Class 10 Mathematics Notes Chapter 7 (Chapter 7) – Examplar Problems (English) Book

Alright class, let's focus on Chapter 7: Coordinate Geometry from your NCERT Exemplar. This chapter is crucial not just for your board exams but forms the foundation for many concepts in higher mathematics and frequently appears in various government entrance exams. We'll cover the essential formulas and concepts, focusing on the type of problems you might encounter.

Chapter 7: Coordinate Geometry - Detailed Notes

1. Introduction & Cartesian System:

• Coordinate Axes: Two perpendicular lines, X'OX (x-axis) and Y'OY (y-axis), intersecting at the Origin O(0, 0).
• Coordinates: Any point P in the plane is represented by an ordered pair (x, y), where 'x' is the abscissa (perpendicular distance from the y-axis) and 'y' is the ordinate (perpendicular distance from the x-axis).
• Quadrants: The axes divide the plane into four quadrants with specific sign conventions for (x, y):
  • Quadrant I: (+, +)
  • Quadrant II: (-, +)
  • Quadrant III: (-, -)
  • Quadrant IV: (+, -)
• Points on Axes: A point on the x-axis is (x, 0). A point on the y-axis is (0, y).

2. Distance Formula:

• Concept: Used to find the distance between two points whose coordinates are known. Derived using the Pythagoras theorem.
• Formula: The distance between two points P(x₁, y₁) and Q(x₂, y₂) is given by:
  PQ = √[(x₂ - x₁)² + (y₂ - y₁)²]
• Distance from Origin: The distance of a point P(x, y) from the origin O(0, 0) is:
  OP = √(x² + y²)
• Key Applications:
  • Finding the length of a line segment.
  • Determining the type of triangle (Scalene, Isosceles, Equilateral, Right-angled) by finding the lengths of its sides. Remember to check the Pythagorean theorem for right-angled triangles (a² + b² = c²).
  • Determining the type of quadrilateral (Parallelogram, Rectangle, Square, Rhombus) by finding side lengths and diagonal lengths.
    • Parallelogram: Opposite sides equal. Diagonals may not be equal.
    • Rectangle: Opposite sides equal AND Diagonals equal.
    • Rhombus: All sides equal. Diagonals may not be equal.
    • Square: All sides equal AND Diagonals equal.
  • Checking for Collinearity: Three points A, B, and C are collinear if the sum of the lengths of any two line segments is equal to the length of the third segment (e.g., AB + BC = AC).

3. Section Formula:

• Concept: Used to find the coordinates of a point that divides the line segment joining two given points in a specific ratio.
• Internal Division 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:
  P(x, y) = [ (m₁x₂ + m₂x₁) / (m₁ + m₂), (m₁y₂ + m₂y₁) / (m₁ + m₂) ]
• Mid-Point Formula (Special Case): If P is the mid-point, the ratio is 1:1 (m₁ = m₂ = 1). The coordinates of the mid-point are:
  P(x, y) = [ (x₁ + x₂) / 2, (y₁ + y₂) / 2 ]
• Key Applications:
  • Finding the coordinates of a point dividing a segment in a given ratio.
  • Finding the ratio in which a given point divides a line segment. Let the ratio be k:1. Substitute the coordinates of the given point into the section formula and solve for k.
  • Finding the ratio in which the x-axis or y-axis divides a line segment.
    • If divided by the x-axis, the y-coordinate of the point of division is 0. Use the y-part of the section formula: (m₁y₂ + m₂y₁) / (m₁ + m₂) = 0 => m₁y₂ + m₂y₁ = 0. Find m₁:m₂.
    • If divided by the y-axis, the x-coordinate of the point of division is 0. Use the x-part of the section formula: (m₁x₂ + m₂x₁) / (m₁ + m₂) = 0 => m₁x₂ + m₂x₁ = 0. Find m₁:m₂.
  • Trisection: Points dividing a line segment into three equal parts. The points divide the segment in the ratios 1:2 and 2:1.
  • Finding coordinates of vertices of a parallelogram if some are given (using the property that diagonals bisect each other, meaning their mid-points coincide).

4. Area of a Triangle:

• Concept: Used to find the area of a triangle when the coordinates of its vertices are known.
• Formula: The area of a triangle with vertices A(x₁, y₁), B(x₂, y₂), and C(x₃, y₃) is given by:
  Area (ΔABC) = ½ | x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂) |
  (Remember the cyclic order 1 -> 2 -> 3 -> 1)
• Important Note: Area is always positive. Take the absolute value of the result obtained from the formula.
• Key Applications:
  • Calculating the area of a given triangle.
  • Checking for Collinearity: Three points A, B, and C are collinear if the area of the triangle formed by them is zero.
    x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂) = 0
  • Finding the area of a quadrilateral by dividing it into two triangles using one diagonal and summing their areas.

5. Centroid of a Triangle (Often useful in competitive exams):

• Concept: The point of intersection of the medians of a triangle. A median connects a vertex to the mid-point of the opposite side.
• Formula: The coordinates of the centroid G(x, y) of a triangle with vertices A(x₁, y₁), B(x₂, y₂), and C(x₃, y₃) are:
  G(x, y) = [ (x₁ + x₂ + x₃) / 3, (y₁ + y₂ + y₃) / 3 ]

Key Takeaways for Government Exams:

• Formula Accuracy: Memorize the formulas precisely, including signs and order.
• Conceptual Clarity: Understand when to use which formula. Distance for lengths/types of shapes, Section for division/ratios, Area for area/collinearity.
• Problem Interpretation: Read questions carefully to identify the given information and what needs to be found. Draw a rough sketch if helpful.
• Calculation Speed & Accuracy: Practice calculations to improve speed and minimize errors, especially with signs and square roots.
• Exemplar Focus: Expect questions that might combine concepts (e.g., finding the area of a triangle whose vertices are mid-points or points of division) or require careful geometric interpretation.

Multiple Choice Questions (MCQs):

1. The distance of the point P(–6, 8) from the origin is:
   (a) 8
   (b) 2√7
   (c) 10
   (d) 6

2. The points A(9, 0), B(9, 6), C(–9, 6) and D(–9, 0) are the vertices of a:
   (a) Square
   (b) Rectangle
   (c) Rhombus
   (d) Trapezium

3. The coordinates of the point which divides the line segment joining the points (5, –2) and (9, 6) in the ratio 3 : 1 are:
   (a) (8, 4)
   (b) (7, 2)
   (c) (6, 3)
   (d) (4, 8)

4. The ratio in which the y-axis divides the line segment joining the points (5, –6) and (–1, –4) is:
   (a) 1 : 5
   (b) 5 : 1
   (c) 1 : 1
   (d) 2 : 3

5. If P(a/3, 4) is the mid-point of the line segment joining the points Q(–6, 5) and R(–2, 3), then the value of 'a' is:
   (a) –4
   (b) –12
   (c) 12
   (d) –6

6. The area of the triangle with vertices A(3, 0), B(7, 0) and C(8, 4) is:
   (a) 14 sq. units
   (b) 28 sq. units
   (c) 8 sq. units
   (d) 6 sq. units

7. If the points A(1, 2), O(0, 0) and C(a, b) are collinear, then:
   (a) a = b
   (b) a = 2b
   (c) 2a = b
   (d) a = -b

8. The points (–4, 0), (4, 0), (0, 3) are the vertices of a/an:
   (a) Right triangle
   (b) Isosceles triangle
   (c) Equilateral triangle
   (d) Scalene triangle

9. The centroid of a triangle whose vertices are (3, –7), (–8, 6) and (5, 10) is:
   (a) (0, 9)
   (b) (0, 3)
   (c) (1, 3)
   (d) (3, 5)

10. The points A(–1, y) and B(5, 7) lie on a circle with centre O(2, –3y). The values of y are:
    (a) 1, -7
    (b) -1, 7
    (c) 2, 7
    (d) -2, -7

Answer Key for MCQs:

1. (c)
2. (b)
3. (a)
4. (b)
5. (b)
6. (c)
7. (c)
8. (b)
9. (b)
10. (b)

Study these concepts thoroughly and practice problems from the Exemplar book. Pay close attention to the application of formulas in different scenarios. Good luck!