Q.Assuming that straight lines work as the plane mirror for a point, find the image of the point $(1, 2)$ in the line $x - 3y + 4 = 0$.

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:

1. $L$ is the perpendicular bisector of segment $PP'$ — that means $L$ passes through the midpoint of $PP'$, and $L$ is perpendicular to $PP'$.
2. $P$ 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. …