Image of a Point in the given Line

A flat, two-dimensional surface, which extends infinitely is a plane. The image of a point in the given line is the reflection of the point over the given line. We use the image of the point to find the reflection of the point which makes our calculations easy.

In this article, we will cover the concept of the Image of a Point in the given Line. This topic falls under the broader category of three-dimensional geometry, which is a crucial chapter in Class 12 Mathematics. This is very important not only for board exams but also for competitive exams, which even include the Joint Entrance Examination Main and other entrance exams: SRM Joint Engineering Entrance, BITSAT, WBJEE, and BCECE. A total of six questions have been asked on this topic in JEE Main from 2013 to 2023 including one in 2020, two in 2021, two in 2022, and one in 2023.

Image of a Point in the given Line

The image of a point in the given Line is the reflection of the point over the given line.

Since $P$ (foot of perpendicular) is the midpoint of $M$ and $N$ (image of a point $M$ in the line), we can get $N$ if $P$ is found out.

Cartesian Form of Image of a Point in the given Line

Let $\mathrm{M}(\alpha, \beta, \gamma)$ be the point and $\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}$ be the equation of line $L$.

Let $P$ be the foot perpendicular from $M$ to line $L$ and let $N$ be the image of the point in the given line, where MP $=P N$

Let the coordinates of $P$ be

$
\left(x_0+a \lambda, y_0+b \lambda, z_0+c \lambda\right)
$
Then, direction ratios of $P L$ are

$
\left(x_0+a \lambda-\alpha, y_0+b \lambda-\beta, z_0+c \lambda-\gamma\right)
$
since, $MP$ is perpendicular to the given line, whose direction ratios are $a, b,$ and $c$ .

$
\begin{aligned}
& \therefore \quad a \cdot\left(x_0+a \lambda-\alpha\right)+b \cdot\left(y_0+b \lambda-\beta\right)+c \cdot\left(z_0+c \lambda-\cdots\right. \\
& \Rightarrow \quad \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}
$

Substituting the value of $\lambda$ we get coordinates of point $P$ (foot of perpendicular)

As $N\left(\alpha^{\prime}, \beta^{\prime}, \gamma^{\prime}\right)$ is image of point $M(\alpha, \beta, \gamma)$
$\therefore$ mid-point of MN is point P

$
\begin{array}{ll}
\therefore & \frac{\alpha+\alpha^{\prime}}{2}=x_0+a \lambda, \quad \frac{\beta+\beta^{\prime}}{2}=y_0+b \lambda, \quad \frac{\gamma+\gamma^{\prime}}{2}=z_0 \\
\therefore & \alpha^{\prime}=2\left(x_0+a \lambda\right)-\alpha, \beta^{\prime}=2\left(y_0+b \lambda\right)-\beta, \gamma^{\prime}=2(
\end{array}
$

Steps to Find an Image of a Point in a Line

Consider the 2 points $M$ and $N$. Let $L$ be a line such that
- There exists a perpendicular line $MN$ to the line $L$ .
- The midpoint of $M N$ is on line $L$ ( $P$ is the midpoint of $M N$ ). Then, the image of the point is either of the points to one another in Line $L$ .

The procedure to find the image of a point in a given plane is as follows:

Step 1: Let the equation of Line $L$ be $\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}$
Step 2: Assume $N\left(\alpha^{\prime}, \beta^{\prime}, \gamma^{\prime}\right)$ is image of point $M(\alpha, \beta, \gamma)$
Step 3: The coordinates of mid point of line MN which is P can be found. Let the coordinates of $P$ be

$
\left(x_0+a \lambda, y_0+b \lambda, z_0+c \lambda\right)$

Then, direction ratios of $P L$ are

$
\left(x_0+a \lambda-\alpha, y_0+b \lambda-\beta, z_0+c \lambda-\gamma\right)
$

Step 4: Line MP is Perpendicular to Line $L$ so we find the value of $\lambda$

Step 5: Substituting the value of $\lambda$ we get coordinates of point $P$ (foot of perpendicular)

Step 6: Find the coordinates of Image $N$

Solved Examples Based on the Image of a Point in the Given Line

Example 1: Let the image of the point $P(1,2,3)$ in the plane $2 x-y+z=9$ be $Q$.If the coordinates of the point $R$ are $(6,10,7)$. then the square of the area of the triangle $P Q R$ is $\qquad$ [JEE MAINS 2023]

Solution:
Let $\mathrm{Q}(\alpha, \beta, \gamma)$ be the image of P , about the plane

$
\begin{aligned}
& 2 \mathrm{x}-\mathrm{y}+\mathrm{z}=9 \\
& \frac{\alpha-1}{2}=\frac{\beta-2}{-1}=\frac{\gamma-3}{1}=2 \\
& \Rightarrow \alpha=5, \beta=0, \gamma=5
\end{aligned}
$

Then area of triangle PQR is $=\frac{1}{2}|\overrightarrow{\mathrm{PQ}} \times \overrightarrow{\mathrm{PR}}|$

$
=|-12 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+21 \hat{\mathrm{k}}|=\sqrt{144+9+441}=\sqrt{594}
$

Square of area $=594$
Hence, the answer is the 594

Example 2: Let the image of the point $\mathrm{P}(1,2,3)$ in the line $\mathrm{L}: \frac{x-6}{3}=\frac{y-1}{2}=\frac{z-2}{3}$ be $\mathrm{Q}_{\text {. }}$.et $\mathrm{R}(\alpha, \beta, \gamma)$ be a point that divides internally the line segment $\mathrm{PQ}_{\text {in the }}$ ratio $1: 3$. Then the value of $22(\alpha+\beta+\gamma)$ is equal to $\qquad$
[JEE MAINS 2022]

Solution

Let $M(3 a+6,2 a+1,3 a+2)$ be the mid point of $P Q$.

$
\begin{aligned}
& \therefore \mathrm{PM} \perp \mathrm{L} \\
& \Rightarrow \overrightarrow{P M} \cdot \vec{b}=0 \\
& \Rightarrow((3 \mathrm{a}+5) \mathrm{i}+(2 \mathrm{a}-1) \mathrm{j}+(3 \mathrm{a}-1) \mathrm{k}) \cdot(3 \mathrm{i}+2 \mathrm{j}+3 \mathrm{k})=0 \\
& \Rightarrow \mathrm{a}=-\frac{5}{11} \\
& \therefore \mathrm{M} \text { is }\left(\frac{51}{11}, \frac{1}{11}, \frac{7}{11}\right)
\end{aligned}
$

As $R$ divides $P Q$ in a ratio of $1: 3$, hence $R$ is the mid-point of $P M$

$
\begin{aligned}
& \mathrm{R}=\left(\frac{62}{22}, \frac{23}{22}, \frac{40}{22}\right) \\
& \therefore 22(\alpha+\beta+\gamma)=62+23+40=125
\end{aligned}
$

Hence, the answer is 125

Example 3: Let $a, b \in R$ If the mirror image of the point $P(a, 6,9)$ with respect to the line $\frac{x-3}{7}=\frac{y-2}{5}=\frac{z-1}{-9}$ is $(20, b,-a-9)$, then $|a+b|$ is equal to:
[JEE MAINS 2021]

Solution

$
\begin{aligned}
& P(a, 6,9) \\
& \frac{x-3}{7}=\frac{y-2}{5}=\frac{z-1}{-9} \\
& Q=(20, b,-a-9)
\end{aligned}
$
Here $Q$ is the mirror image of Point $P$.
Therefore midpoint of $P$ and $Q$ lies on the given line.

Midpoint of PQ is $\frac{20+a}{2}, \frac{b+6}{2}=-\frac{a}{2}$

$
\begin{aligned}
& \frac{\frac{20+a}{2}-3}{7}=\frac{\frac{b+6}{2}-2}{5}=\frac{-\frac{a}{2}-1}{-9} \\
& \frac{14+a}{14}=\frac{b+2}{10}=\frac{a+2}{18} \\
& \Rightarrow a=-56 \text { and } b=-32 \\
& \Rightarrow \quad l a+b l=88
\end{aligned}
$

Hence, the answer is 88

Example 4: If the equation of the plane passing through the mirror image of a point $(2,3,1)_{\text {with respect to the line }} \frac{x+1}{2}=\frac{y-3}{1}=\frac{z+2}{-1}$ and containing the line $\frac{x-2}{3}=\frac{1-y}{2}=\frac{z+1}{1}$ is $\alpha x+\beta y+\gamma z=24$, then $\alpha+\beta+\gamma$ is equal to :

[JEE MAINS 2021]

Solution: Let point A be $(2,3,1)$

$
L_1: \frac{x+1}{2}=\frac{y-3}{1}=\frac{z+2}{-1}=\lambda
$

let any point on line $\mathrm{L}_1$ is $\mathrm{B}(2 \lambda-1, \lambda+3,-\lambda-2)$
Now if $B$ is foot of perpendicular of $A$ in $L_1$, then $A B \perp L_1$

$
2(2 \lambda-3)+1(\lambda)-(-\lambda-3)=0
$

$6 \lambda-3=0 \Rightarrow \lambda=\frac{1}{2}$
Hence $B\left(0, \frac{7}{2},-\frac{5}{2}\right)$
Now image $A^{\prime}(-2,4,-6)$

Now equation of plane containing A'(-2,4,-6) and line

$\mathrm{L}_2: \frac{\mathrm{x}-2}{3}=\frac{\mathrm{y}-1}{-2}=\frac{\mathrm{z}+1}{1}$ is $\left|\begin{array}{ccc}x-2 & y-1 & z+1 \\ 3 & -2 & 1 \\ 4 & -3 & 5\end{array}\right|=0$
$\Rightarrow 7 x+11 y+z=24$
Hence $\alpha=7, \beta=11, y=1$

Hence, the answer is 19

Example 5 : The image of the point $\hat{i}+\hat{j}+\hat{k}$ to the line $r=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}+4 \hat{k})$ will have a position vector

Solution:
$\therefore$ The foot of perpendicular $=$

$
\begin{aligned}
& \qquad \vec{a}-\left(\frac{(\vec{a}-\vec{\alpha}) \cdot \vec{b})}{\left|\vec{b}^2\right|}\right) \vec{b} \\
& \therefore \text { The foot of perpendicular }= \\
& \vec{a}-\vec{\alpha}=\hat{j}+2 \hat{k} \Rightarrow(\vec{a}-\vec{\alpha}) \cdot \vec{b}=11
\end{aligned}
$

$\therefore$ Foot of perpendicular

$
\begin{aligned}
& \therefore \text { Foot of perpendicular }=(\hat{i}+2 \hat{j}+3 \hat{k})-\frac{11(2 \hat{i}+3 \hat{j}+4 \hat{k})}{29} \\
& =\frac{1}{29}(7 \hat{i}+25 \hat{j}+43 \hat{k})
\end{aligned}
$
Hence the image of the point is

$
\begin{aligned}
& =\frac{2}{29}(7 \hat{i}+25 \hat{j}+43 \hat{k})-(\hat{i}+2 \hat{j}+3 \hat{k}) \\
& =\frac{1}{29}(-15 \hat{i}-4 \hat{j}+\hat{k})
\end{aligned}
$

Hence, the answer is $\frac{1}{29}(-15 \hat{i}-4 \hat{j}+\hat{k})$

Summary

The image of a point in a given line is determined by reflecting the point across the line. If the point lies on the line, its image remains unchanged. If the point is above or below the line, its image is found by reflecting it across the line, resulting in a transformation of its coordinates relative to the line. This reflection process is fundamental in geometry and helps determine the relationship between points and lines in a coordinate plane.