A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the reflected ray passes

Miscellaneous Exercise · Q21

(1, 2)

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Concept understanding — Image of Point in Line

Image of a Point in a Line

The idea in plain words

Imagine you're standing in front of a long, straight mirror on the floor. You see your reflection on the other side. Two things are true: the mirror is exactly halfway between you and your reflection, and the line from you to your reflection hits the mirror at a perfect right angle. That's exactly what we mean by the image of a point in a line — the line acts like a mirror, and we're finding where the reflection lands.

The key insight: The line is the perpendicular bisector of the segment joining a point and its image. If you fold the plane along the line, the original point lands right on top of its image.

What it means, precisely

Given a line $L$ (your "mirror") and a point $P$ (the "object"), the image of $P$ in $L$ is the point $P'$ such that:

$L$ is the perpendicular bisector of segment $PP'$ — that means $L$ passes through the midpoint of $PP'$ , and $L$ is perpendicular to $PP'$ .
$P$ and $P'$ lie on opposite sides of $L$ (unless $P$ is on $L$ itself, in which case $P' = P$ — your reflection is you).

Why this works

Think of it this way: the mirror doesn't favour one side over the other. So the distance from you to the mirror must equal the distance from your reflection to the mirror. And because light bounces off a mirror at equal angles, the path from you to the mirror to your eye is symmetric — that symmetry forces the perpendicular condition.

So when you're asked to find the image of a point in a line, you're really being asked: "If I reflect this point across this line, where does it land?"

Step by step — how to find it

Here's the method you'll use in problems:

Step 1: Find the foot of the perpendicular from $P$ to $L$ . Call it $M$ . This is the point on $L$ closest to $P$ — the spot where your perpendicular line would touch the mirror. …

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Reflect the source point $(1,2)$ across the x-axis to $(1,-2)$ ; the straight line from $(1,-2)$ to $(5,3)$ meets the x-axis at $A=\left(\dfrac{13}{5},\,0\right)$ .

Concept: reflection on the x-axis

By the law of reflection, the incident and reflected rays make equal angles with the mirror (the x-axis). Reflecting the source point across the mirror straightens the path: the reflected image of $(1,2)$ is $(1,-2)$ , and the line joining $(1,-2)$ to the target $(5,3)$ crosses the x-axis exactly at the reflection point $A$ .

Step-by-step solution

1. Reflect the source. Reflecting $(1,2)$ across the x-axis flips the $y$ -sign: $P'=(1,-2)$ .

2. Line through $P'(1,-2)$ and $(5,3)$ . Slope:

m=\frac{3-(-2)}{5-1}=\frac{5}{4}.

Equation: $y+2=\dfrac{5}{4}(x-1)$ . …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2025Set 65/1/15 marks
$A'$
(OR)
$P$
›Reveal solutionSolution

Part (a): the foot of the perpendicular from $A(1,6,3)$ is $M(1,3,5)$ , so the image is $A'(1,0,7)$ and line $AA'$ is $x=1,\ \tfrac{y-6}{-3}=\tfrac{z-3}{2}$ .

Part (b): the distance condition gives $(\lambda-1)^2=0$ , so $P(-4,1,-3)$ and line $PQ$ is $\tfrac{x-2}{6}=\tfrac{y-4}{3}=\tfrac{z+1}{2}$ .

Part (a)

Write the line as $\dfrac{x}{1}=\dfrac{y-1}{2}=\dfrac{z-2}{3}=\lambda$ , so a general point is $M(\lambda,2\lambda+1,3\lambda+2)$ and the direction is $(1,2,3)$ .

$\vec{AM}=(\lambda-1,\,2\lambda-5,\,3\lambda-1)$ must be perpendicular to $(1,2,3)$ :
$(\lambda-1)+2(2\lambda-5)+3(3\lambda-1)=14\lambda-14=0\Rightarrow\lambda=1.$
So the foot of the perpendicular is $M(1,3,5)$ . As $M$ is the midpoint of $AA'$ ,
$A'=2M-A=(2-1,\,6-6,\,10-3)=(1,0,7).$
Direction $\vec{AA'}=(0,-6,4)\parallel(0,-3,2)$ , so line $AA'$ is …
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CBSE 2024Set 65/1/15 marks
$P(x, y, z)$
›Reveal solutionSolution
The key idea is that the midpoint of $P$ and its image $P'$ lies on the line, and the line joining $P$ and $P'$ is perpendicular to the given line. Solving these two conditions gives $P = (1, 6, 3)$ .

We are given a line in symmetric form:
$\frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3}$
This line passes through the point $A(0, 1, 2)$ and has direction ratios $(1, 2, 3)$ .

The image of a point $P$ in a line is the point $P'$ such that the line is the perpendicular bisector of segment $PP'$ . That means two things must be true:

The midpoint $M$ of $P$ and $P'$ lies on the given line.

The vector $\overrightarrow{PP'}$ is perpendicular to the direction vector of the line.

We know $P' = (1, 0, 7)$ . Let $P = (x, y, z)$ . We will use these two conditions to find $x, y, z$ .

Step 1: Midpoint lies on the line

The midpoint $M$ of $P$ and $P'$ is:
$M = \left( \frac{x+1}{2}, \frac{y+0}{2}, \frac{z+7}{2} \right)$
Since $M$ lies on the given line, its coordinates must satisfy the line's symmetric equation. That is, there exists some parameter $t$ such that:
$\frac{x+1}{2} = t, \quad \frac{y}{2} = 1 + 2t, \quad \frac{z+7}{2} = 2 + 3t$
From the first equation:
$x + 1 = 2t \quad \Rightarrow \quad x = 2t - 1$
From the second:
$\frac{y}{2} = 1 + 2t \quad \Rightarrow \quad y = 2 + 4t$
From the third:
$\frac{z+7}{2} = 2 + 3t \quad \Rightarrow \quad z + 7 = 4 + 6t \quad \Rightarrow \quad z = -3 + 6t$
So we have expressed $x, y, z$ in terms of a single parameter $t$ :
$P = (2t - 1, \; 2 + 4t, \; -3 + 6t)$

Step 2: Perpendicularity condition

The vector $\overrightarrow{PP'}$ is:
$\overrightarrow{PP'} = (1 - (2t - 1), \; 0 - (2 + 4t), \; 7 - (-3 + 6t))$ $= (2 - 2t, \; -2 - 4t, \; 10 - 6t)$
The direction vector of the line is $\vec{d} = (1, 2, 3)$ . For perpendicularity, the dot product must be zero:
$(2 - 2t)(1) + (-2 - 4t)(2) + (10 - 6t)(3) = 0$
Compute each term:
$2 - 2t - 4 - 8t + 30 - 18t = 0$
Combine constants: $2 - 4 + 30 = 28$

Combine $t$ terms: $-2t - 8t - 18t = -28t$

So: …
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Q.A ray of light passing through the point (1,2)(1, 2)(1,2) reflects on the x-axis at point AAA and the reflected ray passes through the point (5,3)(5, 3)(5,3). Find the coordinates of AAA.

Concept: reflection on the x-axis

Step-by-step solution

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Q.A ray of light passing through the point $(1, 2)$ reflects on the x-axis at point $A$ and the reflected ray passes through the point $(5, 3)$ . Find the coordinates of $A$ .

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