q^2
\end{aligned}
$

Hence, the answer is $4 p^2+q^2$

Example 4: The length of the perpendicular from the origin, on the normal to the curve, $x^2+2 x y-3 y^2=0$ at the point $(2,2)$ is :
[JEE MAINS 2020]

Solution: Perpendicular length from a point $(x 1, y 1)$ to the line $L$ : $A x+B y+C=0$ is
$
\begin{aligned}
& \frac{\left|\mathbf{A} \mathbf{x}_1++\mathbf{B} \mathbf{y}_1+\mathbf{C}\right|}{\sqrt{\mathbf{A}^2+\mathbf{B}^2}} \\
& x^2+2 x y-3 y^2=0 \\
& x^2+3 x y-x y-3 y^2=0 \\
& (x-y)(x+3 y)=0 \\
& x-y=0 \quad x+3 y=0
\end{aligned}
$
$(2,2)$ satisfy $x-y=0$
Normal
$
x+y=\lambda=4
$

Hence the perpendicular distance from the origin
$
=\left|\frac{0+0-4}{\sqrt{2}}\right|=2 \sqrt{2}
$

Example 5: If a variable line, $3 x+4 y-\lambda=0$ is such that the two circles $x^2+y^2-2 x-2 y+1=0$ and $x^2+y^2-18 x-2 y+78=0$ are on opposite sides, then the set of all values of $\lambda$ is the interval

Solution: Given $3 x+4 y-\lambda=0$
$(7-\lambda)(31-\lambda)<0 \quad$ (Since centers are on opposite sides)
$=>\lambda \epsilon(7,31)$ $\qquad$
$\left|\frac{7-\lambda}{5}\right| \geq 1$ and $\left|\frac{7-\lambda}{5}\right| \geq 2$
$
\begin{aligned}
& |7-\lambda| \geq 5 \text { and }|31-\lambda| \geq 10 \\
& =>\lambda \leq 2 \text { or } \lambda \geq 12
\end{aligned}
$
and
$
=>\lambda \leq 21 \text { or } \lambda \geq 41
$
$\qquad$
(1) $\cap(2) \cap(3)$
$
\lambda \in[12,21]
$

Hence, the answer is $[12,21]$.
Summary

The distance of a point from the line in coordinate geometry is a measure of the shortest distance between a point and a line. We can derive the formula by simply using the concept of the area of the triangle. It is also used in various geometrical figures. Understanding the distance between a line and a point helps us to solve various complex problems.