For points P = (x1, y1) and Q = (x2, y2) of the co-ordinate plane, a new distance d(P, Q) is defined

Q: Wait, why do we reject the negative root or other regions? Is the locus not just one single line?

Great question! In Euclidean geometry, the locus of points equidistant from two points is always a single straight line (the perpendicular bisector). However, Manhattan distance is different. It is piecewise linear. When we solve the equation $x + y = |x-3| + |y-2|$, we find that the solution is not a single line but a combination of a segment and a ray. Other regions, like $x \ge 3$ and $y \ge 2$, lead to contradictions like $0 = -5$, which simply means no points in those regions satisfy the condition.

Q: Why can't we just use the standard distance formula $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$?

The standard distance formula is for Euclidean space, where you can move in any direction. This problem explicitly defines a *new* metric: $d(P, Q) = |x_1 - x_2| + |y_1 - y_2|$. This is the 'Taxicab' or 'Manhattan' metric. We must strictly follow the rules of this new space, not the Euclidean one.

Q: How do I know where the critical lines are?

The critical lines are determined by the points where the expressions inside the absolute value bars change sign. In our equation $|x-3| + |y-2|$, the expression $(x-3)$ changes sign at $x=3$, and $(y-2)$ changes sign at $y=2$. These are your boundaries. Think of them as the 'seams' of the geometry where the behavior of the distance function shifts.

Question

For points $P = (x_1, y_1)$ and $Q = (x_2, y_2)$ of the co-ordinate plane, a new distance $d(P, Q)$ is defined by $d(P, Q) = |x_1 - x_2| + |y_1 - y_2|$. Let $O = (0, 0)$ and $A = (3, 2)$. Prove that the set of points in the first quadrant which are equidistant (with respect to the new distance) from $O$ and $A$ consists of the union of a line segment of finite length and an infinite ray. Sketch this set in a labelled diagram.

Sigma Percentile · Accepted Answer

### The Geometry of the Taxicab

Imagine you are standing in a city laid out in a perfect grid, like Manhattan. You want to go from the origin $O(0, 0)$ to a point $A(3, 2)$.

In a standard Euclidean world, you would walk in a straight line, cutting through buildings. But here, you must follow the streets—moving only horizontally and vertically.

This is the essence of the Manhattan distance, defined as:
$$d(P, Q) = |x_1 - x_2| + |y_1 - y_2|$$

Today, we are going to find the set of all points $P(x, y)$ in the first quadrant that are equidistant from $O$ and $A$ under this unique metric. It is a journey that will change how you perceive distance itself.

### Defining the Playing Field

First, let us define our distances. For any point $P(x, y)$ in the first quadrant, the distance to the origin is $d(P, O) = |x - 0| + |y - 0|$.

Since $x \ge 0$ and $y \ge 0$, this simplifies beautifully to $d(P, O) = x + y$.

Now, consider the distance to point $A(3, 2)$. This is $d(P, A) = |x - 3| + |y - 2|$.

Our goal is to find the locus where $d(P, O) = d(P, A)$, which gives us the master equation:
$$x + y = |x - 3| + |y - 2|$$

### The Critical Seams

This equation looks simple, but the absolute value bars are traps. They change their behavior at the 'critical lines' $x = 3$ and $y = 2$.

These lines act as boundaries that divide the first quadrant into four distinct regions. We must explore each region to see if any points there satisfy our condition.

Let us start with the region where $x < 3$ and $y < 2$. In this zone, both $(x - 3)$ and $(y - 2)$ are negative.

Therefore, $|x - 3| = -(x - 3) = 3 - x$ and $|y - 2| = -(y - 2) = 2 - y$. Substituting these into our master equation, we get:
$$x + y = (3 - x) + (2 - y)$$

Simplifying the right side, we have $x + y = 5 - x - y$. Bringing all variables to the left, we get $2x + 2y = 5$, or:
$$x + y = \frac{5}{2}$$

This is a straight line with a slope of $-1$. Because this is only valid for $x < 3$ and $y < 2$, it forms a finite line segment starting from the x-axis at $(\frac{5}{2}, 0)$ and ending at $(0.5, 2)$.

### The Infinite Ray

Now, let us venture into the region where $x < 3$ but $y \ge 2$. Here, $(x - 3)$ is still negative, so $|x - 3| = 3 - x$.

However, $(y - 2)$ is now non-negative, so $|y - 2| = y - 2$. Our equation becomes:
$$x + y = (3 - x) + (y - 2)$$

Look at the magic that happens: the $y$ terms on both sides cancel out! We are left with $x = 1 - x$, which simplifies to $2x = 1$, or:
$$x = \frac{1}{2}$$

This is a vertical line. Since this is valid for all $y \ge 2$, it forms an infinite ray starting at $(0.5, 2)$ and extending upwards to infinity. This is the second part of our locus.

### Conclusion

If you check the other regions, such as $x \ge 3$, you will find that the equations lead to contradictions like $0 = -5$, meaning no points exist there.

Thus, the set of equidistant points is the union of the finite segment $x + y = \frac{5}{2}$ (for $y < 2$) and the infinite ray $x = \frac{1}{2}$ (for $y \ge 2$).

It is a stunning result—a piecewise linear shape that perfectly captures the rigid, grid-like nature of the Manhattan metric. You have just mapped the geometry of a city!

Visualized Solution

Introduction to Manhattan Distance

Distance from Origin $d (P, O)$

Distance from Point $A (3, 2)$

The Equidistant Condition

Analyzing the Critical Lines

Case 1: Region $x < 3$ and $y < 2$

Case 1: Solving the Equation

Case 1: The Finite Line Segment

Case 2: Region $x < 3$ and $y \geq 2$

Case 2: Solving the Equation

Case 2: The Infinite Ray

Checking Other Regions

Final Locus

The Sigma Insight: Various Forms of Equations of a Line

The Geometry of the Taxicab

Defining the Playing Field

The Critical Seams

The Infinite Ray

Conclusion

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