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Coordinate Geometry – CBSE NCERT Study Resources

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Study Materials

9th

9th - Mathematics

Coordinate Geometry

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Overview of the Chapter

This chapter introduces students to the fundamental concepts of Coordinate Geometry, which is a branch of mathematics that combines algebra and geometry. It covers the Cartesian plane, plotting points, and understanding the relationship between algebraic equations and geometric figures.

Coordinate Geometry: A system that uses numerical coordinates to represent points, lines, and shapes on a plane.

Key Concepts

  • Cartesian Plane and Axes
  • Coordinates of a Point
  • Plotting Points on the Plane
  • Distance Between Two Points
  • Section Formula (Midpoint)

Cartesian Plane and Axes

The Cartesian plane consists of two perpendicular number lines intersecting at the origin (0,0). The horizontal line is called the x-axis, and the vertical line is called the y-axis.

Origin: The point (0,0) where the x-axis and y-axis intersect.

Coordinates of a Point

Any point on the Cartesian plane is represented by an ordered pair (x, y), where 'x' is the abscissa (distance from the y-axis) and 'y' is the ordinate (distance from the x-axis).

Plotting Points on the Plane

To plot a point (x, y), move 'x' units along the x-axis and 'y' units parallel to the y-axis. Positive values indicate movement to the right or upward, while negative values indicate movement to the left or downward.

Distance Between Two Points

The distance between two points (x₁, y₁) and (x₂, y₂) is calculated using the formula:

Distance Formula: √[(x₂ - x₁)² + (y₂ - y₁)²]

Section Formula (Midpoint)

The midpoint of a line segment joining two points (x₁, y₁) and (x₂, y₂) is given by:

Midpoint Formula: [(x₁ + x₂)/2, (y₁ + y₂)/2]

All Question Types with Solutions – CBSE Exam Pattern

Explore a complete set of CBSE-style questions with detailed solutions, categorized by marks and question types. Ideal for exam preparation, revision and practice.

Very Short Answer (1 Mark) – with Solutions (CBSE Pattern)

These are 1-mark questions requiring direct, concise answers. Ideal for quick recall and concept clarity.

Question 1:
What is the abscissa of point (5, -3)?
Answer:
5
Question 2:
Name the quadrant where point (-2, 4) lies.
Answer:
II quadrant
Question 3:
Find the distance of point (0, 5) from the x-axis.
Answer:
5 units
Question 4:
What are the coordinates of the origin?
Answer:
(0, 0)
Question 5:
If a point lies on the y-axis, what is its x-coordinate?
Answer:
0
Question 6:
Find the mirror image of (3, -7) over the x-axis.
Answer:
(3, 7)
Question 7:
What is the perpendicular distance of (4, 0) from the y-axis?
Answer:
4 units
Question 8:
In which quadrant do both coordinates have the same sign?
Answer:
I or III quadrant
Question 9:
Find the coordinates where the line x = 3 meets the x-axis.
Answer:
(3, 0)
Question 10:
What is the ordinate of any point on the x-axis?
Answer:
0
Question 11:
If (a, b) lies in IV quadrant, what is the sign of 'a' and 'b'?
Answer:
a (+), b (-)
Question 12:
Find the point where both coordinates are equal and sum is 8.
Answer:
(4, 4)
Question 13:
What is the abscissa of a point in the Cartesian plane?
Answer:

The abscissa is the x-coordinate of a point in the Cartesian plane. It represents the horizontal distance from the y-axis.

Question 14:
Identify the quadrant where the point (-3, 5) lies.
Answer:

The point (-3, 5) lies in Quadrant II because the x-coordinate is negative and the y-coordinate is positive.

Question 15:
What is the distance of the point (4, 0) from the y-axis?
Answer:

The distance of (4, 0) from the y-axis is 4 units, as the y-axis is where x = 0.

Question 16:
If a point lies on the x-axis, what is its y-coordinate?
Answer:

If a point lies on the x-axis, its y-coordinate is always 0.

Question 17:
Find the ordinates of the point (7, -2).
Answer:

The ordinate of (7, -2) is -2, which is the y-coordinate of the point.

Question 18:
What are the coordinates of the origin in the Cartesian plane?
Answer:

The coordinates of the origin are (0, 0), where both the x-axis and y-axis intersect.

Question 19:
If a point has coordinates (0, -5), on which axis does it lie?
Answer:

The point (0, -5) lies on the y-axis because its x-coordinate is 0.

Question 20:
What is the mirror image of the point (2, 3) over the x-axis?
Answer:

The mirror image of (2, 3) over the x-axis is (2, -3), as the y-coordinate changes sign.

Question 21:
Name the quadrant where both x and y coordinates are negative.
Answer:

In Quadrant III, both the x-coordinate and y-coordinate are negative.

Question 22:
What is the perpendicular distance of the point (6, 8) from the x-axis?
Answer:

The perpendicular distance of (6, 8) from the x-axis is 8 units, which is the absolute value of its y-coordinate.

Very Short Answer (2 Marks) – with Solutions (CBSE Pattern)

These 2-mark questions test key concepts in a brief format. Answers are expected to be accurate and slightly descriptive.

Question 1:
If a point lies on the x-axis, what is the value of its ordinate?
Answer:

The ordinate (y-coordinate) of any point on the x-axis is always 0.
This is because the x-axis is defined by the equation y = 0.

Question 2:
Write the coordinates of the origin in the Cartesian plane.
Answer:

The coordinates of the origin are (0, 0).
It is the point where the x-axis and y-axis intersect.

Question 3:
What is the distance of the point (3, 4) from the origin?
Answer:

Using the distance formula:
Distance = √(x² + y²)
= √(3² + 4²)
= √(9 + 16)
= √25
= 5 units.

Question 4:
In which quadrant does the point (-2, 5) lie?
Answer:

The point (-2, 5) lies in the second quadrant.

  • Quadrant II: x is negative, y is positive.

Question 5:
If the abscissa and ordinate of a point are equal and positive, in which quadrant does it lie?
Answer:

The point lies in the first quadrant.

  • Quadrant I: Both x and y are positive.

Question 6:
Find the mirror image of the point (5, -3) across the x-axis.
Answer:

The mirror image across the x-axis changes the sign of the y-coordinate.
New coordinates: (5, 3).

Question 7:
What is the condition for a point to lie on the y-axis?
Answer:

A point lies on the y-axis if its x-coordinate is 0.
Example: (0, 4) lies on the y-axis.

Question 8:
Calculate the midpoint of the line segment joining (1, 2) and (5, 8).
Answer:

Using the midpoint formula:
Midpoint = ((x₁ + x₂)/2, (y₁ + y₂)/2)
= ((1 + 5)/2, (2 + 8)/2)
= (6/2, 10/2)
= (3, 5).

Question 9:
If the distance between (0, 0) and (a, 0) is 6 units, what is the value of a?
Answer:

Using the distance formula:
√((a - 0)² + (0 - 0)²) = 6
√(a²) = 6
|a| = 6
Thus, a = 6 or a = -6.

Short Answer (3 Marks) – with Solutions (CBSE Pattern)

These 3-mark questions require brief explanations and help assess understanding and application of concepts.

Question 1:
Plot the points A(3, 4), B(-2, 5), and C(0, -3) on the Cartesian plane. Identify the quadrant or axis where each point lies.
Answer:

To plot the points on the Cartesian plane:


A(3, 4): 3 units right on the x-axis, 4 units up on the y-axis. Lies in Quadrant I.
B(-2, 5): 2 units left on the x-axis, 5 units up on the y-axis. Lies in Quadrant II.
C(0, -3): On the y-axis (since x = 0), 3 units down. Lies on the negative y-axis.
Question 2:
Find the distance between the points P(5, -2) and Q(-1, 3) using the distance formula.
Answer:

Using the distance formula:
Distance = √[(x₂ - x₁)² + (y₂ - y₁)²]
Here, P(5, -2) and Q(-1, 3).

Step 1: Subtract x-coordinates → (-1 - 5) = -6
Step 2: Subtract y-coordinates → (3 - (-2)) = 5
Step 3: Square both → (-6)² = 36 and (5)² = 25
Step 4: Add and take square root → √(36 + 25) = √61

Thus, the distance is √61 units.

Question 3:
If the point M(a, b) lies on the x-axis, what can you conclude about the value of b? Justify your answer.
Answer:

If a point lies on the x-axis, its y-coordinate must be 0 because the x-axis is defined by the equation y = 0.

Since M(a, b) lies on the x-axis, b = 0.

Thus, the point is M(a, 0), where a can be any real number.

Question 4:
Determine the coordinates of the point which divides the line segment joining A(2, -3) and B(6, 5) in the ratio 1:3 internally.
Answer:

Using the section formula for internal division:
Coordinates = [(m₁x₂ + m₂x₁)/(m₁ + m₂), (m₁y₂ + m₂y₁)/(m₁ + m₂)]
Here, m₁ = 1, m₂ = 3, A(2, -3), and B(6, 5).

Step 1: Calculate x-coordinate → [(1×6 + 3×2)/(1 + 3)] = (6 + 6)/4 = 12/4 = 3
Step 2: Calculate y-coordinate → [(1×5 + 3×(-3))/(1 + 3)] = (5 - 9)/4 = -4/4 = -1

Thus, the coordinates are (3, -1).

Question 5:
Verify whether the points D(1, 5), E(2, 3), and F(-2, -11) are collinear using the area of triangle method.
Answer:

To check collinearity, calculate the area formed by the points:
Area = ½ |x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|
Substitute D(1, 5), E(2, 3), and F(-2, -11).

Step 1: Compute terms → 1(3 - (-11)) + 2(-11 - 5) + (-2)(5 - 3)
Step 2: Simplify → 1(14) + 2(-16) + (-2)(2) = 14 - 32 - 4 = -22
Step 3: Absolute value → ½ |-22| = 11

Since the area is 11 sq. units (≠ 0), the points are non-collinear.

Question 6:
Find the area of the triangle formed by the points D(1, 2), E(4, 5), and F(2, -1) using the area formula.
Answer:

Using the area formula for a triangle:
Area = ½ |x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|
Substitute the values:
= ½ |1(5 - (-1)) + 4((-1) - 2) + 2(2 - 5)|
= ½ |1(6) + 4(-3) + 2(-3)|
= ½ |6 - 12 - 6|
= ½ |-12|
= 6 square units.
Thus, the area is 6 square units.

Long Answer (5 Marks) – with Solutions (CBSE Pattern)

These 5-mark questions are descriptive and require detailed, structured answers with proper explanation and examples.

Question 1:
Plot the points A(3, 4), B(-2, 5), and C(0, -3) on the Cartesian plane. Verify if they lie on a straight line. Explain your reasoning.
Answer:
Introduction

We studied how to plot points on the Cartesian plane using their coordinates. Let's verify if the given points are collinear.


Argument 1
  • Plot points A(3,4), B(-2,5), and C(0,-3) on graph paper.
  • Draw lines AB and BC.

Argument 2

Calculate the slope of AB: (5-4)/(-2-3) = -1/5. Slope of BC: (-3-5)/(0+2) = -4. Since slopes are unequal, points are not collinear.


Conclusion

Our textbook shows that collinear points must have equal slopes. Here, slopes differ, so A, B, C do not lie on a straight line.

Question 2:
Derive the distance formula between two points A(x₁, y₁) and B(x₂, y₂) using the Pythagoras theorem. Show each step clearly.
Answer:
Introduction

We studied the Pythagoras theorem in triangles. Let's use it to derive the distance formula.


Argument 1
  • Plot A(x₁,y₁) and B(x₂,y₂) on the plane.
  • Draw a right triangle with AB as hypotenuse.

Argument 2

Horizontal distance = |x₂ - x₁|. Vertical distance = |y₂ - y₁|. Using Pythagoras: AB² = (x₂-x₁)² + (y₂-y₁)².


Conclusion

Our textbook shows that taking square root gives the distance formula: AB = √[(x₂-x₁)² + (y₂-y₁)²].

Question 3:
A town's map shows the school at (2, 5), library at (-1, 3), and park at (4, -2). Find which two locations are farthest apart. Justify mathematically.
Answer:
Introduction

We can use the distance formula to compare distances between locations on a map.


Argument 1
  • School-Library distance: √[(-1-2)² + (3-5)²] = √(9+4) = √13 units.
  • School-Park distance: √[(4-2)² + (-2-5)²] = √(4+49) = √53 units.

Argument 2

Library-Park distance: √[(4+1)² + (-2-3)²] = √(25+25) = √50 units.


Conclusion

Our textbook shows √53 is largest. Thus, the school and park are farthest apart.

Question 4:
Derive the distance formula between two points P(x₁, y₁) and Q(x₂, y₂) using the Pythagoras theorem. Explain each step.
Answer:
Introduction

We studied the Pythagoras theorem, which helps find distances in right-angled triangles. Here, we use it to derive the distance formula.


Argument 1
  • Plot P(x₁,y₁) and Q(x₂,y₂) on the Cartesian plane.
  • Draw a right-angled triangle with PQ as the hypotenuse.
  • The horizontal leg length = |x₂ - x₁|, and the vertical leg length = |y₂ - y₁|.

Argument 2

Using Pythagoras theorem: PQ² = (x₂ - x₁)² + (y₂ - y₁)². Taking square roots, PQ = √[(x₂ - x₁)² + (y₂ - y₁)²]. This is the distance formula.


Conclusion

Our textbook shows this derivation clearly. The formula helps calculate distances between any two points on the plane.

Question 5:
A town has three landmarks at A(1, 2), B(4, 5), and C(7, 8). Prove that these landmarks are collinear using the section formula.
Answer:
Introduction

We studied the section formula to divide a line segment in a given ratio. Here, we use it to check collinearity of points A(1,2), B(4,5), and C(7,8).


Argument 1
  • Assume B divides AC in ratio k:1.
  • Using section formula: x-coordinate of B = (7k + 1)/(k+1) = 4.
  • Solving, we get k = 1.

Argument 2

Similarly, y-coordinate of B = (8k + 2)/(k+1) = 5. Substituting k=1 gives 5=5, which is true. Thus, B divides AC in 1:1 ratio, proving collinearity.


Conclusion

Our textbook shows that if a point divides two others in the same ratio, they are collinear. Hence, A, B, and C lie on a straight line.

Question 6:
Plot the points A(3, 4), B(-2, 5), and C(0, -3) on the Cartesian plane. Determine the quadrants in which these points lie and justify your answer.
Answer:
Introduction

We studied how to plot points on the Cartesian plane using their coordinates. The plane is divided into four quadrants based on the signs of x and y.


Argument 1
  • Point A(3, 4): Both x and y are positive, so it lies in Quadrant I.
  • Point B(-2, 5): x is negative, y is positive, placing it in Quadrant II.

Argument 2
  • Point C(0, -3): x is zero, so it lies on the y-axis, not in any quadrant.

Conclusion

Our textbook shows that quadrant determination depends on coordinate signs. A and B lie in Quadrants I and II, while C is on the axis.

Question 7:
Derive the distance formula between two points P(x₁, y₁) and Q(x₂, y₂) using the Pythagoras theorem. Explain each step clearly.
Answer:
Introduction

We studied the Pythagoras theorem, which helps calculate distances in right-angled triangles. Here, we apply it to derive the distance formula.


Argument 1
  • Plot P and Q on the plane. Draw horizontal and vertical lines to form a right triangle.
  • The horizontal side length is |x₂ - x₁|, and the vertical side is |y₂ - y₁|.

Argument 2
  • Using Pythagoras theorem: PQ² = (x₂ - x₁)² + (y₂ - y₁)².
  • Take the square root to get PQ = √[(x₂ - x₁)² + (y₂ - y₁)²].

Conclusion

Our textbook shows this derivation clearly. The distance formula is essential for coordinate geometry problems.

Question 8:
A town's map uses a coordinate grid where the school is at (2, -1) and the library at (-3, 4). Calculate the distance between them and explain how this applies in real-life scenarios.
Answer:
Introduction

We studied how coordinate grids model real-world locations. Here, we calculate the distance between two key points.


Argument 1
  • Using the distance formula: √[(-3 - 2)² + (4 - (-1))²] = √[(-5)² + (5)²] = √(25 + 25) = √50 ≈ 7.07 units.

Argument 2
  • In real life, this helps plan routes, like determining the shortest path between two places.
  • Our textbook shows similar applications in urban planning and navigation.

Conclusion

The distance is approximately 7.07 units. Coordinate geometry simplifies real-world distance calculations.

Question 9:
Plot the points A(3, 4), B(-2, 5), and C(0, -3) on the Cartesian plane. Determine the quadrant or axis where each point lies. Explain with steps.
Answer:
Introduction

We studied how to plot points on the Cartesian plane using ordered pairs (x, y). The plane is divided into four quadrants and two axes.


Argument 1
  • Point A(3, 4): x > 0, y > 0 → Lies in Quadrant I.
  • Point B(-2, 5): x < 0, y > 0 → Lies in Quadrant II.

Argument 2
  • Point C(0, -3): x = 0 → Lies on the y-axis.

Conclusion

Our textbook shows similar examples. Plotting helps visualize coordinates in real-life applications like map navigation.

Question 10:
Derive the distance formula between two points P(x₁, y₁) and Q(x₂, y₂) using the Pythagoras theorem. Show steps clearly.
Answer:
Introduction

We studied the Pythagoras theorem for right triangles. The distance formula calculates the length between two points.


Argument 1

Plot P(x₁, y₁) and Q(x₂, y₂). Draw a right triangle with PQ as the hypotenuse. The legs are (x₂ - x₁) and (y₂ - y₁).


Argument 2

Using Pythagoras: PQ² = (x₂ - x₁)² + (y₂ - y₁)². Taking square root gives PQ = √[(x₂ - x₁)² + (y₂ - y₁)²].


Conclusion

Our textbook derives this formula. It is useful in real-life scenarios like finding distances between cities on a map.

Question 11:
Find the coordinates of the midpoint of the line segment joining A(5, -2) and B(-3, 8). Verify using the section formula.
Answer:
Introduction

We studied the midpoint formula, which finds the center point of a line segment. It is a special case of the section formula.


Argument 1

Midpoint M = [(x₁ + x₂)/2, (y₁ + y₂)/2]. For A(5, -2) and B(-3, 8), M = [(5 + (-3))/2, (-2 + 8)/2] = (1, 3).


Argument 2

Using the section formula (m:n = 1:1), M = [(1*-3 + 1*5)/2, (1*8 + 1*-2)/2] = (1, 3). Both methods match.


Conclusion

Our textbook shows similar problems. Midpoints are used in real-life applications like finding the center of a field.

Question 12:
Plot the points A(3, 4), B(-2, 5), C(-1, -6), and D(4, -3) on the Cartesian plane.

Join them in order (A to B to C to D to A) and name the shape formed. Also, calculate its area using the distance formula.

Answer:

To solve this question, follow these steps:


Step 1: Plotting the Points
Draw the Cartesian plane with x-axis and y-axis.
Mark the points as follows:
- A(3, 4): 3 units right on x-axis, 4 units up on y-axis.
- B(-2, 5): 2 units left on x-axis, 5 units up on y-axis.
- C(-1, -6): 1 unit left on x-axis, 6 units down on y-axis.
- D(4, -3): 4 units right on x-axis, 3 units down on y-axis.

Step 2: Joining the Points
Connect A to B, B to C, C to D, and D back to A.
The shape formed is a quadrilateral (specifically, a kite due to its symmetry).

Step 3: Calculating Area Using Distance Formula
First, find the lengths of the diagonals:
- Diagonal AC: Distance between A(3, 4) and C(-1, -6)
Formula: √[(x₂ - x₁)² + (y₂ - y₁)²]
= √[(-1 - 3)² + (-6 - 4)²]
= √[(-4)² + (-10)²]
= √[16 + 100] = √116 ≈ 10.77 units

- Diagonal BD: Distance between B(-2, 5) and D(4, -3)
= √[(4 - (-2))² + (-3 - 5)²]
= √[(6)² + (-8)²]
= √[36 + 64] = √100 = 10 units

Area of kite = (½) × d₁ × d₂
= (½) × 10.77 × 10
≈ 53.85 square units.

Note: The exact area is (½) × √116 × 10, but decimal approximation is acceptable.

Question 13:
Plot the points A(3, 4), B(-2, 5), C(-1, -6), and D(4, -3) on the Cartesian plane. Identify the quadrant or axis in which each point lies and explain how you determined this.
Answer:

To plot the given points on the Cartesian plane, we follow these steps:


Step 1: Understand the coordinates
Each point is given as (x, y), where x is the abscissa (horizontal distance) and y is the ordinate (vertical distance).

Step 2: Plotting the points
  • A(3, 4): Move 3 units right on the x-axis and 4 units up on the y-axis.
  • B(-2, 5): Move 2 units left on the x-axis and 5 units up on the y-axis.
  • C(-1, -6): Move 1 unit left on the x-axis and 6 units down on the y-axis.
  • D(4, -3): Move 4 units right on the x-axis and 3 units down on the y-axis.

Step 3: Identify the quadrant or axis
  • A(3, 4): Both x and y are positive, so it lies in the Quadrant I.
  • B(-2, 5): x is negative, y is positive, so it lies in the Quadrant II.
  • C(-1, -6): Both x and y are negative, so it lies in the Quadrant III.
  • D(4, -3): x is positive, y is negative, so it lies in the Quadrant IV.

Note: If a point lies on the x-axis, its y-coordinate is 0. If it lies on the y-axis, its x-coordinate is 0. None of the given points lie on the axes.

Question 14:
Plot the points A(3, 4), B(6, 7), and C(9, 4) on the Cartesian plane.
(i) Name the type of triangle formed by these points.
(ii) Calculate the area of the triangle ABC.
Answer:

To solve this problem, follow the steps below:



Step 1: Plotting the Points
Draw the Cartesian plane with x-axis and y-axis.
Mark the points:
- A(3, 4): 3 units right on x-axis, 4 units up on y-axis.
- B(6, 7): 6 units right on x-axis, 7 units up on y-axis.
- C(9, 4): 9 units right on x-axis, 4 units up on y-axis.

Step 2: Identifying the Triangle
After plotting, observe the shape:
- Points A and C have the same y-coordinate (4), so AC is a horizontal line.
- Point B is directly above the midpoint of AC, forming an isosceles triangle (since AB = BC).

Step 3: Calculating the Area
Use the formula for area of a triangle:
Area = ½ × base × height

Base (AC) = Distance between A(3, 4) and C(9, 4):
AC = 9 - 3 = 6 units.

Height = Difference in y-coordinates of B and A:
Height = 7 - 4 = 3 units.

Area = ½ × 6 × 3 = 9 square units.

Conclusion: The triangle is isosceles, and its area is 9 square units.

Question 15:
Plot the points A(3, 4), B(-2, 5), C(0, -3), and D(-1, -4) on the Cartesian plane. Identify the quadrant or axis where each point lies and justify your answer with reasoning.
Answer:

To plot the given points on the Cartesian plane, follow these steps:


Step 1: Understand the Cartesian Plane
The Cartesian plane is divided into four quadrants:
  • Quadrant I: (+x, +y)
  • Quadrant II: (-x, +y)
  • Quadrant III: (-x, -y)
  • Quadrant IV: (+x, -y)
Points on the axes do not belong to any quadrant.

Step 2: Plot and Analyze Each Point
A(3, 4):
1. Move 3 units right on the x-axis.
2. Move 4 units up on the y-axis.
Since both coordinates are positive, A(3, 4) lies in Quadrant I.

B(-2, 5):
1. Move 2 units left on the x-axis.
2. Move 5 units up on the y-axis.
Since x is negative and y is positive, B(-2, 5) lies in Quadrant II.

C(0, -3):
1. The x-coordinate is 0, so it lies on the y-axis.
2. Move 3 units down on the y-axis.
Since x = 0, C(0, -3) lies on the negative y-axis (not in any quadrant).

D(-1, -4):
1. Move 1 unit left on the x-axis.
2. Move 4 units down on the y-axis.
Since both coordinates are negative, D(-1, -4) lies in Quadrant III.

Conclusion:
  • A(3, 4): Quadrant I
  • B(-2, 5): Quadrant II
  • C(0, -3): Negative y-axis
  • D(-1, -4): Quadrant III
Question 16:
Plot the points A(3, 4), B(-2, 5), C(-1, -6), and D(4, -3) on the Cartesian plane.

Identify the quadrants in which these points lie and explain how you determined their positions.

Answer:

To plot the given points on the Cartesian plane, follow these steps:


Step 1: Understand the coordinates
Each point is given as (x, y), where x is the abscissa (horizontal distance) and y is the ordinate (vertical distance).

Step 2: Plot the points
  • A(3, 4): Move 3 units right on the x-axis and 4 units up on the y-axis.
  • B(-2, 5): Move 2 units left on the x-axis and 5 units up on the y-axis.
  • C(-1, -6): Move 1 unit left on the x-axis and 6 units down on the y-axis.
  • D(4, -3): Move 4 units right on the x-axis and 3 units down on the y-axis.

Step 3: Identify the quadrants
The Cartesian plane is divided into four quadrants:
  • Quadrant I: (+x, +y) → A(3, 4) lies here.
  • Quadrant II: (-x, +y) → B(-2, 5) lies here.
  • Quadrant III: (-x, -y) → C(-1, -6) lies here.
  • Quadrant IV: (+x, -y) → D(4, -3) lies here.

Explanation: The signs of the coordinates determine the quadrant. If both coordinates are positive, the point is in Quadrant I. If x is negative and y is positive, it’s in Quadrant II. If both are negative, it’s in Quadrant III. If x is positive and y is negative, it’s in Quadrant IV.

Question 17:
Plot the points A(3, 4), B(-2, 5), C(-1, -6)), and D(4, -3) on the Cartesian plane.

Join them in order to form a quadrilateral ABCD and name the type of quadrilateral formed. Justify your answer with mathematical reasoning.

Answer:

To plot the given points on the Cartesian plane:


Step 1: Draw the X-axis (horizontal) and Y-axis (vertical) with a suitable scale.
Step 2: Plot the points as follows:
  • A(3, 4): 3 units right on X-axis, 4 units up on Y-axis.
  • B(-2, 5): 2 units left on X-axis, 5 units up on Y-axis.
  • C(-1, -6): 1 unit left on X-axis, 6 units down on Y-axis.
  • D(4, -3): 4 units right on X-axis, 3 units down on Y-axis.

Step 3: Join the points in order (A → B → C → D → A) to form quadrilateral ABCD.

The quadrilateral formed is a kite. Here's the justification:


1. Check the lengths of sides using the distance formula:
Distance between A(3,4) and B(-2,5):
√[(-2-3)² + (5-4)²] = √[25 + 1] = √26

Distance between B(-2,5) and C(-1,-6):
√[(-1+2)² + (-6-5)²] = √[1 + 121] = √122

Distance between C(-1,-6) and D(4,-3):
√[(4+1)² + (-3+6)²] = √[25 + 9] = √34

Distance between D(4,-3) and A(3,4):
√[(3-4)² + (4+3)²] = √[1 + 49] = √50

2. Observe the pairs of adjacent sides:
AB = √26 and AD = √50 (not equal)
BC = √122 and CD = √34 (not equal)

However, if we calculate the diagonals:
AC = √[(-1-3)² + (-6-4)²] = √[16 + 100] = √116
BD = √[(4+2)² + (-3-5)²] = √[36 + 64] = √100 = 10

The quadrilateral has two distinct pairs of adjacent sides that are equal (though not all sides are equal), and one diagonal (BD) bisects the other (AC) at right angles, which are properties of a kite.

Question 18:
Plot the points A(3, 4), B(-2, 5), C(-1, -6), and D(4, -3) on the Cartesian plane.

Join them in order to form quadrilateral ABCD and name the type of quadrilateral formed. Justify your answer with mathematical reasoning.

Answer:

To solve this question, follow these steps:


Step 1: Plotting the Points
1. Draw the Cartesian plane with x-axis (horizontal) and y-axis (vertical).
2. Locate point A(3, 4) by moving 3 units right on the x-axis and 4 units up on the y-axis.
3. Locate point B(-2, 5) by moving 2 units left on the x-axis and 5 units up on the y-axis.
4. Locate point C(-1, -6) by moving 1 unit left on the x-axis and 6 units down on the y-axis.
5. Locate point D(4, -3) by moving 4 units right on the x-axis and 3 units down on the y-axis.

Step 2: Joining the Points
Connect the points in order: A → B → C → D → A to form quadrilateral ABCD.

Step 3: Identifying the Quadrilateral
To determine the type of quadrilateral, analyze the sides and angles:
  • Calculate the lengths of all sides using the distance formula: √[(x₂ - x₁)² + (y₂ - y₁)²].
    For example, length of AB = √[(-2 - 3)² + (5 - 4)²] = √[25 + 1] = √26.
  • Similarly, calculate BC, CD, and DA.
  • If opposite sides are equal and parallel, it could be a parallelogram or rectangle.
  • Check slopes of sides to verify parallelism (slopes of AB and CD should be equal, slopes of BC and DA should be equal).
  • If diagonals AC and BD are equal in length, it confirms a rectangle.

After calculations, if the quadrilateral satisfies the above conditions, it is a rectangle. If only opposite sides are equal and parallel, it is a parallelogram.


Conclusion: Based on the calculations, quadrilateral ABCD is a parallelogram (or rectangle, if applicable). Justify with the derived measurements and properties.
Question 19:
Plot the points A(3, 4), B(-2, 5), C(0, -3), and D(-1, -4) on the Cartesian plane. Identify the quadrant or axis where each point lies and justify your answer.
Answer:

To plot the given points on the Cartesian plane, follow these steps:


  • Draw the x-axis (horizontal) and y-axis (vertical) intersecting at the origin (0, 0).
  • Mark the points using their coordinates (x, y).

Now, let's identify the quadrant or axis for each point:


  1. A(3, 4):
    x-coordinate = 3 (positive), y-coordinate = 4 (positive).
    Since both coordinates are positive, point A lies in the first quadrant (I).

  2. B(-2, 5):
    x-coordinate = -2 (negative), y-coordinate = 5 (positive).
    Since x is negative and y is positive, point B lies in the second quadrant (II).

  3. C(0, -3):
    x-coordinate = 0, y-coordinate = -3 (negative).
    Since x is 0, point C lies on the y-axis (specifically, the negative y-axis).

  4. D(-1, -4):
    x-coordinate = -1 (negative), y-coordinate = -4 (negative).
    Since both coordinates are negative, point D lies in the third quadrant (III).

Remember: Quadrant I (+, +), Quadrant II (-, +), Quadrant III (-, -), and Quadrant IV (+, -). Points on the x-axis have y = 0, and points on the y-axis have x = 0.

Question 20:
Plot the points A(3, 4), B(-2, 5), C(-1, -6), and D(4, -3) on the Cartesian plane.

Join them in order (A to B to C to D to A) and name the shape formed. Also, calculate its area using the appropriate formula.

Answer:

To solve this question, follow these steps:


Step 1: Plotting the Points
Draw the Cartesian plane with x-axis and y-axis.
Mark the points as follows:
- A(3, 4): 3 units right on x-axis, 4 units up on y-axis.
- B(-2, 5): 2 units left on x-axis, 5 units up on y-axis.
- C(-1, -6): 1 unit left on x-axis, 6 units down on y-axis.
- D(4, -3): 4 units right on x-axis, 3 units down on y-axis.

Step 2: Joining the Points
Connect the points in order: A → B → C → D → A.
The shape formed is a quadrilateral (specifically, a kite due to its symmetry).

Step 3: Calculating Area
Use the shoelace formula for polygons:
Area = ½ |(x₁y₂ + x₂y₃ + x₃y₄ + x₄y₁) - (y₁x₂ + y₂x₃ + y₃x₄ + y₄x₁)|
Substitute the coordinates:
= ½ |(3×5 + (-2)×(-6) + (-1)×(-3) + 4×4) - (4×(-2) + 5×(-1) + (-6)×4 + (-3)×3)|
= ½ |(15 + 12 + 3 + 16) - (-8 - 5 - 24 - 9)|
= ½ |46 - (-46)|
= ½ × 92
= 46 square units.

Thus, the area of quadrilateral ABCD is 46 square units.

Question 21:
Plot the points A(3, 4), B(-2, 5), C(-1, -6), and D(4, -3) on the Cartesian plane.

Identify the quadrant or axis where each point lies and explain how you determined this.

Also, calculate the distance between points A and C using the distance formula.
Answer:

To plot the given points on the Cartesian plane, follow these steps:


Step 1: Plotting the Points
1. A(3, 4): Move 3 units right on the x-axis and 4 units up on the y-axis.
2. B(-2, 5): Move 2 units left on the x-axis and 5 units up on the y-axis.
3. C(-1, -6): Move 1 unit left on the x-axis and 6 units down on the y-axis.
4. D(4, -3): Move 4 units right on the x-axis and 3 units down on the y-axis.

Step 2: Identifying Quadrants/Axes
  • A(3, 4): Lies in Quadrant I because both x and y coordinates are positive.
  • B(-2, 5): Lies in Quadrant II because x is negative and y is positive.
  • C(-1, -6): Lies in Quadrant III because both x and y coordinates are negative.
  • D(4, -3): Lies in Quadrant IV because x is positive and y is negative.

Step 3: Calculating Distance Between A and C
Using the distance formula: d = √[(x₂ - x₁)² + (y₂ - y₁)²]
For A(3, 4) and C(-1, -6):
1. x₁ = 3, y₁ = 4
2. x₂ = -1, y₂ = -6
3. Substitute into the formula:
d = √[(-1 - 3)² + (-6 - 4)²]
d = √[(-4)² + (-10)²]
d = √[16 + 100]
d = √116
d = 2√29 units (simplified form).

Thus, the distance between A and C is 2√29 units.

Case-based Questions (4 Marks) – with Solutions (CBSE Pattern)

These 4-mark case-based questions assess analytical skills through real-life scenarios. Answers must be based on the case study provided.

Question 1:
A farmer plots his rectangular field with vertices at A(2, 3), B(6, 3), C(6, 7), and D(2, 7) on a coordinate plane.
(i) Find the length and width of the field.
(ii) Calculate its area using coordinate geometry.
Answer:
Problem Interpretation

We plot the points and observe the rectangle.


Mathematical Modeling
  • Length (AB) = 6 - 2 = 4 units
  • Width (AD) = 7 - 3 = 4 units

Solution

Area = length × width = 4 × 4 = 16 square units. Our textbook shows similar problems for plotting shapes.

Question 2:
A city map marks a park as a triangle with vertices at P(1, 2), Q(4, 2), and R(4, 6).
(i) Identify the type of triangle formed.
(ii) Derive its area using the formula we studied.
Answer:
Problem Interpretation

We plot the points and analyze the triangle.


Mathematical Modeling
  • PQ (base) = 4 - 1 = 3 units
  • QR (height) = 6 - 2 = 4 units

Solution

It is a right-angled triangle at Q. Area = ½ × base × height = ½ × 3 × 4 = 6 square units, as per NCERT examples.

Question 3:
A farmer plots his rectangular field with vertices at origin (0,0), (5,0), (5,4), and (0,4) on a coordinate plane. Find the area of the field and the length of its diagonal.
Answer:
Problem Interpretation

We studied plotting points in coordinate geometry. The field's vertices form a rectangle.


Mathematical Modeling
  • Length = 5 units (x-coordinate difference)
  • Width = 4 units (y-coordinate difference)

Solution

Area = length × width = 5 × 4 = 20 sq. units. Diagonal = √(5² + 4²) = √41 units.

Question 4:
A delivery robot moves from point A(3,2) to B(7,6) in a straight line. Calculate the distance traveled and midpoint of its path.
Answer:
Problem Interpretation

Our textbook shows how to calculate distance between two points using coordinates.


Mathematical Modeling
  • Distance formula: √[(x₂-x₁)² + (y₂-y₁)²]
  • Midpoint formula: [(x₁+x₂)/2, (y₁+y₂)/2]

Solution

Distance = √[(7-3)² + (6-2)²] = √32 = 4√2 units. Midpoint = [(3+7)/2, (2+6)/2] = (5,4).

Question 5:
A city map marks a school at S(4, 5) and a library at L(8, 9).
(i) Find the midpoint coordinates between S and L.
(ii) If a park is built at this midpoint, how far is it from the school?
Answer:
Problem Interpretation

We use midpoint formula from coordinate geometry. Our textbook shows it as average of x and y coordinates.


Mathematical Modeling
  • Midpoint = [(4+8)/2, (5+9)/2] = (6, 7)
  • Distance (S to park) = √[(6-4)² + (7-5)²]

Solution

Distance = √(4+4) = √8 ≈ 2.83 units. The park is centrally located.

Question 6:
A park has a straight walking path from point A(3, 4) to point B(7, 8). A bench is placed exactly midway between them. Find the coordinates of the bench using the section formula.
Answer:
Problem Interpretation

We need to find the midpoint of the line segment joining points A and B.


Mathematical Modeling

Using the section formula for midpoint (m:n = 1:1), coordinates are calculated as:


Solution
  • x-coordinate = (3+7)/2 = 5
  • y-coordinate = (4+8)/2 = 6

Thus, the bench is at (5, 6). Our textbook shows similar problems.

Question 7:
A farmer plots his field's corners at P(0, 0), Q(5, 0), R(5, 4), and S(0, 4). Verify if the field is rectangular by checking distances between points.
Answer:
Problem Interpretation

We must verify if PQRS forms a rectangle by checking distances and right angles.


Mathematical Modeling

Using distance formula, we calculate:


Solution
  • PQ = √[(5-0)²+(0-0)²] = 5 units
  • QR = √[(5-5)²+(4-0)²] = 4 units
  • Opposite sides are equal and adjacent sides are perpendicular. Thus, it's a rectangle.

Question 8:
A farmer plots his rectangular field on a coordinate plane with vertices at (2, 3), (6, 3), (6, 7), and (2, 7).
  • Find the length and width of the field.
  • Calculate its area using coordinate geometry.
Answer:
Problem Interpretation

We plot the points to visualize the rectangle. Our textbook shows that length and width can be calculated using the distance formula.


Mathematical Modeling
  • Length = Distance between (2, 3) and (6, 3) = √((6-2)² + (3-3)²) = 4 units.
  • Width = Distance between (2, 3) and (2, 7) = √((2-2)² + (7-3)²) = 4 units.

Solution

Area = Length × Width = 4 × 4 = 16 square units.

Question 9:
A city map marks two schools at A(1, 2) and B(5, 6). A new metro station is to be built at the midpoint of AB.
  • Find the coordinates of the metro station.
  • Verify your answer using the section formula.
Answer:
Problem Interpretation

We studied that the midpoint divides a line segment into two equal parts. The midpoint formula is derived from the section formula.


Mathematical Modeling
  • Midpoint coordinates = ((1+5)/2, (2+6)/2) = (3, 4).
  • Using section formula (m:n = 1:1), we get the same result.

Solution

The metro station should be built at (3, 4). This matches our verification.

Question 10:
A farmer plots his rectangular field with vertices at A(2, 3), B(6, 3), C(6, 7), and D(2, 7) on a coordinate grid.
(i) Find the length and width of the field.
(ii) Calculate its area using coordinate geometry methods.
Answer:
Problem Interpretation

We plot the points to visualize the rectangle. Our textbook shows that length and width can be calculated using distance formula.


Mathematical Modeling
  • Length AB = √[(6-2)² + (3-3)²] = 4 units
  • Width AD = √[(2-2)² + (7-3)²] = 4 units

Solution

Area = length × width = 4 × 4 = 16 square units. This matches real-life land measurement methods.

Question 11:
A delivery drone starts at origin (0,0) and moves to point P(5,12) in a straight path.
(i) Plot this movement on graph paper.
(ii) Find the distance covered by the drone using coordinate geometry formula.
Answer:
Problem Interpretation

We studied how to calculate distance between two points. The drone's path forms the hypotenuse of a right triangle.


Mathematical Modeling
  • Horizontal distance (x-coordinate) = 5 units
  • Vertical distance (y-coordinate) = 12 units

Solution

Using distance formula: √(5² + 12²) = √(25 + 144) = 13 units. This application shows how coordinate geometry helps in navigation systems.

Question 12:

A farmer has a rectangular field with vertices at A(2, 3), B(6, 3), C(6, 7), and D(2, 7) on the coordinate plane. He wants to divide the field into two equal parts by building a fence along one of its diagonals. Help the farmer by answering the following:

  • Calculate the length of both diagonals.
  • Which diagonal should he choose to divide the field into two equal parts? Justify your answer.
Answer:

To solve the problem, we first identify the coordinates of the vertices:


A(2, 3), B(6, 3), C(6, 7), and D(2, 7).

The two diagonals are AC and BD.


Step 1: Calculate the length of diagonal AC
Using the distance formula: √[(x₂ - x₁)² + (y₂ - y₁)²]
AC = √[(6 - 2)² + (7 - 3)²] = √[16 + 16] = √32 = 4√2 units.

Step 2: Calculate the length of diagonal BD
BD = √[(6 - 2)² + (3 - 7)²] = √[16 + 16] = √32 = 4√2 units.

Both diagonals are equal in length (4√2 units). Since the diagonals of a rectangle bisect each other and divide the rectangle into two congruent triangles, the farmer can choose either diagonal to divide the field into two equal parts.

Question 13:

A delivery boy starts his journey from point P(1, 2) and delivers packages to points Q(4, 6) and R(7, 2) before returning to P. Plot these points on the coordinate plane and answer:

  • What type of triangle is formed by points P, Q, and R?
  • Calculate the area of the triangle.
Answer:

First, plot the points on the coordinate plane:


P(1, 2), Q(4, 6), and R(7, 2).

To determine the type of triangle, we calculate the lengths of all sides using the distance formula.


Step 1: Calculate PQ
PQ = √[(4 - 1)² + (6 - 2)²] = √[9 + 16] = √25 = 5 units.

Step 2: Calculate QR
QR = √[(7 - 4)² + (2 - 6)²] = √[9 + 16] = √25 = 5 units.

Step 3: Calculate PR
PR = √[(7 - 1)² + (2 - 2)²] = √[36 + 0] = √36 = 6 units.

Since two sides (PQ and QR) are equal (5 units each), the triangle is an isosceles triangle.


Step 4: Calculate the area
Using the formula for the area of a triangle with coordinates (x₁, y₁), (x₂, y₂), (x₃, y₃):
Area = ½ |x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|
= ½ |1(6 - 2) + 4(2 - 2) + 7(2 - 6)|
= ½ |4 + 0 - 28| = ½ × 24 = 12 square units.
Question 14:

A farmer has a rectangular field with vertices at A(2, 3), B(6, 3), C(6, 7), and D(2, 7) on the coordinate plane. He wants to divide the field into two equal triangular plots by drawing a diagonal from A to C.

(i) Find the area of the rectangular field.

(ii) Verify if the diagonal divides the field into two triangles of equal area.

Answer:

Solution:

(i) Area of the rectangle:


Length = Distance between A(2, 3) and B(6, 3) = 6 - 2 = 4 units
Breadth = Distance between A(2, 3) and D(2, 7) = 7 - 3 = 4 units
Area = Length × Breadth = 4 × 4 = 16 square units.

(ii) Verification of equal areas:


Diagonal AC divides the rectangle into triangles ABC and ADC.
Area of ABC = ½ × base × height = ½ × 4 (AB) × 4 (BC) = 8 square units.
Area of ADC = ½ × base × height = ½ × 4 (AD) × 4 (DC) = 8 square units.

Since both triangles have equal areas, the diagonal divides the field equally.

Question 15:

A delivery boy starts his journey from point P(0, 0) and delivers packages to three locations: Q(3, 4), R(6, 0), and S(3, -4).

(i) Plot these points on the coordinate plane and name the shape formed by connecting all the points in order.

(ii) Calculate the area of the shape formed.

Answer:

Solution:

(i) Shape formed:


Plotting the points:
P(0, 0), Q(3, 4), R(6, 0), and S(3, -4).
Connecting them in order forms a kite (since PQ = PS and QR = SR).

(ii) Area of the kite:


Length of diagonal PR = Distance between P(0, 0) and R(6, 0) = 6 units.
Length of diagonal QS = Distance between Q(3, 4) and S(3, -4) = 8 units.
Area of kite = ½ × product of diagonals = ½ × 6 × 8 = 24 square units.
Question 16:
A farmer plots the corners of his rectangular field in the coordinate plane as A(2, 3), B(6, 3), C(6, 7), and D(2, 7).

(i) Find the length and width of the field using the distance formula.

(ii) Calculate the area of the field.

Answer:

Solution:

(i) To find the length and width:


Length AB = Distance between A(2, 3) and B(6, 3):
√[(6 - 2)² + (3 - 3)²] = √(16 + 0) = 4 units

Width AD = Distance between A(2, 3) and D(2, 7):
√[(2 - 2)² + (7 - 3)²] = √(0 + 16) = 4 units

(ii) Area of the field:


Area = Length × Width = 4 × 4 = 16 square units

Key Concept: The distance formula helps calculate the side lengths, and area is derived from the product of adjacent sides in a rectangle.

Question 17:
A city planner marks three important locations on a grid: Park at P(1, 4), School at S(5, 4), and Library at L(5, 8).

(i) Plot these points and determine if they form a right-angled triangle.

(ii) Find the perimeter of triangle PSL.

Answer:

Solution:

(i) Checking for right angle:


Calculate distances:
PS = √[(5 - 1)² + (4 - 4)²] = √16 = 4 units
SL = √[(5 - 5)² + (8 - 4)²] = √16 = 4 units
PL = √[(5 - 1)² + (8 - 4)²] = √32 ≈ 5.66 units

Check Pythagoras theorem: PS² + SL² = 16 + 16 = 32 = PL²
Since PS² + SL² = PL², ∠PSL is 90°. Hence, it is a right-angled triangle.

(ii) Perimeter of ΔPSL:


Perimeter = PS + SL + PL = 4 + 4 + 5.66 ≈ 13.66 units

Key Concept: The Pythagorean theorem verifies right angles, and perimeter is the sum of all sides.

Question 18:
A farmer plots the corners of his rectangular field on a coordinate plane. The coordinates of three corners are (2, 3), (2, 7), and (6, 3).

(a) Find the coordinates of the fourth corner.
(b) Calculate the area of the field.
Answer:

(a) To find the fourth corner, observe that the given points form a rectangle. The missing corner will have the same x-coordinate as (6, 3) and the same y-coordinate as (2, 7).
Thus, the fourth corner is (6, 7).

(b) To calculate the area:
Length along the x-axis = 6 - 2 = 4 units.
Width along the y-axis = 7 - 3 = 4 units.
Area = Length × Width = 4 × 4 = 16 square units.

Question 19:
A delivery robot moves from point A (1, 2) to point B (5, 6) in a straight line.

(a) Plot these points on a graph and draw the path.
(b) Calculate the distance between A and B using the distance formula.
Answer:

(a) Plotting the points:
Point A (1, 2) is 1 unit right on the x-axis and 2 units up on the y-axis.
Point B (5, 6) is 5 units right on the x-axis and 6 units up on the y-axis.
Draw a straight line connecting A and B.

(b) Using the distance formula:
Distance = √[(x₂ - x₁)² + (y₂ - y₁)²]
= √[(5 - 1)² + (6 - 2)²]
= √[16 + 16]
= √32
= 4√2 units.

Question 20:
A farmer plots the corners of his rectangular field on a coordinate plane with vertices at A(2, 3), B(6, 3), C(6, 7), and D(2, 7).

Help the farmer verify if his field is indeed a rectangle by checking the properties of its sides and diagonals using coordinate geometry.

Answer:

To verify if the given points form a rectangle, we need to check two properties:

  • Opposite sides must be equal and parallel.
  • Diagonals must be equal in length.

Step 1: Calculate the lengths of all sides using the distance formula.
Distance between A(2, 3) and B(6, 3): √[(6-2)² + (3-3)²] = √16 = 4 units.
Distance between B(6, 3) and C(6, 7): √[(6-6)² + (7-3)²] = √16 = 4 units.
Distance between C(6, 7) and D(2, 7): √[(2-6)² + (7-7)²] = √16 = 4 units.
Distance between D(2, 7) and A(2, 3): √[(2-2)² + (3-7)²] = √16 = 4 units.

Step 2: Verify opposite sides are parallel.
Slope of AB: (3-3)/(6-2) = 0 (horizontal line).
Slope of CD: (7-7)/(2-6) = 0 (horizontal line).
Slope of BC: (7-3)/(6-6) = undefined (vertical line).
Slope of DA: (3-7)/(2-2) = undefined (vertical line).
Since AB || CD (both horizontal) and BC || DA (both vertical), opposite sides are parallel.

Step 3: Check diagonals.
Distance between A(2, 3) and C(6, 7): √[(6-2)² + (7-3)²] = √32 ≈ 5.66 units.
Distance between B(6, 3) and D(2, 7): √[(2-6)² + (7-3)²] = √32 ≈ 5.66 units.

Since all sides are equal, opposite sides are parallel, and diagonals are equal, the field is indeed a rectangle.

Question 21:
A delivery boy starts from his house at H(0, 0) and delivers packages to three locations: P(3, 4), Q(6, 0), and R(3, -4).

Using coordinate geometry, determine whether these points lie on a straight line (collinear) or form a triangle.

Answer:

To check if the points are collinear or form a triangle, we calculate the area formed by them. If the area is zero, they are collinear; otherwise, they form a triangle.


Step 1: Use the area formula for three points.
Area = ½ |x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|
Substitute H(0, 0), P(3, 4), and Q(6, 0):
Area = ½ |0(4 - 0) + 3(0 - 0) + 6(0 - 4)| = ½ |0 + 0 - 24| = 12 square units.

Step 2: Check another combination (e.g., P, Q, R).
Area = ½ |3(0 - (-4)) + 6(-4 - 4) + 3(4 - 0)| = ½ |12 - 48 + 12| = ½ |-24| = 12 square units.

Since the area is non-zero in both cases, the points do not lie on a straight line and form a triangle. The delivery boy's path is triangular.

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