Coordinates and graphs

To reason with algebra, treat the infinite line as a one-dimensional number line. Place point $A$ at $0$ and put point $B$ at a position $d$ units away.

• Coordinate of $A=0$
• Coordinate of $B=d$ (where $d\ne 0$ because $A$ and $B$ are different points)

Now let a point $P$ on this same line have coordinate $x$. We want $P$ to be equally far from $A$ and $B$, which is a standard idea in Coordinate geometry basics.

On a number line, the distance between positions $x_1$ and $x_2$ is $|x_1-x_2|$.

So if $P$ (at $x$) is the same distance from $A$ and $B$, we must have: $|x-0|=|x-d|$ This becomes: $|x|=|x-d|$

An equation like $|a|=|b|$ means one of two things:

1. $a=b$
2. $a=-b$

Apply this to $|x|=|x-d|$:

Case 1: $x=x-d$ Subtract $x$ from both sides: $0=-d$. But $d\ne 0$, so this case cannot happen.

Case 2: $x=-(x-d)$ $\begin{array}{rl}x& =-x+d\\ 2x& =d\\ x& =\frac{d}{2}\end{array}$

We obtained a single valid coordinate: $x=\frac{d}{2}$.

That location is exactly halfway between $0$ and $d$, so it represents the midpoint of segment $AB$. Since the other case produced no solution, there cannot be any second point on the full line that also works.

Because the algebra shows there is no alternative point on the line with equal distances to $A$ and $B$, the option claiming the statement is incorrect must be removed.

This confirms that $x=d/2$ is the unique point on the infinite line that is equidistant from $A$ and $B$, so the statement is True.

In geometry, the locus of points that are the same distance from two distinct points $A$ and $B$ is the perpendicular bisector of segment $AB$.

Any point $P$ with $PA=PB$ must lie somewhere on that perpendicular bisector. This is a key fact from Coordinate geometry basics.

The question restricts us to points on the infinite line through $A$ and $B$.

The perpendicular bisector is perpendicular to $AB$, so it crosses the line containing $AB$. In a plane, two non-parallel lines meet at exactly one point.

$\text{Intersection Point}=\text{Midpoint }M$

Since every equidistant point lies on the perpendicular bisector and that bisector touches line $AB$ only at $M$, there cannot be any other equidistant point on line $AB$. So the statement cannot be false.

The locus reasoning shows the midpoint is the unique point on the line through $A$ and $B$ that is equally distant from both endpoints, so the statement is True.

To check whether any point besides the midpoint could be equally far from $A$ and $B$, divide the line into three parts:

1. Region 1: points to the left of $A$.
2. Region 2: points between $A$ and $B$.
3. Region 3: points to the right of $B$.

This ensures we consider every possible position on the line, linking to Coordinate geometry basics.

In Region 1 (left of $A$), any point $P$ is nearer to $A$ than to $B$.

If the distance from $P$ to $A$ is $d$, then the distance from $P$ to $B$ is $d+AB$. Because $AB>0$, we cannot have $d=d+AB$.

So no point in Region 1 is equidistant from $A$ and $B$.

In Region 3 (right of $B$), any point $P$ is nearer to $B$ than to $A$.

If the distance from $P$ to $B$ is $d$, then the distance from $P$ to $A$ is $d+AB$. Since $AB>0$, it is impossible for these to match ($d\ne d+AB$).

Therefore, Region 3 also has no equidistant point.

In Region 2 (between $A$ and $B$), suppose $P$ is $x$ units from $A$. Then it is $AB-x$ units from $B$.

Equidistant means: $\begin{array}{rl}x& =AB-x\\ 2x& =AB\\ x& =\frac{AB}{2}\end{array}$

This value $x=\frac{AB}{2}$ corresponds to exactly one point: the midpoint $M$.

Regions 1 and 3 give no solutions, and Region 2 gives exactly one solution (the midpoint). So the statement is correct.