Q.If the image of the point $P(x, y, z)$ in the line $\frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3}$ is $P'(1, 0, 7)$, then find the coordinates of point P.

Concept: Image of a point in a line — the line is the perpendicular bisector of the segment joining a point and its image. So the midpoint of $P$ and $P'$ lies on the line, and the segment $PP'$ is perpendicular to the line.

Step 1 – Midpoint condition

Let $P = (x, y, z)$. Midpoint $M$ of $P$ and $P'(1, 0, 7)$ is

Since $M$ lies on the line $\frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3} = \lambda$, we have

Step 2 – Perpendicularity condition

Direction ratios of the line: $(1, 2, 3)$. Direction ratios of $PP'$: $(1-x, 0-y, 7-z)$. For perpendicularity,

Step 3 – Solve …

We are given a line in symmetric form:

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:

1. The midpoint $M$ of $P$ and $P'$ lies on the given line.
2. 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:

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:

From the first equation:

From the second:

From the third:

So we have expressed $x, y, z$ in terms of a single parameter $t$:

Step 2: Perpendicularity condition

The vector $\overrightarrow{PP'}$ is:

The direction vector of the line is $\vec{d} = (1, 2, 3)$. For perpendicularity, the dot product must be zero:

Compute each term:

Combine constants: $2 - 4 + 30 = 28$

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

So: …

Method: Image (reflection) of a point in a line

Use this whenever a point is reflected in a line and you must find either the image or the original point — the line behaves like a mirror.

Steps

Step 1: Use the two defining conditions.

The line is the perpendicular bisector of the segment joining a point and its image, so (i) the MIDPOINT of the two points lies on the line, and (ii) the segment joining them is PERPENDICULAR to the line's direction.

Step 2: Write a general point on the line. …

Common Mistakes

Mistake 1: Writing the general point on the line as $(t,2t,3t)$.

Why it's wrong: the line $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$ passes through $(0,1,2)$, so its general point is $(t,\,1+2t,\,2+3t)$; dropping the constants describes a different line. Correct approach: keep the $+1$ and $+2$.

Mistake 2: Using only one of the two mirror conditions. …