Distance of a Point From a Line: Definition and Examples

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 $[12,21]$.