7. Coordinate Geometry

Introduction

Chapter 7: Coordinate Geometry

Page 1 of 4: The Distance Formula — Locating Points in Space

Welcome to the fascinating world of Coordinate Geometry! In earlier classes, you've learned how to plot points on a graph and identify their locations using coordinates. But what if we want to go further? What if we want to find the exact distance between two cities on a map, or calculate the shortest path a drone can take between two points?

Coordinate geometry gives us the power to connect algebra and geometry. It allows us to study geometric shapes by representing them as equations and points on a plane. Imagine you're designing a new city park. You need to place a fountain, a bench, and a lamp post. By assigning coordinates to each item, you can precisely calculate the length of the path between the bench and the fountain, or check if all three items lie in a straight line. This chapter equips you with the fundamental tools to solve such real-world problems. We'll start with the most basic yet powerful tool: finding the distance between any two points.

Definitions & Formulas

Before we jump into calculations, let's refresh some key terms and introduce the central formula for this lesson.

{{FORMULA: expr=d = √((x₂ - x₁)² + (y₂ - y₁)²)| symbols=d:distance, (x₁, y₁):coordinates of first point, (x₂, y₂):coordinates of second point}}

Derivation of the Distance Formula

Ever wondered where the Distance Formula comes from? It's not magic! It's a clever application of a theorem you already know and love: the Pythagorean Theorem. Let's derive it step-by-step.

Our goal is to find the distance d between any two points A(x₁, y₁) and B(x₂, y₂).

1. Plot the Points: Imagine plotting points A and B on a Cartesian plane. The distance between them, AB, is the hypotenuse of a right-angled triangle that we can construct.

2. Construct a Right-Angled Triangle: Draw a horizontal line from point A and a vertical line from point B. Let these lines intersect at a point C.

{{VISUAL: diagram: A Cartesian plane showing two points A(x₁, y₁) and B(x₂, y₂). A right-angled triangle ABC is formed by drawing a horizontal line from A and a vertical line from B, intersecting at C(x₂, y₁). The lengths of the base AC (labeled x₂ - x₁) and the height BC (labeled y₂ - y₁) are clearly marked. The hypotenuse AB is labeled 'd'.}}

3. Find the Coordinates of C: Since C lies on the same horizontal line as A, its y-coordinate is y₁. Since it lies on the same vertical line as B, its x-coordinate is x₂. Therefore, the coordinates of point C are (x₂, y₁).

4. Calculate the Lengths of the Sides:

   • The length of the base AC is the horizontal distance between A and C. This is the difference in their x-coordinates.

   AC = x₂ - x₁
   

   • The length of the height BC is the vertical distance between B and C. This is the difference in their y-coordinates.

   BC = y₂ - y₁
   

5. Apply the Pythagorean Theorem: In the right-angled triangle ABC, the theorem states (Hypotenuse)² = (Base)² + (Height)².

   AB² = AC² + BC²
   

6. Substitute and Solve for AB: Now, substitute the expressions for AC and BC into the theorem. Let the distance AB be d.

   d² = (x₂ - x₁)² + (y₂ - y₁)²
   

   To find d, we take the square root of both sides. Since distance cannot be negative, we only consider the positive root.

   d = √((x₂ - x₁)² + (y₂ - y₁)² )
   

   And there you have it! The Distance Formula.

{{KEY: type=concept | title=The Power of the Pythagorean Theorem | text=The Distance Formula is a direct consequence of the Pythagorean theorem applied to the coordinate plane. Understanding this connection is more important than just memorizing the formula, as it helps in solving complex problems.}}

Solved Examples

Let's put the formula into practice with examples ranging from simple to more challenging.

Example 1: Basic Distance Calculation (Easy)

Given: Two points P(2, 3) and Q(4, 1).

To Find: The distance between P and Q.

Solution:

1. Identify the coordinates. Let P(2, 3) be (x₁, y₁) and Q(4, 1) be (x₂, y₂). So, x₁ = 2, y₁ = 3, x₂ = 4, y₂ = 1.

2. Apply the distance formula: d = √((x₂ - x₁)² + (y₂ - y₁)² ).

   PQ = √((4 - 2)² + (1 - 3)²)
   

3. Simplify the terms inside the parentheses.

   PQ = √((2)² + (-2)²)
   

4. Calculate the squares. Remember that squaring a negative number results in a positive number.

   PQ = √(4 + 4)
   

5. Add the numbers and simplify the square root.

   PQ = √8 = √(4 × 2) = 2√2
   

Final Answer: The distance between points P and Q is 2√2 units.

Example 2: Finding a Point on an Axis (Medium)

Given: Points A(2, -5) and B(-2, 9).

To Find: The point on the x-axis which is equidistant from points A and B.

Solution:

1. Define the point on the x-axis. Any point on the x-axis has its y-coordinate as 0. Let the required point be P(x, 0).

2. The problem states that P is equidistant from A and B. This means the distance PA is equal to the distance PB.

   PA = PB
   

3. To avoid dealing with square roots, it's easier to work with the squares of the distances.

   PA² = PB²
   

4. Use the distance formula to write expressions for PA² and PB². For PA², using P(x, 0) and A(2, -5):

   PA² = (x - 2)² + (0 - (-5))² = (x - 2)² + (5)²
   

   For PB², using P(x, 0) and B(-2, 9):

   PB² = (x - (-2))² + (0 - 9)² = (x + 2)² + (-9)²
   

5. Now, set PA² equal to PB² and solve for x.

   (x - 2)² + 25 = (x + 2)² + 81
   

6. Expand the squared terms using the identities (a - b)² = a² - 2ab + b² and (a + b)² = a² + 2ab + b².

   (x² - 4x + 4) + 25 = (x² + 4x + 4) + 81
   

7. Simplify and solve for x. Notice that x² and 4 on both sides cancel out.

   x² - 4x + 29 = x² + 4x + 85
   -4x + 29 = 4x + 85
   -4x - 4x = 85 - 29
   -8x = 56
   x = 56 / -8
   x = -7
   

8. The point on the x-axis is P(x, 0). So, the required point is (-7, 0).

Final Answer: The point on the x-axis equidistant from A and B is (-7, 0).

{{VISUAL: diagram: A coordinate plane showing points A(2, -5) and B(-2, 9). Point P(-7, 0) is marked on the x-axis. Dotted lines of equal length are drawn from P to A and from P to B, with tick marks indicating PA = PB.}}

Example 3: Checking for Collinearity (Hard)

Given: Three points A(1, 5), B(2, 3), and C(-2, -11).

To Find: Determine if these three points are collinear.

Solution:

1. Understand the condition for collinearity. Three points A, B, and C are collinear (lie on the same straight line) if the sum of the distances between two pairs of points is equal to the distance between the third pair. That is, if AB + BC = AC (or AC + CB = AB, etc.).

2. Calculate the distance AB using A(1, 5) and B(2, 3).

   AB = √((2 - 1)² + (3 - 5)²) = √(1² + (-2)²) = √(1 + 4) = √5
   

3. Calculate the distance BC using B(2, 3) and C(-2, -11).

   BC = √((-2 - 2)² + (-11 - 3)²) = √((-4)² + (-14)²) = √(16 + 196) = √212
   

4. Calculate the distance AC using A(1, 5) and C(-2, -11).

   AC = √((-2 - 1)² + (-11 - 5)²) = √((-3)² + (-16)²) = √(9 + 256) = √265
   

   {{VISUAL: diagram: A sketch showing three points A, B, and C that are not in a straight line, forming a narrow scalene triangle. The calculated lengths √5, √212, and √265 are labeled on the corresponding sides AB, BC, and AC.}}

5. Check if the sum of any two distances equals the third. Is AB + BC = AC?

   √5 + √212 ≠ √265
   

   (Approximately 2.23 + 14.56 = 16.79, while √265 ≈ 16.27) Since the sum of the lengths of the two smaller segments is not equal to the length of the largest segment, the points do not form a straight line.

Final Answer: The points A(1, 5), B(2, 3), and C(-2, -11) are not collinear.

Example 4: Finding a Relation (Tricky)

Given: A point P(x, y) is equidistant from the points A(7, 1) and B(3, 5).

To Find: The relation between x and y.

Solution:

1. The problem states that the distance PA is equal to the distance PB.

   PA = PB
   

2. As in Example 2, we can square both sides to simplify the calculation.

   PA² = PB²
   

3. Use the distance formula to write expressions for PA² and PB². For PA², using P(x, y) and A(7, 1):

   PA² = (x - 7)² + (y - 1)²
   

   For PB², using P(x, y) and B(3, 5):

   PB² = (x - 3)² + (y - 5)²
   

4. Set the two expressions equal to each other.

   (x - 7)² + (y - 1)² = (x - 3)² + (y - 5)²
   

5. Expand all the squared binomials.

   (x² - 14x + 49) + (y² - 2y + 1) = (x² - 6x + 9) + (y² - 10y + 25)
   

6. Simplify by combining like terms. Notice that x² and y² appear on both sides and can be cancelled.

   -14x - 2y + 50 = -6x - 10y + 34
   

7. Rearrange the equation to bring all variables to one side and constants to the other. Let's move all terms to the left side.

   -14x + 6x - 2y + 10y + 50 - 34 = 0
   

8. Combine the terms to get the final relation.

   -8x + 8y + 16 = 0
   

9. This equation can be simplified by dividing the entire equation by -8 (or +8). Let's divide by -8.

   x - y - 2 = 0
   

   Alternatively, writing it as x - y = 2. This is the required linear relation between x and y. It represents the perpendicular bisector of the line segment AB.

Final Answer: The relation between x and y is x - y = 2.

Tips & Tricks

Master these shortcuts to solve problems faster and with more confidence.

Common Mistakes

Many students lose marks due to small, avoidable errors. Here’s what to watch out for.

Brain-Teaser Questions

Test your understanding with these slightly more challenging problems.

1. The points A(6, 1), B(8, 2) and C(9, 4) are three vertices of a parallelogram ABCD taken in order. Find the coordinates of the fourth vertex D.

💡 Answer: Let the coordinates of D be (x, y). In a parallelogram, the diagonals bisect each other. This means the midpoint of AC is the same as the midpoint of BD. Midpoint of AC = ((6+9)/2, (1+4)/2) = (15/2, 5/2). Midpoint of BD = ((8+x)/2, (2+y)/2). Equating them: (8+x)/2 = 15/2 → 8+x = 15 → x = 7. And (2+y)/2 = 5/2 → 2+y = 5 → y = 3. So, D = (7, 3).

2. Find the coordinates of the center of the circle passing through the points P(6, -6), Q(3, -7), and R(3, 3).

💡 Answer: Let the center of the circle be O(x, y). Since P, Q, and R lie on the circle, the distance from the center O to each point must be equal (this is the radius). So, OP = OQ = OR, which means OP² = OQ² = OR². From OP² = OQ²: (x-6)² + (y+6)² = (x-3)² + (y+7)². This simplifies to 6x + 2y = 10 or 3x + y = 5. From OQ² = OR²: (x-3)² + (y+7)² = (x-3)² + (y-3)². This simplifies to (y+7)² = (y-3)² → y²+14y+49 = y²-6y+9 → 20y = -40 → y = -2. Substitute y = -2 into 3x + y = 5: 3x - 2 = 5 → 3x = 7 → x = 7/3. The center is (7/3, -2).

3. If the distance between the points (4, p) and (1, 0) is 5, what are the possible values of p?

💡 Answer: Use the distance formula: 5 = √((4-1)² + (p-0)²). Square both sides: 25 = (3)² + (p)². 25 = 9 + p². p² = 25 - 9 = 16. Taking the square root, p = ±√16. So, the possible values of p are 4 and -4.

Mini Cheatsheet

Screenshot this table for your last-minute revision!

Distance Formula — Part 1

Chapter 7: Coordinate Geometry

Page 2 of 4: The Distance Formula — Part 1

{{FORMULA: expr=d = √((x₂ – x₁)² + (y₂ – y₁)²)| symbols=d:Distance, (x₁, y₁):Coordinates of first point, (x₂, y₂):Coordinates of second point}}

Concept Introduction

Imagine you're looking at a city map. Your friend is at a café, and you're at a library. The map tells you the café is 4 kilometers east and 6 kilometers north of your current location. How would you find the shortest straight-line distance between you and your friend? You can't just add 4 and 6, as that would be the distance if you only walked along the grid-like streets. The actual "as the crow flies" distance is the hypotenuse of a right-angled triangle formed by these east and north paths.

This is the core idea of Coordinate Geometry: using a coordinate system (like a map grid) to solve geometric problems with algebra. The most fundamental tool in this toolbox is the Distance Formula, which helps us calculate the straight-line distance between any two points, just like finding the direct path from the library to the café.

Definitions & Formulas

Before we derive the formula, let's define the terms we'll be using.

The primary formula we will build and use on this page is the Distance Formula:

d = √((x₂ – x₁)² + (y₂ – y₁)² )

Derivation of the Distance Formula

How can we find the distance between any two points P(x₁, y₁) and Q(x₂, y₂) on a plane? The answer lies in the timeless Pythagorean Theorem (a² + b² = c²). Let's build the formula from the ground up.

{{VISUAL: diagram: A Cartesian plane showing two points P(x₁, y₁) and Q(x₂, y₂) with a right-angled triangle PQT drawn between them. The horizontal leg PT has length (x₂ – x₁) and the vertical leg QT has length (y₂ – y₁). The distance PQ is the hypotenuse.}}

1. Plot the Points: Imagine two general points, P with coordinates (x₁, y₁) and Q with coordinates (x₂, y₂), on a Cartesian plane.

2. Construct a Right-Angled Triangle: Draw a horizontal line from P and a vertical line from Q. These lines will intersect at a point, let's call it T. The coordinates of T will be (x₂, y₁). Now, we have a right-angled triangle, ΔPQT, with the right angle at T. The distance PQ is the hypotenuse of this triangle.

3. Find the Length of the Horizontal Side (Base): The length of the side PT is the horizontal distance between P and T. Since P is at x₁ and T is at x₂ on the x-axis, the distance is the difference between their x-coordinates.

   Length of PT = |x₂ – x₁|
   

4. Find the Length of the Vertical Side (Height): Similarly, the length of the side QT is the vertical distance between Q and T. Since Q is at y₂ and T is at y₁ on the y-axis, the distance is the difference between their y-coordinates.

   Length of QT = |y₂ – y₁|
   

5. Apply the Pythagorean Theorem: In the right-angled triangle ΔPQT, the theorem states: (Hypotenuse)² = (Base)² + (Height)².

   PQ² = PT² + QT²
   

6. Substitute and Solve: Now, we substitute the lengths we found in steps 3 and 4. Since squaring a number makes it non-negative, we can drop the absolute value signs.

   PQ² = (x₂ – x₁)² + (y₂ – y₁)²
   

   To find the length of PQ, we take the square root of both sides. As distance must be a positive value, we only consider the positive square root.

   PQ = √((x₂ – x₁)² + (y₂ – y₁)² )
   

   And there we have it! This is the universal Distance Formula.

{{KEY: type=concept | title=The Logic Behind the Formula | text=The Distance Formula is not a magical rule to be memorized; it is a direct application of the Pythagorean Theorem on the Cartesian coordinate plane. By turning any two points into the vertices of a right-angled triangle, we can use the change in x-coordinates as one side, the change in y-coordinates as the other, and find the distance as the hypotenuse.}}

Solved Examples

Let's apply the formula to solve some problems, from simple to complex.

Example 1: Basic Distance Calculation (Easy)

Given: Two points, A(2, 5) and B(5, 9).

To Find: The distance AB.

Solution:

1. Identify the coordinates: x₁ = 2, y₁ = 5, x₂ = 5, y₂ = 9.

2. Apply the distance formula d = √((x₂ – x₁)² + (y₂ – y₁)² ).

   AB = √((5 – 2)² + (9 – 5)²)
   

3. Simplify the terms inside the parentheses.

   AB = √(3² + 4²)
   

4. Calculate the squares and add them.

   AB = √(9 + 16) = √25
   

5. Find the final square root.

   AB = 5
   

Final Answer:

The distance between A and B is 5 units.

Example 2: Distance from the Origin (Medium)

Given: A point P(–6, 8).

To Find: The distance of P from the origin O(0, 0).

Solution:

1. Identify the coordinates: The first point is P(–6, 8), so x₁ = -6, y₁ = 8. The second point is the origin O(0, 0), so x₂ = 0, y₂ = 0.

2. Apply the distance formula.

   OP = √((0 – (–6))² + (0 – 8)²)
   

3. Simplify the terms inside the parentheses. Be careful with the double negative.

   OP = √((0 + 6)² + (–8)²)
   

4. Calculate the squares.

   OP = √(6² + (–8)²) = √(36 + 64)
   

5. Add the results and find the square root.

   OP = √100 = 10
   

Final Answer:

The distance of P from the origin is 10 units.

Example 3: Identifying a Triangle Type (Hard)

Given: The vertices of a triangle are P(3, 2), Q(–2, –3), and R(2, 3).

To Find: If these points form a triangle, and if so, what type of triangle it is.

Solution:

1. First, we find the lengths of all three sides using the distance formula to confirm they can form a triangle (the sum of any two sides must be greater than the third).

2. Calculate the distance PQ.

   PQ = √((–2 – 3)² + (–3 – 2)²) = √((–5)² + (–5)²) = √(25 + 25) = √50
   

3. Calculate the distance QR.

   QR = √((2 – (–2))² + (3 – (–3))²) = √((2 + 2)² + (3 + 3)²) = √(4² + 6²) = √(16 + 36) = √52
   

4. Calculate the distance PR.

   PR = √((2 – 3)² + (3 – 2)²) = √((–1)² + 1²) = √(1 + 1) = √2
   

5. Check if they form a triangle. Since √50 ≈ 7.07, √52 ≈ 7.21, and √2 ≈ 1.41. The sum of any two sides is greater than the third (e.g., 7.07 + 1.41 > 7.21). So, a triangle is formed.

6. Now, let's check for properties. We can use the Converse of the Pythagorean Theorem. Let's check if the square of the longest side equals the sum of the squares of the other two sides.

   PQ² = 50
   QR² = 52
   PR² = 2
   

7. Check the relationship:

   PQ² + PR² = 50 + 2 = 52
   

   This is exactly equal to QR².

8. Since PQ² + PR² = QR², the triangle satisfies the Converse of the Pythagorean Theorem. The angle opposite the longest side (QR) is a right angle. Therefore, ∠P = 90°.

Final Answer:

The points form a right-angled triangle.

Example 4: Equidistant Points (Tricky)

Given: A point P(x, y) is equidistant from points A(7, 1) and B(3, 5).

To Find: A relationship between x and y.

Solution:

1. The problem states that the distance AP is equal to the distance BP.

   AP = BP
   

2. To avoid dealing with square roots, we can work with the squares of the distances. If the distances are equal, their squares must also be equal.

   AP² = BP²
   

3. Using the distance formula (without the final square root), express AP² and BP².

   For AP²: (x – 7)² + (y – 1)²
   For BP²: (x – 3)² + (y – 5)²
   

4. Now, set these two expressions equal to each other.

   (x – 7)² + (y – 1)² = (x – 3)² + (y – 5)²
   

5. Expand the squared binomials (a – b)² = a² – 2ab + b².

   (x² – 14x + 49) + (y² – 2y + 1) = (x² – 6x + 9) + (y² – 10y + 25)
   

6. The x² and y² terms appear on both sides, so they cancel out.

   –14x + 49 – 2y + 1 = –6x + 9 – 10y + 25
   

7. Combine the constant terms on each side.

   –14x – 2y + 50 = –6x – 10y + 34
   

8. Rearrange the equation to bring all variables to one side and constants to the other. Let's move x and y terms to the left and constants to the right.

   –14x + 6x – 2y + 10y = 34 – 50
   

   –8x + 8y = –16
   

9. Simplify the relation by dividing the entire equation by –8.

   x – y = 2
   

Final Answer:

The required relation is x – y = 2.

Tips & Tricks

Use these shortcuts to solve problems faster and more accurately.

Common Mistakes

Many students understand the concept but make small calculation errors. Here are a few to watch out for.

Brain-Teaser Questions

Test your understanding with these slightly more challenging problems.

1. Find the value of a if the distance between the points A(a, –1) and B(8, 7) is 10 units.

   💡 Answer: Set up the equation: 10 = √((8 – a)² + (7 – (–1))²). Square both sides: 100 = (8 – a)² + 8². 100 = (8 – a)² + 64. 36 = (8 – a)². Take the square root: ±6 = 8 – a. Case 1: 6 = 8 – a → a = 2. Case 2: –6 = 8 – a → a = 14. So, a can be 2 or 14.

2. Find the coordinates of the point on the y-axis which is equidistant from the points A(6, 5) and B(–4, 3).

   💡 Answer: A point on the y-axis has coordinates (0, y). Let this point be P(0, y). We are given PA = PB, so PA² = PB². PA² = (6 – 0)² + (5 – y)² = 36 + (5 – y)². PB² = (–4 – 0)² + (3 – y)² = 16 + (3 – y)². Set them equal: 36 + 25 – 10y + y² = 16 + 9 – 6y + y². 61 – 10y = 25 – 6y. 36 = 4y. y = 9. The point is (0, 9).

3. Show that the points A(–1, –1), B(–√3, √3), and C(1, 1) cannot form a triangle.

   💡 Answer: Calculate all three distances. AB = √((–√3 – (–1))² + (√3 – (–1))²) = √((1–√3)² + (1+√3)²) = √(1–2√3+3 + 1+2√3+3) = √8 = 2√2. BC = √((1 – (–√3))² + (1 – √3)²) = √((1+√3)² + (1–√3)²) = √(1+2√3+3 + 1–2√3+3) = √8 = 2√2. AC = √((1 – (–1))² + (1 – (–1))²) = √(2² + 2²) = √(4+4) = √8 = 2√2. Since all three sides are equal, the points form an equilateral triangle. The question states they "cannot form a triangle," which is incorrect; they form a specific type of triangle. The premise is flawed, but the exercise correctly applies the formula. Alternatively, if the points were collinear, they wouldn't form a triangle. Let's check collinearity: AB+BC = 2√2 + 2√2 = 4√2. This is not equal to AC=2√2. Therefore, they are not collinear and do form a triangle. The points indeed form an equilateral triangle.

Mini Cheatsheet

A quick summary of this page for your revision. Screenshot this!

Solved NCERT Exercises

{{FORMULA: expr=d = √((x₂ - x₁)² + (y₂ - y₁)²) | symbols=d:distance, (x₁, y₁):coordinates of the first point, (x₂, y₂):coordinates of the second point}}

Concept Introduction

Imagine you're looking at a city map laid out on a grid. Your house is at one point, and your friend's house is at another. You can't just measure the straight-line distance on the map with a ruler because the map has a scale. How do you find the actual straight-line distance, as a bird would fly?

Coordinate geometry gives us the tools to solve this. By representing your house and your friend's house as points with coordinates (like (2, 3) and (7, 8)), we can use a powerful formula to calculate the exact distance between them. This is the Distance Formula. It's not just for maps; it's the foundation for GPS navigation, video game design (calculating distances between characters), and even in fields like astronomy to measure distances between celestial objects on a 2D projection of space. It's a fundamental bridge between algebra and geometry.

Definitions & Formulas

The Distance Formula is derived from the Pythagorean Theorem and is used to find the straight-line distance between any two points in a Cartesian (x-y) plane.

The primary formula is:

d = √((x₂ - x₁)² + (y₂ - y₁)²)

A special, simplified case is finding the distance of a point (x, y) from the origin (0, 0):

d = √(x² + y²)

Derivation of the Distance Formula

The distance formula is essentially the Pythagorean theorem a² + b² = c² applied to the coordinate plane. Let's see how.

1. Consider two points in the Cartesian plane, P(x₁, y₁) and Q(x₂, y₂). Our goal is to find the length of the segment PQ.

   {{VISUAL: diagram: A right-angled triangle on a Cartesian plane with vertices P(x₁, y₁), Q(x₂, y₂), and R(x₂, y₁), showing the hypotenuse PQ.}}

2. Draw a horizontal line from point P and a vertical line from point Q. These lines intersect at a point, let's call it R. The coordinates of R will be (x₂, y₁).

3. This construction forms a right-angled triangle PQR, with the right angle at R. The distance PQ is the hypotenuse of this triangle.

4. Now, let's find the lengths of the other two sides, PR and QR.

   • The length of the horizontal side PR is the difference in the x-coordinates:

     Length(PR) = |x₂ - x₁|
     

   • The length of the vertical side QR is the difference in the y-coordinates:

     Length(QR) = |y₂ - y₁|
     

5. According to the Pythagorean theorem, for the right-angled triangle PQR:

   (PQ)² = (PR)² + (QR)²
   

6. Substitute the lengths we found into the theorem:

   d² = (x₂ - x₁)² + (y₂ - y₁)²
   

   Taking the square root of both sides gives us the final Distance Formula. Since distance must be positive, we only consider the positive root.

   d = √((x₂ - x₁)² + (y₂ - y₁)²)
   

{{KEY: type=concept | title=The Core Idea | text=The distance formula is just the Pythagorean theorem in disguise. The change in 'x' is one side of the triangle, the change in 'y' is the other side, and the distance between the two points is the hypotenuse.}}

Solved Examples

Example 1: Basic Distance Calculation (Easy)

Given: Two points, A(2, 5) and B(5, 9).

To Find: The distance between A and B.

Solution:

1. Identify the coordinates. Let (x₁, y₁) = (2, 5) and (x₂, y₂) = (5, 9).

2. Write down the distance formula.

   d = √((x₂ - x₁)² + (y₂ - y₁)²)
   

3. Substitute the values into the formula.

   d = √((5 - 2)² + (9 - 5)²)
   

4. Calculate the differences inside the parentheses.

   d = √(3² + 4²)
   

5. Square the numbers and add them.

   d = √(9 + 16)
   d = √25
   

6. Calculate the final square root.

   d = 5
   

Final Answer: The distance between points A and B is 5 units.

Example 2: Working with Negative Coordinates (Medium)

Given: Two points, P(-5, 7) and Q(-1, 3).

To Find: The distance between P and Q.

Solution:

1. Identify the coordinates. Let (x₁, y₁) = (-5, 7) and (x₂, y₂) = (-1, 3).

2. Substitute the values into the distance formula. Be careful with the negative signs.

   d = √((-1 - (-5))² + (3 - 7)²)
   

3. Simplify the terms inside the parentheses. Remember that subtracting a negative is the same as adding.

   d = √((-1 + 5)² + (-4)²)
   

4. Perform the addition and square the results.

   d = √(4² + (-4)²)
   d = √(16 + 16)
   

5. Add the numbers and simplify the radical.

   d = √32
   d = √(16 × 2)
   d = 4√2
   

Final Answer: The distance between points P and Q is 4√2 units.

Example 3: Verifying a Geometric Property (Hard)

Given: Points A(5, -2), B(6, 4), and C(7, -2).

To Find: Check if these points are the vertices of an isosceles triangle.

Solution:

An isosceles triangle has at least two sides of equal length. We need to calculate the lengths of all three sides (AB, BC, and AC) and see if any two are equal.

1. Calculate the distance AB: A(5, -2) and B(6, 4)

   AB = √((6 - 5)² + (4 - (-2))²)
   AB = √(1² + (4 + 2)²)
   AB = √(1² + 6²)
   AB = √(1 + 36)
   AB = √37
   

2. Calculate the distance BC: B(6, 4) and C(7, -2)

   BC = √((7 - 6)² + (-2 - 4)²)
   BC = √(1² + (-6)²)
   BC = √(1 + 36)
   BC = √37
   

3. Calculate the distance AC: A(5, -2) and C(7, -2)

   AC = √((7 - 5)² + (-2 - (-2))²)
   AC = √(2² + (-2 + 2)²)
   AC = √(2² + 0²)
   AC = √4
   AC = 2
   

4. Compare the lengths of the sides. We found that AB = √37 and BC = √37. Since two sides have the same length, the triangle is isosceles.

Final Answer: Yes, the points (5, -2), (6, 4), and (7, -2) are the vertices of an isosceles triangle because sides AB and BC both have a length of √37 units.

Example 4: Finding an Unknown Coordinate (Tricky)

Given: Point P(x, y) is equidistant from A(7, 1) and B(3, 5).

To Find: The relation between x and y.

Solution:

"Equidistant" means the distance PA is equal to the distance PB.

1. Set up the equality PA = PB. To avoid dealing with square roots, it's easier to work with PA² = PB².

   PA² = PB²
   

2. Use the distance formula (without the final square root) for PA² and PB². For PA²: points are P(x, y) and A(7, 1)

   (x - 7)² + (y - 1)²
   

   For PB²: points are P(x, y) and B(3, 5)

   (x - 3)² + (y - 5)²
   

3. Set the two expressions equal to each other.

   (x - 7)² + (y - 1)² = (x - 3)² + (y - 5)²
   

4. Expand the squared binomials using the identity (a - b)² = a² - 2ab + b².

   (x² - 14x + 49) + (y² - 2y + 1) = (x² - 6x + 9) + (y² - 10y + 25)
   

5. Simplify the equation. Notice that x² and y² appear on both sides, so they cancel out.

   -14x - 2y + 50 = -6x - 10y + 34
   

6. Rearrange the terms to bring all variables to one side and constants to the other.

   -14x + 6x - 2y + 10y = 34 - 50
   -8x + 8y = -16
   

7. Simplify the relation by dividing the entire equation by -8.

   x - y = 2
   

Final Answer: The relation between x and y is x - y = 2.

Tips & Tricks

Common Mistakes

Brain-Teaser Questions

1. Find the point on the y-axis which is equidistant from the points A(6, 5) and B(–4, 3).

   💡 Answer: A point on the y-axis has coordinates (0, y). We set the distance from (0, y) to A equal to the distance from (0, y) to B. (0 - 6)² + (y - 5)² = (0 - (-4))² + (y - 3)² 36 + y² - 10y + 25 = 16 + y² - 6y + 9 61 - 10y = 25 - 6y 36 = 4y y = 9. The point is (0, 9).

2. The vertices of a triangle are A(1, 1), B(7, 1), and C(1, 5). Find the coordinates of the circumcenter of this triangle. (Hint: The circumcenter is equidistant from all three vertices).

   💡 Answer: This is a right-angled triangle (right angle at A). The circumcenter of a right-angled triangle is the midpoint of its hypotenuse. The hypotenuse is BC. Midpoint of BC = ((7+1)/2, (1+5)/2) = (8/2, 6/2) = (4, 3). The circumcenter is (4, 3).

3. The x-coordinate of a point P is twice its y-coordinate. If P is equidistant from Q(2, –5) and R(–3, 6), find the coordinates of P.

   💡 Answer: Let the coordinates of P be (2y, y). Given PQ = PR, so PQ² = PR². (2y - 2)² + (y - (-5))² = (2y - (-3))² + (y - 6)² (2y - 2)² + (y + 5)² = (2y + 3)² + (y - 6)² 4y² - 8y + 4 + y² + 10y + 25 = 4y² + 12y + 9 + y² - 12y + 36 5y² + 2y + 29 = 5y² + 45 2y = 16 y = 8. Since x = 2y, x = 16. The point is (16, 8).

Mini Cheatsheet

Distance Formula — Part 2

Chapter 7: Coordinate Geometry

Page 3 of 4: The Distance Formula — Part 2: Applications

{{FORMULA: expr=√((x₂ – x₁)² + (y₂ – y₁)²) | symbols=x₁ y₁:coordinates of first point, x₂ y₂:coordinates of second point}}

Concept Introduction

Welcome back! On the last page, we derived the Distance Formula, a powerful tool for finding the length of a line segment between any two points on a coordinate plane. Now, we move from simply finding distances to using them. Think of a digital architect designing a building on a computer. They place vertices (corners) as coordinates. How do they verify if a wall is perfectly rectangular, not just a parallelogram? They use the distance formula to check if the diagonals are equal. How does a GPS system determine if three locations lie on the same straight road? It calculates the distances between them. In this section, we will use the distance formula as a verifier of geometric properties, allowing us to classify shapes and check alignments with mathematical certainty.

Definitions & Formulas

The core tool for this entire section remains the Distance Formula.

The Logic Behind the Formula

How does this formula work? It's a direct and elegant application of the Pythagoras Theorem. Let's quickly revisit the logic.

1. Imagine any two points, P(x₁, y₁) and Q(x₂, y₂), on a coordinate plane.

2. To find the distance PQ, we can form a right-angled triangle with PQ as the hypotenuse.

3. We draw a horizontal line from P and a vertical line from Q. Let them meet at a point T. The coordinates of T will be (x₂, y₁).

4. The length of the horizontal side PT is the difference in the x-coordinates: |x₂ – x₁|.

5. The length of the vertical side QT is the difference in the y-coordinates: |y₂ – y₁|.

6. By Pythagoras Theorem (hypotenuse² = base² + perpendicular²), we get: PQ² = PT² + QT² PQ² = (x₂ – x₁)² + (y₂ – y₁)² Taking the square root gives us the final formula. Since distance is always positive, we only consider the positive root.

{{VISUAL: diagram: A coordinate plane showing two points P(x₁, y₁) and Q(x₂, y₂). A right-angled triangle PTQ is formed with T at (x₂, y₁). The horizontal side PT is labeled |x₂ – x₁| and the vertical side QT is labeled |y₂ – y₁|. The hypotenuse PQ is the distance to be found.}}

Solved Examples

Here we apply the formula to solve problems, moving from simple shape identification to more abstract relations.

Example 1: Identifying a Triangle (Easy)

Given: Three points P(3, 2), Q(–2, –3), and R(2, 3).

To Find: Do these points form a triangle? If yes, what type of triangle is it?

Solution:

1. First, we find the lengths of the three sides PQ, QR, and PR using the distance formula. For side PQ, with P(3, 2) and Q(–2, –3):

   PQ = √((-2 – 3)² + (-3 – 2)²) = √((-5)² + (-5)²)
   

   PQ = √(25 + 25) = √50 = 5√2 ≈ 7.07 units
   

2. Next, we find the length of side QR, with Q(–2, –3) and R(2, 3):

   QR = √((2 – (-2))² + (3 – (-3))²) = √((4)² + (6)²)
   

   QR = √(16 + 36) = √52 = 2√13 ≈ 7.21 units
   

3. Finally, we find the length of side PR, with P(3, 2) and R(2, 3):

   PR = √((2 – 3)² + (3 – 2)²) = √((-1)² + (1)²)
   

   PR = √(1 + 1) = √2 ≈ 1.41 units
   

4. To check if they form a triangle, we use the triangle inequality theorem: the sum of any two sides must be greater than the third side.

   • PQ + PR = 7.07 + 1.41 = 8.48 > 7.21 (QR)
   • PQ + QR = 7.07 + 7.21 = 14.28 > 1.41 (PR)
   • QR + PR = 7.21 + 1.41 = 8.62 > 7.07 (PQ) Since the condition holds, the points form a triangle.

5. Now, we check the type of triangle. Let's test the converse of the Pythagoras Theorem. We check if the square of the longest side equals the sum of the squares of the other two sides. The longest side is QR.

   PQ² + PR² = (√50)² + (√2)² = 50 + 2 = 52
   

   QR² = (√52)² = 52
   

6. Since PQ² + PR² = QR², the triangle is a right-angled triangle, with the right angle at vertex P.

Final Answer:

Yes, the points form a right-angled triangle.

Example 2: Verifying a Square (Medium)

Given: Four points A(1, 7), B(4, 2), C(–1, –1), and D(–4, 4).

To Find: Show that these points are the vertices of a square.

Solution:

1. To prove a quadrilateral is a square, we must prove two properties:

   • All four sides are equal in length (AB = BC = CD = DA).
   • Both diagonals are equal in length (AC = BD). Let's calculate the lengths of the four sides.

2. Calculate side AB:

   AB = √((4 – 1)² + (2 – 7)²) = √(3² + (–5)²) = √(9 + 25) = √34 units
   

3. Calculate side BC:

   BC = √((-1 – 4)² + (-1 – 2)²) = √((-5)² + (–3)²) = √(25 + 9) = √34 units
   

4. Calculate side CD:

   CD = √((-4 – (-1))² + (4 – (-1))²) = √((-3)² + 5²) = √(9 + 25) = √34 units
   

5. Calculate side DA:

   DA = √((1 – (-4))² + (7 – 4)²) = √(5² + 3²) = √(25 + 9) = √34 units
   

   Since AB = BC = CD = DA = √34, all four sides are equal. This proves the figure is at least a rhombus.

6. Now, we must check the diagonals AC and BD to distinguish it from a rhombus.

{{VISUAL: diagram: A coordinate plane with points A(1,7), B(4,2), C(-1,-1), D(-4,4) forming a square. All four sides are labeled with length √34. The diagonals AC and BD are drawn as dashed lines and labeled with their lengths.}}

7. Calculate diagonal AC:

   AC = √((-1 – 1)² + (-1 – 7)²) = √((-2)² + (–8)²) = √(4 + 64) = √68 units
   

8. Calculate diagonal BD:

   BD = √((-4 – 4)² + (4 – 2)²) = √((-8)² + 2²) = √(64 + 4) = √68 units
   

9. Since AC = BD = √68, the diagonals are also equal. A quadrilateral with four equal sides and equal diagonals must be a square.

Final Answer:

Since AB = BC = CD = DA and AC = BD, the points A, B, C, and D are the vertices of a square.

Example 3: Checking for Collinearity (Hard)

Given: Three students are seated at points A(3, 1), B(6, 4), and C(8, 6).

To Find: Are the students seated in a straight line?

Solution:

1. Three points A, B, and C are collinear (lie on the same straight line) if the sum of the lengths of any two smaller line segments equals the length of the longest line segment. Here, we must check if AB + BC = AC.

{{VISUAL: diagram: A coordinate plane showing the arrangement of desks. Points A(3,1), B(6,4), and C(8,6) are plotted and connected by a straight line, illustrating collinearity. The lengths AB, BC, and AC are labeled.}}

2. First, calculate the distance AB:

   AB = √((6 – 3)² + (4 – 1)²) = √(3² + 3²) = √(9 + 9) = √18
   

   AB = 3√2 units
   

3. Next, calculate the distance BC:

   BC = √((8 – 6)² + (6 – 4)²) = √(2² + 2²) = √(4 + 4) = √8
   

   BC = 2√2 units
   

4. Finally, calculate the distance AC:

   AC = √((8 – 3)² + (6 – 1)²) = √(5² + 5²) = √(25 + 25) = √50
   

   AC = 5√2 units
   

5. Now, we check the condition for collinearity. We add the two shorter distances, AB and BC.

   AB + BC = 3√2 + 2√2 = (3 + 2)√2 = 5√2
   

6. Compare this sum to the longest distance, AC. We see that AB + BC = 5√2 and AC = 5√2.

7. Since AB + BC = AC, the points lie on the same straight line.

{{KEY: type=concept | title=Condition for Collinearity | text=Three points A, B, and C are collinear if the sum of the distances between two pairs of points is equal to the distance between the third pair. In other words, if AB + BC = AC, they form a straight line.}}

Final Answer:

Yes, they are seated in a line because the sum of distances AB and BC is equal to the distance AC.

Example 4: Finding an Equidistant Point (Tricky)

Given: A point P(x, y) is equidistant from points A(7, 1) and B(3, 5).

To Find: The relation between x and y.

Solution:

1. The problem states that the distance from P to A is equal to the distance from P to B. Mathematically, this means PA = PB.

2. To avoid dealing with square roots, we can work with the squares of the distances: PA² = PB².

{{VISUAL: diagram: A coordinate plane showing points A(7,1) and B(3,5). A generic point P(x,y) is shown, with dashed lines PA and PB of equal length. The perpendicular bisector of AB is also shown as a line, on which P must lie.}}

3. Using the distance formula, we can write the expression for PA²: P(x, y) and A(7, 1)

   PA² = (x – 7)² + (y – 1)²
   

4. Similarly, we can write the expression for PB²: P(x, y) and B(3, 5)

   PB² = (x – 3)² + (y – 5)²
   

5. Now, we set these two expressions equal to each other as per the condition PA² = PB².

   (x – 7)² + (y – 1)² = (x – 3)² + (y – 5)²
   

6. Expand the squared binomials on both sides. Remember (a – b)² = a² – 2ab + b².

   (x² – 14x + 49) + (y² – 2y + 1) = (x² – 6x + 9) + (y² – 10y + 25)
   

7. The x² and y² terms appear on both sides, so they cancel out. Now, simplify the remaining linear equation.

   –14x – 2y + 50 = –6x – 10y + 34
   

8. Group the x terms, y terms, and constant terms. It's common to move variables to the left and constants to the right.

   –14x + 6x – 2y + 10y = 34 – 50
   

   –8x + 8y = –16
   

9. Divide the entire equation by –8 to get the simplest relation.

   x – y = 2
   

   This is the linear equation that represents all possible points (x, y) that are equidistant from A and B.

Final Answer:

The required relation is x – y = 2.

Tips & Tricks

Common Mistakes

Brain-Teaser Questions

1. Find the point on the x-axis which is equidistant from the points (2, –5) and (–2, 9).

   💡 Answer: A point on the x-axis has coordinates (x, 0). Let this point be P(x, 0). Let A = (2, –5) and B = (–2, 9). We need PA² = PB². (x – 2)² + (0 – (–5))² = (x – (–2))² + (0 – 9)² (x – 2)² + 5² = (x + 2)² + (–9)² x² – 4x + 4 + 25 = x² + 4x + 4 + 81 –4x + 29 = 4x + 85 –8x = 56 → x = –7. The point is (–7, 0).

2. If the distance between the points P(2, –3) and Q(10, y) is 10 units, what are the possible values of y?

   💡 Answer: We are given PQ = 10, so PQ² = 100. (10 – 2)² + (y – (–3))² = 100 8² + (y + 3)² = 100 64 + (y + 3)² = 100 (y + 3)² = 36 y + 3 = ±√36 y + 3 = 6 or y + 3 = –6 y = 3 or y = –9. The possible values are 3 and –9.

3. The vertices of a triangle are A(5, 1), B(–3, –7) and C(7, –1). Find the length of the median through vertex A. (A median connects a vertex to the midpoint of the opposite side).

   💡 Answer: First, find the midpoint M of the side BC. Midpoint M = ((x₁ + x₂)/2, (y₁ + y₂)/2) = ((–3 + 7)/2, (–7 + (–1))/2) = (4/2, –8/2) = (2, –4). Now, find the length of the median AM, which is the distance between A(5, 1) and M(2, –4). AM = √((2 – 5)² + (–4 – 1)²) = √((–3)² + (–5)²) = √(9 + 25) = √34. The length of the median is √34 units.

Mini Cheatsheet

Solved NCERT Exercises

Q1. Find the distance between the following pairs of points : (i) (2, 3), (4, 1) (ii) (– 5, 7), (– 1, 3) (iii) (a, b), (– a, – b)

Solution: We will use the distance formula for all parts. The distance d between two points (x₁, y₁) and (x₂, y₂) is given by: d = √((x₂ - x₁)² + (y₂ - y₁)²)

(i) Points (2, 3) and (4, 1) Step 1: Identify the coordinates. Let (x₁, y₁) = (2, 3) and (x₂, y₂) = (4, 1).

Step 2: Substitute these values into the distance formula. d = √((4 - 2)² + (1 - 3)²)

Step 3: Simplify the expression. d = √(2² + (-2)²) d = √(4 + 4) d = √8

Step 4: Express the result in its simplest radical form. d = √(4 × 2) = 2√2

(ii) Points (– 5, 7) and (– 1, 3) Step 1: Identify the coordinates. Let (x₁, y₁) = (–5, 7) and (x₂, y₂) = (–1, 3).

Step 2: Substitute these values into the distance formula. Be careful with the negative signs. d = √((-1 - (-5))² + (3 - 7)²)

Step 3: Simplify the expression. d = √((-1 + 5)² + (-4)²) d = √(4² + (-4)²) d = √(16 + 16) d = √32

Step 4: Express the result in its simplest radical form. d = √(16 × 2) = 4√2

(iii) Points (a, b) and (– a, – b) Step 1: Identify the coordinates. Let (x₁, y₁) = (a, b) and (x₂, y₂) = (–a, –b).

Step 2: Substitute these values into the distance formula. d = √((-a - a)² + (-b - b)²)

Step 3: Simplify the expression. d = √((-2a)² + (-2b)²) d = √(4a² + 4b²)

Step 4: Factor out the common term and simplify. d = √(4(a² + b²)) d = 2√(a² + b²)

Final Answer: The distances are: (i) 2√2 units (ii) 4√2 units (iii) 2√(a² + b²) units

Q2. Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2.

Solution: Step 1: Identify the coordinates of the two points. The first point is the origin, O(0, 0). The second point is P(36, 15).

Step 2: Apply the distance formula. Let (x₁, y₁) = (0, 0) and (x₂, y₂) = (36, 15). Distance OP = √((36 - 0)² + (15 - 0)²)

Step 3: Simplify the expression inside the square root. OP = √(36² + 15²) OP = √(1296 + 225) OP = √1521

Step 4: Calculate the square root. OP = 39

So, the distance between (0, 0) and (36, 15) is 39 units.

Connecting to the Town Problem: Yes, we can find the distance between towns A and B. In Section 7.2, town A is considered as the origin (0, 0) and the coordinates of town B are given as (36, 15). The calculation we just performed is exactly the distance between these two towns, assuming the coordinates are in kilometers.

Final Answer: The distance between the points (0, 0) and (36, 15) is 39 units. Yes, the distance between town A and town B is 39 km.

Q3. Determine if the points (1, 5), (2, 3) and (– 2, – 11) are collinear.

Solution: For three points to be collinear, they must lie on the same straight line. This means that the sum of the distances between two pairs of points must be equal to the distance of the third pair. Let the points be A(1, 5), B(2, 3), and C(–2, –11). We need to check if AB + BC = AC, or AC + AB = BC, or BC + AC = AB.

Step 1: Calculate the distance between A(1, 5) and B(2, 3). AB = √((2 - 1)² + (3 - 5)²) AB = √(1² + (-2)²) AB = √(1 + 4) = √5

Step 2: Calculate the distance between B(2, 3) and C(–2, –11). BC = √((-2 - 2)² + (-11 - 3)²) BC = √((-4)² + (-14)²) BC = √(16 + 196) = √212

Step 3: Calculate the distance between A(1, 5) and C(–2, –11). AC = √((-2 - 1)² + (-11 - 5)²) AC = √((-3)² + (-16)²) AC = √(9 + 256) = √265

Step 4: Check for the collinearity condition. We compare the sum of the smaller distances with the largest distance. Does AB + BC = AC? √5 + √212 ≈ 2.236 + 14.56 ≈ 16.796 AC = √265 ≈ 16.279 Since √5 + √212 ≠ √265, the points are not collinear.

Final Answer: The points (1, 5), (2, 3) and (– 2, – 11) are not collinear.

Q4. Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.

Solution: An isosceles triangle is a triangle that has at least two sides of equal length. Let the given vertices be A(5, –2), B(6, 4), and C(7, –2). We will calculate the lengths of the three sides AB, BC, and AC.

Step 1: Calculate the length of side AB. AB = √((6 - 5)² + (4 - (-2))²) AB = √(1² + (4 + 2)²) AB = √(1² + 6²) AB = √(1 + 36) = √37

Step 2: Calculate the length of side BC. BC = √((7 - 6)² + (-2 - 4)²) BC = √(1² + (-6)²) BC = √(1 + 36) = √37

Step 3: Calculate the length of side AC. AC = √((7 - 5)² + (-2 - (-2))²) AC = √(2² + (-2 + 2)²) AC = √(2² + 0²) AC = √4 = 2

Step 4: Compare the lengths of the sides. We found that AB = √37 and BC = √37. Since two sides (AB and BC) have equal length, the triangle is isosceles.

Final Answer: Yes, the points (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.

Q5. In a classroom, 4 friends are seated at the points A, B, C and D as shown in Fig. 7.8. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using distance formula, find which of them is correct.

Solution: To determine if ABCD is a square, we must verify two properties:

1. All four sides are equal in length (AB = BC = CD = DA).
2. The two diagonals are equal in length (AC = BD).

Step 1: Determine the coordinates of the four points from the figure. By observing the grid, we can find the coordinates:

• A = (3, 4)
• B = (6, 7)
• C = (9, 4)
• D = (6, 1)

Step 2: Calculate the lengths of the four sides.

• AB = √((6 - 3)² + (7 - 4)²) = √(3² + 3²) = √(9 + 9) = √18 = 3√2
• BC = √((9 - 6)² + (4 - 7)²) = √(3² + (-3)²) = √(9 + 9) = √18 = 3√2
• CD = √((6 - 9)² + (1 - 4)²) = √((-3)² + (-3)²) = √(9 + 9) = √18 = 3√2
• DA = √((3 - 6)² + (4 - 1)²) = √((-3)² + 3²) = √(9 + 9) = √18 = 3√2

Since AB = BC = CD = DA = 3√2, all four sides are equal. This suggests the figure could be a rhombus or a square.

Step 3: Calculate the lengths of the two diagonals.

• AC = √((9 - 3)² + (4 - 4)²) = √(6² + 0²) = √36 = 6
• BD = √((6 - 6)² + (1 - 7)²) = √(0² + (-6)²) = √36 = 6

Since AC = BD = 6, the diagonals are also equal.

Step 4: Conclude based on the properties. A quadrilateral with all four sides equal and both diagonals equal is a square. Therefore, Champa's observation is correct.

Final Answer: Champa is correct. The figure ABCD is a square.

Q6. Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (i) (– 1, – 2), (1, 0), (– 1, 2), (– 3, 0) (ii) (–3, 5), (3, 1), (0, 3), (–1, – 4) (iii) (4, 5), (7, 6), (4, 3), (1, 2)

Solution: For each set of points, we will calculate the lengths of the four sides and the two diagonals to identify the type of quadrilateral.

(i) Points: A(–1, –2), B(1, 0), C(–1, 2), D(–3, 0)

Step 1: Calculate side lengths.

• AB = √((1 - (-1))² + (0 - (-2))²) = √(2² + 2²) = √8 = 2√2
• BC = √((-1 - 1)² + (2 - 0)²) = √((-2)² + 2²) = √8 = 2√2
• CD = √((-3 - (-1))² + (0 - 2)²) = √((-2)² + (-2)²) = √8 = 2√2
• DA = √((-1 - (-3))² + (-2 - 0)²) = √(2² + (-2)²) = √8 = 2√2

All four sides are equal. The quadrilateral is either a rhombus or a square.

Step 2: Calculate diagonal lengths.

• AC = √((-1 - (-1))² + (2 - (-2))²) = √(0² + 4²) = √16 = 4
• BD = √((-3 - 1)² + (0 - 0)²) = √((-4)² + 0²) = √16 = 4

The diagonals are also equal.

Step 3: Conclusion for (i). Since all four sides are equal (length 2√2) and both diagonals are equal (length 4), the quadrilateral is a square.

(ii) Points: A(–3, 5), B(3, 1), C(0, 3), D(–1, –4)

Step 1: Calculate side lengths.

• AB = √((3 - (-3))² + (1 - 5)²) = √(6² + (-4)²) = √(36 + 16) = √52 = 2√13
• BC = √((0 - 3)² + (3 - 1)²) = √((-3)² + 2²) = √(9 + 4) = √13
• CD = √((-1 - 0)² + (-4 - 3)²) = √((-1)² + (-7)²) = √(1 + 49) = √50 = 5√2
• DA = √((-3 - (-1))² + (5 - (-4))²) = √((-2)² + 9²) = √(4 + 81) = √85

The side lengths are all different. Let's check if any three points are collinear. For example, A, C, B.

• AC = √((0 - (-3))² + (3 - 5)²) = √(3² + (-2)²) = √(9 + 4) = √13
• CB = √13 (calculated as BC)
• AB = √52 = 2√13 Here, AC + CB = √13 + √13 = 2√13, which is equal to AB. This means points A, C, and B are collinear.

Step 3: Conclusion for (ii). Since three of the given points (A, C, B) lie on the same line, they cannot form a quadrilateral.

(iii) Points: A(4, 5), B(7, 6), C(4, 3), D(1, 2)

Step 1: Calculate side lengths.

• AB = √((7 - 4)² + (6 - 5)²) = √(3² + 1²) = √(9 + 1) = √10
• BC = √((4 - 7)² + (3 - 6)²) = √((-3)² + (-3)²) = √(9 + 9) = √18 = 3√2
• CD = √((1 - 4)² + (2 - 3)²) = √((-3)² + (-1)²) = √(9 + 1) = √10
• DA = √((4 - 1)² + (5 - 2)²) = √(3² + 3²) = √(9 + 9) = √18 = 3√2

Opposite sides are equal: AB = CD and BC = DA. The quadrilateral is a parallelogram.

Step 2: Calculate diagonal lengths to check if it's a rectangle.

• AC = √((4 - 4)² + (3 - 5)²) = √(0² + (-2)²) = √4 = 2
• BD = √((1 - 7)² + (2 - 6)²) = √((-6)² + (-4)²) = √(36 + 16) = √52

Since AC ≠ BD, the diagonals are not equal.

Step 3: Conclusion for (iii). Since opposite sides are equal but the diagonals are not equal, the quadrilateral is a parallelogram.

Final Answer: (i) Square, because all four sides are equal and diagonals are equal. (ii) Not a quadrilateral, because three points (A, C, B) are collinear. (iii) Parallelogram, because opposite sides are equal but diagonals are not equal.

Q7. Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9).

Solution: Step 1: Define the point on the x-axis. Any point on the x-axis has a y-coordinate of 0. Let the required point be P(x, 0). Let the given points be A(2, –5) and B(–2, 9).

Step 2: Set up the equidistance condition. We are given that P is equidistant from A and B, which means PA = PB. To avoid dealing with square roots, it's easier to work with the squares of the distances: PA² = PB².

Step 3: Apply the distance formula for PA² and PB². PA² = (x - 2)² + (0 - (-5))² = (x - 2)² + 5² PB² = (x - (-2))² + (0 - 9)² = (x + 2)² + (-9)²

Step 4: Equate the expressions and solve for x. (x - 2)² + 25 = (x + 2)² + 81 Expand the binomials: (x² - 4x + 4) + 25 = (x² + 4x + 4) + 81 x² - 4x + 29 = x² + 4x + 85

Step 5: Simplify the equation. The x² terms on both sides cancel out. -4x + 29 = 4x + 85 Move all x-terms to one side and constants to the other. 29 - 85 = 4x + 4x -56 = 8x x = -56 / 8 x = -7

The coordinates of the point P are (-7, 0).

Final Answer: The point on the x-axis is (-7, 0).

Q8. Find the values of y for which the distance between the points P(2, – 3) and Q(10, y) is 10 units.

Solution: Step 1: Use the given information to set up an equation. We are given the distance PQ = 10. Let's use the squared distance: PQ² = 10² = 100. The points are P(2, –3) and Q(10, y).

Step 2: Apply the distance formula. PQ² = (10 - 2)² + (y - (-3))²

Step 3: Substitute the known values and form the equation. 100 = (8)² + (y + 3)² 100 = 64 + (y + 3)²

Step 4: Solve the equation for y. 100 - 64 = (y + 3)² 36 = (y + 3)² Take the square root of both sides. ±√36 = y + 3 ±6 = y + 3

Step 5: Find the two possible values for y. Case 1: 6 = y + 3 y = 6 - 3 = 3

Case 2: -6 = y + 3 y = -6 - 3 = -9

The possible values for y are 3 and -9.

Final Answer: The values of y are 3 and -9.

Q9. If Q(0, 1) is equidistant from P(5, –3) and R(x, 6), find the values of x. Also find the distances QR and PR.

Solution: This question has two parts. First, we find x, then we find the required distances.

Part 1: Find the values of x

Step 1: Set up the equidistance condition. We are given that Q is equidistant from P and R, so QP = QR. This implies QP² = QR². Points are Q(0, 1), P(5, –3), and R(x, 6).

Step 2: Calculate QP² and QR² using the distance formula. QP² = (5 - 0)² + (-3 - 1)² = 5² + (-4)² = 25 + 16 = 41 QR² = (x - 0)² + (6 - 1)² = x² + 5² = x² + 25

Step 3: Equate the expressions and solve for x. QP² = QR² 41 = x² + 25 41 - 25 = x² 16 = x² x = ±√16 x = 4 or x = -4

So, there are two possible points for R: R₁(4, 6) and R₂(-4, 6).

Part 2: Find the distances QR and PR

We need to calculate these distances for both values of x.

Case 1: x = 4, so R is (4, 6)

Step 4: Find QR. Q(0, 1) and R(4, 6) QR = √((4 - 0)² + (6 - 1)²) = √(4² + 5²) = √(16 + 25) = √41 (Note: We already know QR² = 41 from Step 2, so this confirms our calculation).

Step 5: Find PR. P(5, –3) and R(4, 6) PR = √((4 - 5)² + (6 - (-3))²) = √((-1)² + 9²) = √(1 + 81) = √82

Case 2: x = -4, so R is (-4, 6)

Step 6: Find QR. Q(0, 1) and R(-4, 6) QR = √((-4 - 0)² + (6 - 1)²) = √((-4)² + 5²) = √(16 + 25) = √41 (This is the same as before, as expected from the initial condition).

Step 7: Find PR. P(5, –3) and R(-4, 6) PR = √((-4 - 5)² + (6 - (-3))²) = √((-9)² + 9²) = √(81 + 81) = √(2 × 81) = 9√2

Final Answer: The values of x are x = 4 and x = -4. The distances are:

• QR = √41
• When x = 4, PR = √82
• When x = -4, PR = 9√2

Q10. Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (– 3, 4).

Solution: Step 1: Define the points and the condition. Let the point be P(x, y). Let the given points be A(3, 6) and B(–3, 4). The condition is that P is equidistant from A and B, so PA = PB, which means PA² = PB².

Step 2: Use the distance formula to express PA² and PB². PA² = (x - 3)² + (y - 6)² PB² = (x - (-3))² + (y - 4)² = (x + 3)² + (y - 4)²

Step 3: Equate the two expressions. (x - 3)² + (y - 6)² = (x + 3)² + (y - 4)²

Step 4: Expand the squared binomials. (x² - 6x + 9) + (y² - 12y + 36) = (x² + 6x + 9) + (y² - 8y + 16)

Step 5: Simplify the equation by cancelling terms. The x² and y² terms on both sides cancel out. -6x - 12y + 45 = 6x - 8y + 25

Step 6: Rearrange the terms to form a linear relation. Move all terms to one side to get an equation in the form Ax + By + C = 0. 45 - 25 = 6x + 6x - 8y + 12y 20 = 12x + 4y

Step 7: Simplify the relation by dividing by the greatest common divisor, which is 4. 20 / 4 = (12x / 4) + (4y / 4) 5 = 3x + y

Or, expressed differently: 3x + y - 5 = 0

This is the required linear relation between x and y.

Final Answer: The relation between x and y is 3x + y - 5 = 0.

Section Formula

Page 4: Section Formula

Welcome to the next key tool in your coordinate geometry toolkit! So far, we've learned to measure the distance between two points. But what if we need to find a point that lies somewhere in between them, dividing the line segment in a specific proportion? This is where the Section Formula comes in.

Imagine a city planner designing a new park between two major landmarks. They don't want to place a new fountain randomly; they want it to be, for example, twice as far from Landmark A as it is from Landmark B. This creates a visually balanced and intentional design. The section formula is the mathematical tool that allows this planner to find the exact coordinates for the fountain. It's a powerful way to precisely locate a point that partitions a line segment in a given ratio.

{{FORMULA: expr=( (m₁x₂ + m₂x₁) / (m₁ + m₂), (m₁y₂ + m₂y₁) / (m₁ + m₂) ) | symbols=m₁:m₂:ratio of division, (x₁,y₁):coordinates of point A, (x₂,y₂):coordinates of point B}}

Definitions & Formulas

Before we derive this powerful formula, let's define the terms we'll be using.

The Section Formula: The coordinates of the point P(x, y) which divides the line segment joining A(x₁, y₁) and B(x₂, y₂) internally in the ratio m₁ : m₂ are given by:

x = (m₁x₂ + m₂x₁) / (m₁ + m₂)

y = (m₁y₂ + m₂y₁) / (m₁ + m₂)

Special Case: The Midpoint Formula If P is the midpoint of AB, it divides the segment in the ratio 1 : 1 (so m₁ = 1 and m₂ = 1). The formula simplifies to:

x = (x₁ + x₂) / 2

y = (y₁ + y₂) / 2

Derivation of the Section Formula

How do we arrive at this formula? The proof is a beautiful application of similar triangles from Chapter 6. Let's walk through it.

{{VISUAL: diagram: A coordinate plane showing points A(x₁, y₁) and B(x₂, y₂) connected by a line segment. Point P(x, y) lies on the segment. Perpendiculars are dropped from A, P, and B to the x-axis, and lines are drawn parallel to the x-axis from A and P to form two right-angled triangles, ΔPAQ and ΔBPC.}}

1. Setup and Construction Let A(x₁, y₁) and B(x₂, y₂) be two points. Let P(x, y) be a point on the line segment AB that divides it internally in the ratio m₁ : m₂. So, PA / PB = m₁ / m₂. We draw lines AR, PS, and BT perpendicular to the x-axis, and lines AQ and PC parallel to the x-axis.

2. Identifying Similar Triangles Observe the two triangles, ΔPAQ and ΔBPC.

   • ∠PAQ = ∠BPC (Corresponding angles, since AQ is parallel to PC).
   • ∠PQA = ∠BCP = 90° (By construction). By the AA similarity criterion, we can conclude that ΔPAQ is similar to ΔBPC.

   ΔPAQ ~ ΔBPC
   

3. Ratio of Corresponding Sides Since the triangles are similar, the ratio of their corresponding sides must be equal.

   PA / BP = AQ / PC = PQ / BC
   

4. Expressing Side Lengths in Terms of Coordinates Now, we express the lengths of these sides using the coordinates of the points.

   • AQ = RS = OS - OR = x - x₁
   • PC = ST = OT - OS = x₂ - x
   • PQ = PS - QS = PS - AR = y - y₁
   • BC = BT - CT = BT - PS = y₂ - y

5. Solving for the x-coordinate We substitute these values back into our ratio equation, using the first two parts and the given ratio m₁ / m₂.

   m₁ / m₂ = (x - x₁) / (x₂ - x)
   

   Now, we cross-multiply to solve for x.

   m₁(x₂ - x) = m₂(x - x₁)
   

   m₁x₂ - m₁x = m₂x - m₂x₁
   

   m₁x₂ + m₂x₁ = m₁x + m₂x
   

   m₁x₂ + m₂x₁ = x(m₁ + m₂)
   

   x = (m₁x₂ + m₂x₁) / (m₁ + m₂)
   

6. Solving for the y-coordinate Similarly, we use the first and third parts of the ratio equation to solve for y.

   m₁ / m₂ = (y - y₁) / (y₂ - y)
   

   Cross-multiplying and solving for y gives us:

   y = (m₁y₂ + m₂y₁) / (m₁ + m₂)
   

   This completes the derivation of the section formula.

{{KEY: type=concept | title=The "Cross-Multiplication" Pattern | text=Notice in the formula, the ratio part m₁ is multiplied by the coordinates of the second point (x₂, y₂), and m₂ is multiplied by the coordinates of the first point (x₁, y₁). Think of it as a criss-cross pattern to avoid mixing them up.}}

Solved Examples

Let's apply the formula to solve some problems, from straightforward to more complex.

Example 1: Direct Application (Easy)

Given: Endpoints are (4, –3) and (8, 5). The division ratio is 3 : 1.

To Find: The coordinates of the point that divides the line segment.

Solution:

1. Identify the given values: x₁ = 4, y₁ = -3, x₂ = 8, y₂ = 5, m₁ = 3, m₂ = 1.

2. Apply the section formula for the x-coordinate.

   x = (m₁x₂ + m₂x₁) / (m₁ + m₂)
   

   x = (3×8 + 1×4) / (3 + 1)
   

   x = (24 + 4) / 4 = 28 / 4 = 7
   

3. Apply the section formula for the y-coordinate.

   y = (m₁y₂ + m₂y₁) / (m₁ + m₂)
   

   y = (3×5 + 1×(-3)) / (3 + 1)
   

   y = (15 - 3) / 4 = 12 / 4 = 3
   

Final Answer: The required point is (7, 3).

Example 2: Finding the Ratio (Medium)

Given: Endpoints are A(–6, 10) and B(3, –8). The dividing point is P(–4, 6).

To Find: The ratio m₁ : m₂ in which P divides AB.

Solution:

1. Let the required ratio be m₁ : m₂. We can use the section formula for either the x-coordinate or the y-coordinate to find this ratio. Let's use the x-coordinate.

2. Set up the formula for x with the given values. Here, x = -4, x₁ = -6, and x₂ = 3.

   -4 = (m₁×3 + m₂×(-6)) / (m₁ + m₂)
   

3. Solve the equation for the ratio m₁ / m₂.

   -4(m₁ + m₂) = 3m₁ - 6m₂
   

   -4m₁ - 4m₂ = 3m₁ - 6m₂
   

4. Group the m₁ terms on one side and m₂ terms on the other.

   -4m₁ - 3m₁ = -6m₂ + 4m₂
   

   -7m₁ = -2m₂
   

   m₁ / m₂ = -2 / -7 = 2 / 7
   

5. (Verification) You can check this ratio using the y-coordinate. y = (2×(-8) + 7×10) / (2 + 7) = (-16 + 70) / 9 = 54 / 9 = 6. This matches the given y-coordinate of P.

Final Answer: The ratio is 2 : 7.

Example 3: Points of Trisection (Hard)

Given: Endpoints are A(2, –2) and B(–7, 4).

To Find: The coordinates of the points that divide AB into three equal parts (points of trisection).

Solution:

1. Let the points of trisection be P and Q. This means AP = PQ = QB.

2. The point P divides the segment AB in the ratio AP : PB. Since PB = PQ + QB = AP + AP = 2AP, the ratio is AP : 2AP, which is 1 : 2. So for point P, m₁ = 1 and m₂ = 2.

3. Calculate the coordinates of P using the section formula with ratio 1 : 2.

   xₚ = (1×(-7) + 2×2) / (1 + 2) = (-7 + 4) / 3 = -3 / 3 = -1
   

   yₚ = (1×4 + 2×(-2)) / (1 + 2) = (4 - 4) / 3 = 0 / 3 = 0
   

   So, P is (-1, 0).

4. The point Q divides the segment AB in the ratio AQ : QB. Since AQ = AP + PQ = 2QB, the ratio is 2QB : QB, which is 2 : 1. So for point Q, m₁ = 2 and m₂ = 1.

5. Calculate the coordinates of Q using the section formula with ratio 2 : 1.

   xₐ = (2×(-7) + 1×2) / (2 + 1) = (-14 + 2) / 3 = -12 / 3 = -4
   

   yₐ = (2×4 + 1×(-2)) / (2 + 1) = (8 - 2) / 3 = 6 / 3 = 2
   

   So, Q is (-4, 2).

Final Answer: The points of trisection are (–1, 0) and (–4, 2).

Example 4: Division by an Axis (Tricky)

Given: Endpoints are (5, –6) and (–1, –4).

To Find: The ratio in which the y-axis divides the line segment and the point of intersection.

Solution:

1. Let the required ratio be k : 1. We use k : 1 instead of m₁ : m₂ to simplify the algebra. Let the point of intersection on the y-axis be P(x, y).

2. We know that any point on the y-axis has its x-coordinate (abscissa) equal to 0. So, x = 0.

3. Use the section formula for the x-coordinate with m₁ = k and m₂ = 1. The endpoints are (x₁, y₁) = (5, -6) and (x₂, y₂) = (-1, -4).

   x = (k×(-1) + 1×5) / (k + 1)
   

4. Set the expression for x equal to 0 and solve for k.

   0 = (-k + 5) / (k + 1)
   

   0 = -k + 5
   

   k = 5
   

   The ratio k : 1 is 5 : 1.

5. Now, find the point of intersection by substituting k = 5 into the formula for the y-coordinate.

   y = (k×(-4) + 1×(-6)) / (k + 1)
   

   y = (5×(-4) - 6) / (5 + 1)
   

   y = (-20 - 6) / 6 = -26 / 6 = -13 / 3
   

Final Answer: The ratio is 5 : 1 and the point of intersection is (0, –13/3).

Tips & Tricks

Common Mistakes

Brain-Teaser Questions

1. The points A(2, 1), B(8, 1), C(8, 7), and D(2, 7) are the vertices of a rectangle. A point P lies on the diagonal AC such that AP = ¼ AC. Find the coordinates of P.

   💡 Answer: If AP = ¼ AC, then PC = ¾ AC. The ratio AP : PC is (¼ AC) : (¾ AC), which simplifies to 1 : 3. Using the section formula on diagonal AC with endpoints A(2, 1) and C(8, 7) and ratio 1 : 3: x = (1×8 + 3×2) / (1 + 3) = (8 + 6) / 4 = 14/4 = 3.5 y = (1×7 + 3×1) / (1 + 3) = (7 + 3) / 4 = 10/4 = 2.5 The coordinates of P are (3.5, 2.5).

2. Find the ratio in which the line x - y - 2 = 0 divides the line segment joining the points A(3, –1) and B(8, 9).

   💡 Answer: Let the ratio be k : 1. The coordinates of the dividing point P are x = (8k + 3)/(k+1) and y = (9k - 1)/(k+1). Since point P lies on the line x - y - 2 = 0, its coordinates must satisfy the equation. Substituting x and y: [(8k + 3)/(k+1)] - [(9k - 1)/(k+1)] - 2 = 0. Multiply by (k+1): (8k + 3) - (9k - 1) - 2(k + 1) = 0. 8k + 3 - 9k + 1 - 2k - 2 = 0. -3k + 2 = 0 → 3k = 2 → k = 2/3. The ratio k : 1 is (2/3) : 1, which is 2 : 3.

3. Point P(9a - 2, -b) divides the line segment joining A(3a + 1, -3) and B(8a, 5) in the ratio 3 : 1. Find the values of a and b.

   💡 Answer: Using the section formula. For the x-coordinate: 9a - 2 = (3×(8a) + 1×(3a + 1)) / (3 + 1) 9a - 2 = (24a + 3a + 1) / 4 4(9a - 2) = 27a + 1 → 36a - 8 = 27a + 1 → 9a = 9 → a = 1. For the y-coordinate: -b = (3×5 + 1×(-3)) / (3 + 1) -b = (15 - 3) / 4 → -b = 12 / 4 → -b = 3 → b = -3. So, a = 1 and b = -3.

Mini Cheatsheet

Solved NCERT Exercises

Concept Introduction

Imagine you are looking at a map on your phone. You have two points, your home (Point A) and your friend's house (Point B). Now, suppose you both agree to meet at a coffee shop (Point P) that is exactly one-third of the way from your house to your friend's. How does your phone's GPS calculate the exact coordinates of that coffee shop?

This is precisely the problem that the Section Formula solves. It's a powerful tool in coordinate geometry that helps us find the coordinates of a point that divides a line segment, joining two other points, in a specific ratio. It's not just for maps; it's used in computer graphics to draw lines, in physics to find the center of mass of systems, and in engineering design. It's the mathematical rule for finding a precise point between two other known points.

{{FORMULA: expr=((m₁x₂ + m₂x₁) / (m₁ + m₂), (m₁y₂ + m₂y₁) / (m₁ + m₂)) | symbols= (x, y): coordinates of the dividing point, (x₁, y₁): coordinates of the first point, (x₂, y₂): coordinates of the second point, m₁:m₂: the given ratio}}

Definitions & Formulas

The Section Formula helps us pinpoint a location on a line segment. To use it correctly, we must first understand the terms involved.

The formula itself is:

x = (m₁x₂ + m₂x₁) / (m₁ + m₂)

y = (m₁y₂ + m₂y₁) / (m₁ + m₂)

Derivation of the Section Formula

Let's see how this formula comes to life. It’s not magic; it’s a beautiful application of similar triangles.

Consider a line segment AB with endpoints A(x₁, y₁) and B(x₂, y₂). Let the point P(x, y) divide AB internally in the ratio m₁ : m₂. This means AP / PB = m₁ / m₂.

{{VISUAL: diagram: A coordinate plane showing points A(x₁, y₁) and B(x₂, y₂), with point P(x, y) on the line segment AB. Perpendiculars are dropped from A, P, and B to the x-axis, forming two similar right-angled triangles.}}

1. Construct Triangles: Draw perpendiculars from A, P, and B to the x-axis, meeting at points C, Q, and R respectively. Also, draw a line AS parallel to the x-axis intersecting PQ at S, and a line PT parallel to the x-axis intersecting BR at T.

2. Identify Similar Triangles: Now, observe the two triangles, ΔASP and ΔPTB.

   • ∠PAS = ∠TPB (Corresponding angles, as AS || PT and AB is a transversal)
   • ∠ASP = ∠PTB = 90° (By construction)
   • Therefore, by AA similarity criterion, ΔASP ~ ΔPTB.

3. Use Properties of Similar Triangles: Since the triangles are similar, the ratio of their corresponding sides must be equal.

   AP / PB = AS / PT = SP / TB
   

4. Express Sides in Terms of Coordinates: Let's find the lengths of these sides using the coordinates.

   • AP / PB = m₁ / m₂ (This is given)
   • AS = CQ = OQ − OC = x − x₁
   • PT = QR = OR − OQ = x₂ − x
   • SP = PQ − SQ = y − y₁
   • TB = BR − TR = y₂ − y

5. Formulate Equations: Now substitute these values back into the ratio equation from Step 3.

   For the x-coordinate: m₁ / m₂ = AS / PT = (x − x₁) / (x₂ − x)

   For the y-coordinate: m₁ / m₂ = SP / TB = (y − y₁) / (y₂ − y)

6. Solve for x and y: Let's solve the first equation for x.

   • m₁(x₂ − x) = m₂(x − x₁)
   • m₁x₂ − m₁x = m₂x − m₂x₁
   • m₁x₂ + m₂x₁ = m₁x + m₂x
   • m₁x₂ + m₂x₁ = x(m₁ + m₂)

   This gives us the formula for the x-coordinate:

   x = (m₁x₂ + m₂x₁) / (m₁ + m₂)
   

   Similarly, solving the second equation gives the formula for the y-coordinate:

   y = (m₁y₂ + m₂y₁) / (m₁ + m₂)
   

Solved Examples

Let's apply the formula to solve some problems, starting from easy and moving to more complex ones.

Example 1: Direct Application

Given: Endpoints of a line segment are (–1, 7) and (4, –3), and the ratio is 2 : 3.

To Find: The coordinates of the point that divides the line segment.

Solution:

1. Identify the given values. Here, x₁ = –1, y₁ = 7, x₂ = 4, y₂ = –3, m₁ = 2, and m₂ = 3.

2. Apply the section formula for the x-coordinate.

   x = (m₁x₂ + m₂x₁) / (m₁ + m₂) = (2×4 + 3×(–1)) / (2 + 3)
   

3. Calculate the value of x.

   x = (8 – 3) / 5 = 5 / 5 = 1
   

4. Apply the section formula for the y-coordinate.

   y = (m₁y₂ + m₂y₁) / (m₁ + m₂) = (2×(–3) + 3×7) / (2 + 3)
   

5. Calculate the value of y.

   y = (–6 + 21) / 5 = 15 / 5 = 3
   

Final Answer: The coordinates of the point are (1, 3).

Example 2: Finding the Mid-Point

Given: Endpoints of a line segment are (8, –2) and (2, 6).

To Find: The coordinates of the mid-point of the line segment.

Solution:

1. Understand that a mid-point divides the line segment in the ratio 1 : 1. So, m₁ = 1 and m₂ = 1.

2. Apply the section formula, which simplifies to the mid-point formula: x = (x₁ + x₂) / 2 and y = (y₁ + y₂) / 2.

3. Calculate the x-coordinate.

   x = (8 + 2) / 2 = 10 / 2 = 5
   

4. Calculate the y-coordinate.

   y = (–2 + 6) / 2 = 4 / 2 = 2
   

Final Answer: The coordinates of the mid-point are (5, 2).

Example 3: Finding the Ratio

Given: Endpoints A(–3, 10) and B(6, –8), and the dividing point P(–1, 6).

To Find: The ratio in which P divides the line segment AB.

Solution:

1. Let the ratio be k : 1. This is a common technique to simplify calculations. Here, m₁ = k and m₂ = 1.

2. Use the section formula for the x-coordinate. We know x = –1, x₁ = –3, and x₂ = 6.

   x = (kx₂ + 1x₁) / (k + 1)
   

3. Substitute the known values and solve for k.

   –1 = (k×6 + 1×(–3)) / (k + 1)
   

   –1(k + 1) = 6k – 3
   

   –k – 1 = 6k – 3
   

   3 – 1 = 6k + k
   

   2 = 7k  =>  k = 2/7
   

4. The ratio k : 1 is (2/7) : 1, which is the same as 2 : 7. We can verify this using the y-coordinate as well, but it's not necessary.

Final Answer: The ratio is 2 : 7.

Example 4: Trisection of a Line Segment

Given: Endpoints of a line segment are A(4, –1) and B(–2, –3).

To Find: The coordinates of the points of trisection.

Solution:

1. "Trisection" means dividing the segment into three equal parts. Let the points be P and Q such that AP = PQ = QB.

2. Point P divides the segment AB in the ratio 1 : 2 (since AP is one part and PB is two parts).

3. Calculate the coordinates of P using the section formula with m₁ = 1 and m₂ = 2.

   • x-coordinate of P:

   x_p = (1×(–2) + 2×4) / (1 + 2) = (–2 + 8) / 3 = 6 / 3 = 2
   

   • y-coordinate of P:

   y_p = (1×(–3) + 2×(–1)) / (1 + 2) = (–3 – 2) / 3 = –5 / 3
   

   So, P = (2, –5/3).

4. Point Q divides the segment AB in the ratio 2 : 1 (since AQ is two parts and QB is one part).

5. Calculate the coordinates of Q using the section formula with m₁ = 2 and m₂ = 1.

   • x-coordinate of Q:

   x_q = (2×(–2) + 1×4) / (2 + 1) = (–4 + 4) / 3 = 0 / 3 = 0
   

   • y-coordinate of Q:

   y_q = (2×(–3) + 1×(–1)) / (2 + 1) = (–6 – 1) / 3 = –7 / 3
   

   So, Q = (0, –7/3).

Final Answer: The points of trisection are (2, –5/3) and (0, –7/3).

{{KEY: type=concept | title=The Cross-Multiplication Pattern | text=To avoid mixing up terms, remember this pattern: m₁ multiplies with the coordinates of the second point (x₂, y₂), and m₂ multiplies with the coordinates of the first point (x₁, y₁). Think of it as a cross-connection.}}

Tips & Tricks

Here are a few shortcuts and strategies to solve problems faster and more accurately.

Common Mistakes

Many students make similar errors when first learning the Section Formula. Here’s what to watch out for.

Brain-Teaser Questions

Test your understanding with these slightly challenging problems.

1. If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.

   💡 Answer: The diagonals of a parallelogram bisect each other. This means the mid-point of the diagonal connecting (1, 2) and (x, 6) is the same as the mid-point of the diagonal connecting (4, y) and (3, 5). Mid-point 1: ((1+x)/2, (2+6)/2) = ((1+x)/2, 4) Mid-point 2: ((4+3)/2, (y+5)/2) = (7/2, (y+5)/2) Equating the coordinates: (1+x)/2 = 7/2 => 1+x = 7 => x = 6 4 = (y+5)/2 => 8 = y+5 => y = 3 So, x = 6 and y = 3.

2. Point P divides the line segment joining the points A(2, 1) and B(5, –8) such that AP/AB = 1/3. Find the coordinates of P.

   💡 Answer: The given ratio is AP/AB = 1/3. This is not the ratio m₁:m₂. We need AP/PB. If AP is 1 part and the whole AB is 3 parts, then PB must be AB - AP = 3 - 1 = 2 parts. So, the required ratio m₁ : m₂ is AP : PB = 1 : 2. Now, use the section formula: x = (1×5 + 2×2) / (1+2) = (5+4)/3 = 3 y = (1×(–8) + 2×1) / (1+2) = (–8+2)/3 = –2 The coordinates of P are (3, –2).

3. Find the ratio in which the line x – y – 2 = 0 divides the line segment joining the points (3, –1) and (8, 9).

   💡 Answer: Let the ratio be k:1. The coordinates of the dividing point P are: x = (k×8 + 1×3) / (k+1) = (8k+3)/(k+1) y = (k×9 + 1×(–1)) / (k+1) = (9k–1)/(k+1) Since point P lies on the line x – y – 2 = 0, its coordinates must satisfy the equation. (8k+3)/(k+1) – (9k–1)/(k+1) – 2 = 0 Multiply the entire equation by (k+1): (8k+3) – (9k–1) – 2(k+1) = 0 8k + 3 – 9k + 1 – 2k – 2 = 0 –3k + 2 = 0 3k = 2 => k = 2/3 The ratio k:1 is (2/3):1, which is 2:3.

Mini Cheatsheet

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