Distance of a Point From a Line - Definition and Examples

What is the Distance of a Point From a Line?

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

Perpendicular Distance of a Point From a Line

The shortest distance of a point from a line is the length of the perpendicular drawn from the point to the line.

Let $L$ be $a$ line in the plane and let $M$ be any point not on the line. Then, we define distance $d$ from M to L as the length of the line segment $\overrightarrow{M P}$, where P is a point on L such that $\overrightarrow{M P}$ is perpendicular to L .

Derivation of Distance of a Point From a Line

The equation of line L is $\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}$
Let $P$ be the foot of the perpendicular drawn from the point $M(\alpha, \beta, \gamma)$ on the line $L$.
Let the coordinates of P be $\left(x_0+a \lambda, y_0+b \lambda, z_0+c \lambda\right)$
Then the direction ratio of MP are $\left(x_0+a \lambda-\alpha, y_0+b \lambda-\beta, z_0+c \lambda-\gamma\right)$.
The direction ratio of line $L$ is $(a, b, c)$

Since MP is perpendicular to line L ,

$
\begin{aligned}
& a\left(x_0+a \lambda-\alpha\right)+b\left(y_0+b \lambda-\beta\right)+c\left(z_0+c \lambda-\gamma\right)=0 \\
& \lambda=\frac{a\left(\alpha-x_0\right)+b\left(\beta-y_0\right)+c\left(\gamma-z_0\right)}{a^2+b^2+c^2}
\end{aligned}
$
Put the value of $\lambda$ in $\left(x_0+a \lambda, y_0+b \lambda, z_0+c \lambda\right)$.

we get the foot of the perpendicular. Now, we can get distance MP using the distance formula.

Steps To Calculate Distance between the Point And the Line

Consider the equation of line L is $\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}$

Step 1: Consider a point $\mathrm{P}\left(x_0+a \lambda, y_0+b \lambda, z_0+c \lambda\right)$ on line L such that P is the foot of the perpendicular drawn from the point $M(\alpha, \beta, \gamma)$ on line $L$.

Step 2: Find the direction ratio of the $MP$
Step 3: Since MP is perpendicular to line $L$, Find the value of $\lambda$
Step 4: Find the coordinates Of $P$
Step 5: Using the distance formula calculate the distance between a point and the line

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 a line $L$ pass through the point $P(2,3,1)$ and be parallel to the line $x+3 y-2 z-2=0=x-y+2 z$. If the distance of $L$ from the point $(5,3,8)$ is $\alpha$, then $3 \alpha^2$ is equal to $\qquad$
[JEE MAINS 2023]

Solution:
The Direction ratio of the line

$
\begin{aligned}
\left|\begin{array}{ccc}
\mathrm{i} & \mathrm{j} & \mathrm{k} \\
1 & 3 & -2 \\
1 & -1 & 2
\end{array}\right| & =\mathrm{i}(6-2)-\mathrm{j}(2+2)+\mathrm{k}(-1-3) \\
& =4 \mathrm{i}-4 \mathrm{j}-4 \mathrm{k}
\end{aligned}
$
Equation of line $L$

$\begin{aligned} & \frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}-3}{-1}=\frac{\mathrm{z}-1}{-1}=\lambda(\alpha \mathrm{d}) \\ & \text { Let } \mathrm{M}(\lambda+2,-\lambda+3,-\lambda+1) \\ & \text { DR's of MQ is }<\lambda+2-5,-\lambda+3-3,-\lambda+1-8> \\ & <\lambda-3,-\lambda \cdot-\lambda>7> \\ & \because \mathrm{L} \perp \mathrm{MQ} \\ & \Rightarrow(\lambda-3)(1)+(-\lambda)(-1)+(-\lambda-7)(-1)=0 \\ & \Rightarrow \lambda-3+\lambda+\lambda+7=0 \\ & \Rightarrow 3 \lambda=-4 \Rightarrow \lambda=-\frac{4}{3} \\ & \therefore \mathrm{M}\left(-\frac{4}{3}+2, \frac{+4}{3}+3, \frac{4}{3}+1\right)=\left(\frac{2}{3}, \frac{13}{3}, \frac{7}{3}\right) \\ & \mathrm{MQ}=\alpha \\ & \therefore 3 \alpha^2=3 \times\left(\left(5-\frac{2}{3}\right)^2+\left(3-\frac{13}{3}\right)^2+\left(8-\frac{7}{5}\right)^2\right) \\ & =3\left(\frac{169}{9}+\frac{16}{9}+\frac{289}{9}\right)^2 \Rightarrow \frac{474}{9}=158\end{aligned}$

Hence, the answer is 158

Example 2: Let $y(x)=(1+x)\left(1+x^2\right)\left(1+x^4\right)\left(1+x^8\right)\left(1+x^{16}\right)$. Then $y^{\prime}-y^{\prime \prime}$ at $x=-1$ is equal to
[JEE MAINS 2023]

Solution
$
\begin{aligned}
& f(x)=y=\frac{(1-x)(1+x)\left(1+x^2\right)\left(1+x^4\right)\left(1+x^8\right)\left(1+x^{16}\right)}{(1-x)} \\
& f(x)=y=\frac{\left(1-x^{32}\right)}{1-x} \Rightarrow f(-1)=0 \\
& (1-x) y=1-x^{32}
\end{aligned}
$

differentiate both sides

$
(1-x) y^{\prime}+y(-1)=-32 x^{31} \quad x=-1 \Rightarrow y^{\prime}=16
$

differentiate both sides

$
\begin{aligned}
& (1-x) y^{\prime}+y^{\prime}(-1)-y^{\prime}=-(32)(31) x 30 \\
& \text { Put } x=-1 \\
& 2 y^{\prime \prime}-2 y^{\prime}=-(32)(31) \\
& y^{\prime \prime}-y^{\prime}=-(16)(31) \\
& y^{\prime}-y^{\prime \prime}=496
\end{aligned}
$

Hence, the answer is 496

Example 3: Let Q and R be two points on the line $\frac{\mathrm{x}+1}{2}=\frac{\mathrm{y}+2}{3}=\frac{\mathrm{z}-1}{2}$ at a distance $\sqrt{26}$ from the point $P(4,2,7)$. Then the square of the area of the triangle $P Q R$ is
[JEE MAINS 2022]

Solution:
Let $(2 \lambda-1,3 \lambda-2,2 \lambda+1)$ be any point on the line.
$
\begin{aligned}
& \Rightarrow(2 \lambda-5)^2+(3 \lambda-4)^2+(2 \lambda-6)^2=26 \\
& \Rightarrow \lambda=1,3 \\
& \mathrm{Q}(1,1,3) ; \mathrm{R}(5,7,7) ; \mathrm{P}(4,2,7)
\end{aligned}
$
Area of triangle $\mathrm{PQR}=\frac{1}{2}|\mathrm{P} \vec{Q} \times \mathrm{P} \vec{R}|$

$
=\sqrt{153}
$

Hence, the answer is 153

Example 4: If two distinct points $\mathrm{Q}, \mathrm{R}$ lie on the line of intersection of the planes $-x+2 y-z=0$ and $3 x-5 y+2 z=0$ and $P Q=P R=\sqrt{18}$ where the point $P$ is $(1,-2,3)$, then the area of the triangle $PQR$ is equal to
[JEE MAINS 2022]

Solution: Let $\mathrm{z}=\mathrm{k}$

$
\begin{aligned}
& -x+2 y=k---(1) \\
& 3 x-5 y=-2 k---(2) \\
& \frac{(1) \times 3 \Rightarrow-3 \mathrm{x}+6 \mathrm{y}=3 \mathrm{k}---(3)}{(2)+(3) \Rightarrow \mathrm{y}=\mathrm{k} ; \quad \mathrm{x}=\mathrm{k}}
\end{aligned}
$

Also, both lines pass through $(0,0,0)$ So their line of intersection is

$
\frac{\mathrm{x}}{1}=\frac{\mathrm{y}}{1}=\frac{\mathrm{z}}{1}=\mathrm{k}
$
Let the Foot of Perpendicular from P to Line QR is

$
\begin{aligned}
& \mathrm{S}(\mathrm{k}, \mathrm{k}, \mathrm{k}) . \quad \mathrm{P}(1,-2,3) \\
& \overrightarrow{P S}=(\mathrm{k}-1, \mathrm{k}+2, \mathrm{k}-3) . \quad \overrightarrow{P S} \cdot \vec{b}=0 \\
& \Rightarrow \mathrm{k}-1+\mathrm{k}+2+\mathrm{k}-3=0 \Rightarrow \mathrm{k}=2 / 3 \\
& \mathrm{~S}\left(\frac{2}{3}, \frac{2}{3}, \frac{2}{3}\right), \quad|\overrightarrow{P S}|=\sqrt{\frac{1}{9}+\frac{64}{9}+\frac{49}{9}}=\frac{\sqrt{114}}{3} \\
& \mathrm{RS}=\mathrm{QS}=\sqrt{18-\frac{114}{9}}=\sqrt{\frac{48}{9}}=\frac{4}{\sqrt{3}}
\end{aligned}
$
Area of $\triangle \mathrm{PQR}=2 \times \frac{1}{2} \times \mathrm{RS} \times \mathrm{PS}=\frac{\sqrt{114}}{3} \times \frac{4}{\sqrt{3}}=\frac{4}{3} \sqrt{38}$

Hence, the answer is $\frac{4}{3} \sqrt{38}$

Example 5: If the length of the perpendicular from the point $(\beta, 0, \beta),(\beta \neq 0)$ to the line, $\frac{x}{1}=\frac{y-1}{0}=\frac{z+1}{-1}$ is $\sqrt{\frac{3}{2}}$,then $\beta$ is equal to
[JEE MAINS 2019]

Solution:
Let point $P(\beta, 0, \beta)$ given that length of the perpendicular distance from $P$ to the line is $\sqrt{\frac{3}{2}}$.

$\frac{x}{1}=\frac{y-1}{0}=\frac{z+1}{-1}$

$
R=(\lambda, 1,-\lambda-1)
$
Direction ratio of $P R=(\lambda-\beta, 1,-\lambda-\beta-1)$
PR is perpendicular to the line

$
\begin{aligned}
& \Rightarrow>(\lambda-\beta)(1)+(1) 0+(-1)(-\lambda-\beta-1)=0 \\
& =>\lambda-\beta+\lambda+\beta+1=0 \\
& =>\lambda=\frac{1}{2} \\
& P R=\sqrt{(\lambda-\beta)^2+1^2+(-\lambda-1-\beta)^2}=\sqrt{\frac{3}{2}} \\
& \quad=>2 \beta^2+2 \beta=0 \\
& \quad=>\beta=0, \beta=-1
\end{aligned}
$

Hence, the answer is -1