In short

The equation |x - a| = |x - b| is asking a very specific question: for what value of x is the distance to a the same as the distance to b? On the number line, there is exactly one such point — the midpoint of a and b. Algebraically, x = \dfrac{a + b}{2}. As long as a \neq b, this equation always has exactly one solution. (If a = b, the equation becomes |x - a| = |x - a|, which is true for every real x.)

You have already seen the equation |x - c| = r in Absolute Value — Equations — it asks for the points that are exactly r units from c, and the answer is two points, one on each side. Now consider what looks like a small twist: instead of a fixed distance r on the right, there is another absolute value.

|x - a| = |x - b|

This is no longer "be a fixed distance from one point". It is "be the same distance from two points". The right-hand side is no longer a number — it changes with x. So what is this equation really asking, and how many answers does it have?

The geometric reading

Strip away the algebra for a moment and just read the equation in plain English. The expression |x - a| is the distance between x and a on the number line. The expression |x - b| is the distance between x and b. The equation says these two distances are equal.

Picture two villages on a long straight highway — say Pune at position a and Mumbai at position b. You are standing at some point x on the highway, and someone asks: where do you have to stand so that you are exactly as far from Pune as from Mumbai?

There is only one honest answer: stand exactly halfway between them. Take one step closer to Pune and you are closer to Pune than to Mumbai. Take one step the other way and the imbalance flips. The equidistant point is unique, and it is the midpoint.

Midpoint of a and b on the number line is equidistant from bothA horizontal number line. Two points labelled a and b are marked, with a on the left and b on the right. The midpoint, labelled (a plus b) over 2, is circled exactly halfway between them. Two arrows above the line, one from the midpoint to a and one from the midpoint to b, are labelled with equal lengths showing the equal distances. a b (a + b) ⁄ 2 equal distance equal distance
The unique point on the number line that is equidistant from $a$ and $b$ is the midpoint $(a+b)/2$. Move left and you are closer to $a$; move right and you are closer to $b$. Only the centre balances both.

That is the entire geometric story. The equation |x - a| = |x - b| is a verbal description of "find the midpoint", and the midpoint is unique on a line.

Why: the number line is one-dimensional, so once you fix the requirement "distance to a equals distance to b", there is no wiggle room left — exactly one point satisfies it.

The algebraic derivation

Geometry tells you the answer should be the midpoint. Algebra can confirm it without ever drawing a picture. Start with the equation:

|x - a| = |x - b|

Both sides are non-negative (absolute values always are), so squaring is safe — it does not introduce any sign-flipping mischief. Why: for any real u, v \ge 0, the statement u = v is equivalent to u^2 = v^2. Squaring two non-negative numbers preserves equality in both directions. And since |u|^2 = u^2 for every real u, squaring the absolute values just removes the bars:

(x - a)^2 = (x - b)^2

Expand both sides.

x^2 - 2ax + a^2 = x^2 - 2bx + b^2

The x^2 terms on both sides are identical and cancel, leaving:

-2ax + a^2 = -2bx + b^2

Collect the x terms on one side and the constants on the other:

2bx - 2ax = b^2 - a^2

Factor the left side and use the difference-of-squares identity on the right:

2(b - a) \, x = (b - a)(b + a)

If a \neq b, you may divide both sides by 2(b - a):

x = \frac{(b - a)(b + a)}{2(b - a)} = \frac{a + b}{2}

That is the midpoint, exactly as the geometry promised. Why: (b - a)(b + a) over 2(b - a) — the (b - a) factor cancels (which is legal precisely because a \neq b), leaving (a + b)/2.

So whenever the two anchor points a and b are different, the equation has exactly one solution: the average of a and b.

What happens if a = b?

Look back at the algebra. The step 2(b - a) x = (b - a)(b + a) becomes 0 \cdot x = 0 when a = b — a statement that is true for every real x. The cancellation that gave us the midpoint is no longer legal, because dividing by zero is forbidden.

Geometrically, this is also obvious. If a = b, then the two "villages" are actually the same village. The equation reduces to:

|x - a| = |x - a|

which is trivially true no matter what x is. Every point on the number line is equidistant from a point and itself — the distance is just |x - a| on both sides. So the solution set is all of \mathbb{R}.

This is not a defect of the equation; it is the equation telling you something honest. When you ask "what is equidistant from a and b?" and the two are the same point, the answer is "everything".

Three worked examples

Example 1: Solve $|x - 2| = |x - 8|$

The equation asks for the point equidistant from 2 and 8 on the number line.

Step 1. Apply the midpoint rule.

x = \frac{2 + 8}{2} = \frac{10}{2} = 5

Why: 2 and 8 are six units apart, so their midpoint is three units from each.

Step 2. Verify by plugging back in.

|5 - 2| = |3| = 3 \qquad |5 - 8| = |-3| = 3

Both sides equal 3. The point x = 5 is genuinely equidistant from 2 and 8.

Result. x = 5.

Example 2: Solve $|x + 1| = |x - 7|$

The plus sign in |x + 1| is hiding an a. Rewrite it so that both terms look the same:

|x - (-1)| = |x - 7|

Now a = -1 and b = 7.

Step 1. Apply the midpoint rule.

x = \frac{-1 + 7}{2} = \frac{6}{2} = 3

Why: -1 and 7 are eight units apart on the number line, and their midpoint sits four units from each.

Step 2. Verify.

|3 + 1| = |4| = 4 \qquad |3 - 7| = |-4| = 4

Both sides equal 4. Confirmed.

Result. x = 3.

Example 3: What does the same question look like in higher dimensions?

Forget the number line for a moment. Suppose a and b are two points in a plane — say two friends standing at fixed spots on a cricket field — and you want all the points equidistant from both. That set is no longer a single point. It is the perpendicular bisector of the segment joining a and b — an entire straight line passing through the midpoint at right angles to the segment ab.

In three dimensions, the same equidistant condition gives a whole plane — the perpendicular bisector plane of the segment ab. In n dimensions, you get an (n-1)-dimensional hyperplane.

The pattern is consistent. The condition "equidistant from two fixed points" removes one degree of freedom. On the line (one dimension), one minus one equals zero — a single point. In the plane (two dimensions), two minus one equals one — a line. In space (three dimensions), three minus one equals two — a plane.

This connects directly to what you will study in CBSE Class 11 coordinate geometry — the perpendicular bisector of a segment is exactly the locus of points equidistant from the two endpoints. The one-dimensional version you just solved is the simplest case of that idea: a "hyperplane" in 1D is just a single point.

Summary

For real numbers a and b:

Both the geometry (equidistance on a line forces the midpoint) and the algebra (squaring both sides reduces to a linear equation in x) point to the same conclusion. And the same equidistance question, asked in two or three dimensions, generalises to perpendicular bisector lines and planes — a thread you will pick up again in coordinate geometry.

References

  1. NCERT, Mathematics Textbook for Class 11, Chapter 10: Straight Lines — perpendicular bisector as the locus of equidistant points.
  2. Khan Academy, Midpoint of a Segment — geometric and coordinate definitions.
  3. Wikipedia, Bisection — Perpendicular bisector — the higher-dimensional generalisation of the equidistance condition.
  4. Wikipedia, Absolute value — definition and the metric property |a - b| as distance.