If the sum of the distances of a moving point in a plane from the axes is 1, then find the locus of the point.

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Q. If the sum of the distances of a moving point in a plane from the axes is 1, then find the locus of the point.

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Concept understanding — Locus of Point

The Locus of a Point: From Intuition to Precision

Imagine you're walking in a park, but you must always stay exactly 5 metres away from a fountain at the centre. As you walk, your path traces out a circle. That circle is the locus of your position — the set of all points that satisfy the rule "distance from fountain = 5 m".

Now think of a different rule: you must always be equally far from two trees. Your path becomes the perpendicular bisector of the line joining those trees — a straight line.

Every geometric shape you know — circle, line, parabola, ellipse — is really just a locus. A circle is the set of points at a fixed distance from a centre. A line is the set of points that satisfy a linear equation. The word "locus" (plural: loci) simply means "place" or "path" in Latin.

The Precise Definition

Important

Locus of a point is the set of all positions (points) that satisfy a given geometric condition or a set of conditions.

In coordinate geometry, a locus is represented by an equation in $x$ and $y$ (or $x$ , $y$ , $z$ in 3D). Every point $(x, y)$ that satisfies the condition lies on the locus; every point that does not satisfy it lies off the locus.

How to Find the Equation of a Locus

The process is mechanical. Suppose a point $P(x, y)$ moves so that its distance from a fixed point $A(2, 3)$ is always 5 units.

Write the condition in words: Distance $PA = 5$ .
Translate into algebra: $\sqrt{(x-2)^2 + (y-3)^2} = 5$ .
Simplify: Square both sides: $(x-2)^2 + (y-3)^2 = 25$ .

That's it. The locus is a circle with centre $(2, 3)$ and radius 5.

A Slightly Harder Example

Find the locus of a point $P(x, y)$ that moves so that its distance from $A(1, 0)$ is twice its distance from $B(4, 0)$ .

Step 1 — Condition: $PA = 2 \cdot PB$ .

Step 2 — Algebra:

\sqrt{(x-1)^2 + y^2} = 2 \sqrt{(x-4)^2 + y^2}

Step 3 — Square and simplify:

(x-1)^2 + y^2 = 4[(x-4)^2 + y^2]

x^2 - 2x + 1 + y^2 = 4(x^2 - 8x + 16 + y^2)

x^2 - 2x + 1 + y^2 = 4x^2 - 32x + 64 + 4y^2

0 = 3x^2 - 30x + 3y^2 + 63

x^2 - 10x + y^2 + 21 = 0

Step 4 — Complete the square:

(x^2 - 10x + 25) + y^2 = 4

(x-5)^2 + y^2 = 4

The locus is a circle with centre $(5, 0)$ and radius 2.

Tip

When the condition involves distances in a ratio, the locus is often a circle (called the Apollonius circle). If the ratio is 1:1, the locus is the perpendicular bisector — a straight line.

Common Loci You Must Know

Condition	Locus	Equation (standard form)
Fixed distance from a point	Circle	$(x-h)^2 + (y-k)^2 = r^2$
Equal distances from two points	Perpendicular bisector	Linear equation
Fixed distance from a line	Pair of parallel lines	$
Sum of distances from two fixed points is constant	Ellipse	$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

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A point whose coordinate-sum of absolute distances from both axes equals 1 traces four line segments forming a square with vertices at $(1,0)$ , $(0,1)$ , $(-1,0)$ , $(0,-1)$ ; the locus is $|x| + |y| = 1$ .

Understanding the problem

The distance of a point $P(x, y)$ from the $x$ -axis is $|y|$ , and its distance from the $y$ -axis is $|x|$ . We're told the sum of these distances is always 1, so we need to find all points satisfying

|x| + |y| = 1.

This single equation encodes four different linear relationships depending on the signs of $x$ and $y$ . The absolute values force us to consider each quadrant separately.

Breaking down by quadrant

The plane divides naturally into four regions based on the signs of the coordinates. In each region, the absolute values simplify differently.

1. First quadrant ( $x \geq 0$ , $y \geq 0$ )

Here $|x| = x$ and $|y| = y$ , so the equation becomes

x + y = 1.

This is the line segment from $(1, 0)$ to $(0, 1)$ .

2. Second quadrant ( $x \leq 0$ , $y \geq 0$ )

Now $|x| = -x$ and $|y| = y$ , giving

-x + y = 1 \quad \Rightarrow \quad y = x + 1.

This is the line segment from $(0, 1)$ to $(-1, 0)$ .

3. Third quadrant ( $x \leq 0$ , $y \leq 0$ )

Both absolute values flip: $|x| = -x$ and $|y| = -y$ , so

-x - y = 1 \quad \Rightarrow \quad x + y = -1.

This is the line segment from $(-1, 0)$ to $(0, -1)$ .

4. Fourth quadrant ( $x \geq 0$ , $y \leq 0$ )

Here $|x| = x$ and $|y| = -y$ , yielding

x - y = 1 \quad \Rightarrow \quad y = x - 1.

This is the line segment from $(0, -1)$ to $(1, 0)$ .

Visualizing the locus …

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Q.If the sum of the distances of a moving point in a plane from the axes is 1, then find the locus of the point.

Understanding the problem

Breaking down by quadrant

Visualizing the locus …

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