Orienting Yourself: The Use of Coordinates

Introduction to Coordinate Systems

**Definition**: A **coordinate system** is a structured framework using numbered grid lines that enables us to describe exact physical locations of points or objects using numbers.

**Historical Context** (Exam-Important):

The **Sindhu-Sarasvatī Civilization** (ancient India) used grid-based city planning with streets running North-South and East-West at uniform 10-metre intervals—the first practical coordinate system

**Baudhāyana** (c. 800 CE) developed geometric coordinate concepts leading to the Baudhāyana-Pythagoras Theorem

**Ujjayinī** was marked as the central longitude meridian from which all locations were measured (4th century BCE)

**Āryabhaṭa** (499 CE) replaced Greek chords with sines for easier coordinate calculations

**Brahmagupta** (628 CE) formalized **zero** and **negative numbers**, making the four-quadrant Cartesian plane possible

**René Descartes** (1637 CE) formalized that any point in 2-D plane could be defined by exactly two numbers representing distances from two perpendicular axes

**Key Insight**: Without the Indian mathematical concepts of zero and negative numbers, the modern coordinate system would be impossible.

The 2-D Cartesian Coordinate System

**Definition**: The **2-D Cartesian Coordinate System** uses two perpendicular lines to mark points in two-dimensional space (2-D space).

**Components of the Coordinate System**:

**x-axis**: The horizontal line

**y-axis**: The vertical line

**Origin O**: The point of intersection of x-axis and y-axis; coordinates are (0, 0)

**Coordinate axes** (plural of axis): Help locate any point in 2-D space using coordinates

**Sign Conventions**:

Distances to the **right** of O or **upward** from O are **positive (+)**

Distances to the **left** of O or **downward** from O are **negative (−)**

**Standard Notation for Points**:

A point P with coordinates x and y is written as:

**P(x, y)** or **P = (x, y)**

The order of coordinates is crucial: **(x-coordinate first, y-coordinate second)**

**Example 1**:

Point B(4.5, 0) lies on the x-axis, 4.5 units to the right of origin

Point G(0, −4.5) lies on the y-axis, 4.5 units below origin

Point H(0, 4) lies on the y-axis, 4 units above origin

**Points on the Axes**:

A point **P(x, 0)** lies on the **x-axis**:

If x > 0, then P is to the right of O

If x < 0, then P is to the left of O

A point **P(0, y)** lies on the **y-axis**:

If y > 0, then P is above O

If y < 0, then P is below O

**Exam Rule**: On the x-axis, the **y-coordinate is always 0**. On the y-axis, the **x-coordinate is always 0**.

The Four Quadrants

**Definition**: The coordinate axes divide the **Cartesian plane** (also called **coordinate plane** or **xy-plane**) into **four parts called quadrants**, numbered I, II, III, and IV (in counterclockwise order).

**Sign Rules for Each Quadrant**:

**Quadrant I**: Both x and y coordinates are **positive (+, +)**

**Quadrant II**: x-coordinate is **negative**, y-coordinate is **positive (−, +)**

**Quadrant III**: Both x and y coordinates are **negative (−, −)**

**Quadrant IV**: x-coordinate is **positive**, y-coordinate is **negative (+, −)**

**Meaning of Coordinates**:

For a point **P(x, y)**:

**x-coordinate** (or abscissa): The perpendicular distance of P from the y-axis, measured along the x-axis

**y-coordinate** (or ordinate): The perpendicular distance of P from the x-axis, measured along the y-axis

**Example 2**:

Point S(3, −5): x-coordinate = 3 (3 units right of y-axis), y-coordinate = −5 (5 units below x-axis), lies in **Quadrant IV**

Point Q(−5, 3): x-coordinate = −5 (5 units left of y-axis), y-coordinate = 3 (3 units above x-axis), lies in **Quadrant II**

**Important Observation About Order**:

The point **(x, y)** is different from **(y, x)** unless **x = y**

**(x, y) = (y, x)** if and only if **x = y**

This means coordinates are an **ordered pair**

Distance Between Two Points in the 2-D Plane

**Problem**: Finding the straight-line distance between any two points in the coordinate plane.

**Method Using Baudhāyana-Pythagoras Theorem**:

**Formula for Distance Between Two Points**:

If A(x₁, y₁) and B(x₂, y₂) are two points in the coordinate plane, then the distance between them is:

**d = √[(x₂ − x₁)² + (y₂ − y₁)²]**

**Derivation**:

Consider two points A(x₁, y₁) and B(x₂, y₂). Drop a perpendicular from B to meet a horizontal line through A at point F. This creates a right-angled triangle AFB where:

The horizontal distance (base) = |x₂ − x₁|

The vertical distance (height) = |y₂ − y₁|

The hypotenuse AB = distance between A and B

By the **Baudhāyana-Pythagoras Theorem**:

**AB² = (x₂ − x₁)² + (y₂ − y₁)²**

Therefore: **AB = √[(x₂ − x₁)² + (y₂ − y₁)²]**

**Step-by-Step Example**:

**Example 3**: Find the distance between A(3, 4) and D(7, 1)

Step 1: Identify coordinates: (x₁, y₁) = (3, 4) and (x₂, y₂) = (7, 1)

Step 2: Calculate horizontal shift:

x₂ − x₁ = 7 − 3 = 4

Step 3: Calculate vertical shift:

y₂ − y₁ = 1 − 4 = −3

Step 4: Apply distance formula:

d = √[(4)² + (−3)²]

d = √[16 + 9]

d = √25

d = 5 units

**Important Notes About the Formula**:

It makes **no difference whether (x₂ − x₁) and (y₂ − y₁) are positive or negative** because we square them

The formula works for all quadrants, including when coordinates are negative

The distance is always **positive** (it's a length)

The formula is **symmetric**: distance from A to B equals distance from B to A

**Example 4**: Distance with negative coordinates

Find distance between A'(−3, 4) and D'(−7, 1)

x₂ − x₁ = −7 − (−3) = −7 + 3 = −4

y₂ − y₁ = 1 − 4 = −3

d = √[(−4)² + (−3)²] = √[16 + 9] = √25 = 5 units

(Same as Example 3 because reflection preserves distances)

**Special Cases**:

**Distance between points on x-axis**: A(x₁, 0) and B(x₂, 0)

d = √[(x₂ − x₁)² + 0²] = |x₂ − x₁|

**Distance between points on y-axis**: A(0, y₁) and B(0, y₂)

d = √[0² + (y₂ − y₁)²] = |y₂ − y₁|

**Distance from origin**: Distance from O(0, 0) to P(x, y)

d = √[x² + y²]

Midpoint of a Line Segment

**Definition**: The **midpoint M** of a line segment joining points A(x₁, y₁) and B(x₂, y₂) is the point that divides the segment into two equal parts.

**Formula for Midpoint**:

**M = ((x₁ + x₂)/2, (y₁ + y₂)/2)**

**How to Use**:

Add the x-coordinates and divide by 2 to get the x-coordinate of midpoint

Add the y-coordinates and divide by 2 to get the y-coordinate of midpoint

**Example 5**: Find the midpoint of segment joining S(−3, 0) and T(3, 0)

M = ((−3 + 3)/2, (0 + 0)/2) = (0, 0)

The origin is the midpoint! (Symmetric about origin)

**Example 6**: Find the midpoint of segment joining A(2, 3) and B(4, 5)

M = ((2 + 4)/2, (3 + 5)/2) = (3, 4)

**Application**: To check if M is the midpoint of ST, verify that:

x-coordinate of M = (x-coordinate of S + x-coordinate of T)/2

y-coordinate of M = (y-coordinate of S + y-coordinate of T)/2

**Exam Tip**: If given that M is the midpoint of AB and coordinates of A and M, you can find B using:

x-coordinate of B = 2(x-coordinate of M) − x-coordinate of A

y-coordinate of B = 2(y-coordinate of M) − y-coordinate of A

Points of Trisection

**Definition**: The **points of trisection** of a line segment AB are two points P and Q that divide the segment into three equal parts, with P closer to A and Q closer to B.

**Method to Find Trisection Points**:

If A(x₁, y₁) and B(x₂, y₂) are endpoints, then:

**P (point of trisection closer to A)**:

P divides AB in ratio 1:2

**P = ((2x₁ + x₂)/3, (2y₁ + y₂)/3)**

**Q (point of trisection closer to B)**:

Q divides AB in ratio 2:1

**Q = ((x₁ + 2x₂)/3, (y₁ + 2y₂)/3)**

**Example 7**: Find trisection points of A(4, 7) and B(16, −2)

For P (closer to A):

P = ((2(4) + 16)/3, (2(7) + (−2))/3) = ((8 + 16)/3, (14 − 2)/3) = (24/3, 12/3) = (8, 4)

For Q (closer to B):

Q = ((4 + 2(16))/3, (7 + 2(−2))/3) = ((4 + 32)/3, (7 − 4)/3) = (36/3, 3/3) = (12, 1)

**Verification**:

Distance AP = distance PQ = distance QB? Yes: both are 1/3 of total distance AB

Collinearity of Three Points

**Definition**: Three points are **collinear** if they lie on the same straight line.

**Method to Check Without Plotting**:

For three points A, B, and C to be collinear, they must satisfy:

**Distance(A, B) + Distance(B, C) = Distance(A, C)**

OR

**Distance(A, C) + Distance(C, B) = Distance(A, B)**

(One point must lie between the other two)

**Alternative Method Using Slopes** (for higher grades):

If slope of AB = slope of BC, then A, B, C are collinear.

**Example 8**: Check if M(−3, −4), A(0, 0), and G(6, 8) are collinear

Step 1: Calculate distance MA

MA = √[(0−(−3))² + (0−(−4))²] = √[9 + 16] = √25 = 5

Step 2: Calculate distance AG

AG = √[(6−0)² + (8−0)²] = √[36 + 64] = √100 = 10

Step 3: Calculate distance MG

MG = √[(6−(−3))² + (8−(−4))²] = √[81 + 144] = √225 = 15

Step 4: Check if MA + AG = MG

5 + 10 = 15 ✓

**Yes, the points are collinear** (A lies between M and G)

Points on a Circle

**Definition**: A point P lies on a circle with center C and radius r if **Distance(P, C) = r**.

**Method to Check if Point is on a Circle**:

1. Find distance from point to center

2. Compare with radius:

If distance = radius → point is **on the circle**

If distance < radius → point is **inside the circle**

If distance > radius → point is **outside the circle**

**Example 9**: Check if A(1, −8), B(−4, 7), C(−7, −4) lie on a circle with center O(0, 0)

Distance OA = √[1² + (−8)²] = √[1 + 64] = √65

Distance OB = √[(−4)² + 7²] = √[16 + 49] = √65

Distance OC = √[(−7)² + (−4)²] = √[49 + 16] = √65

All three distances equal √65, so **all three points lie on a circle with center O(0, 0) and radius √65**.

Common Mistakes to Avoid

**Reversing coordinates**: Remember (x, y) ≠ (y, x) unless x = y

**Sign errors**: Negative coordinates often cause mistakes; use |difference| in distance formula

**Order in distance formula**: It doesn't matter which point you call (x₁, y₁) or (x₂, y₂) because differences are squared

**Forgetting to square in distance formula**: The formula uses squared differences, not simple differences

**Confusing axes**: x-axis is horizontal, y-axis is vertical

**Quadrant locations**: Double-check signs of both coordinates for correct quadrant identification

**Midpoint vs. endpoints**: Midpoint has average coordinates; it's not the same as either endpoint

Key Formulas Summary

**Distance Formula**: d = √[(x₂ − x₁)² + (y₂ − y₁)²]

**Midpoint Formula**: M = ((x₁ + x₂)/2, (y₁ + y₂)/2)

**Trisection (Point P closer to A)**: P = ((2x₁ + x₂)/3, (2y₁ + y₂)/3)

**Trisection (Point Q closer to B)**: Q = ((x₁ + 2x₂)/3, (y₁ + 2y₂)/3)

**Collinearity Check**: Distance(A, B) + Distance(B, C) = Distance(A, C)

**Circle Check**: If Distance(P, Center) = r, then P is on the circle