Class 10 Mathematics Notes Chapter 7 (Coordinate geometry) – Mathematics Book

Read here detailed notes of Chapter 7: Coordinate Geometry from your NCERT Class 10 Mathematics textbook. This chapter is crucial not just for your board exams but also forms the foundation for many concepts in higher mathematics and is frequently tested in government exams. We'll break it down systematically.

Chapter 7: Coordinate Geometry - Detailed Notes

1. Introduction & Recap

• Coordinate Geometry: It's the branch of geometry where we use a coordinate system (like the Cartesian system) to study geometric shapes. We represent points as ordered pairs (x, y) and use algebraic methods to solve geometric problems.
• Cartesian Plane:
  • Two perpendicular lines: X-axis (horizontal) and Y-axis (vertical).
  • Intersection point: Origin (0, 0).
  • Coordinates: An ordered pair (x, y) representing a point's position.
    • x-coordinate (Abscissa): Perpendicular distance from the Y-axis.
    • y-coordinate (Ordinate): Perpendicular distance from the X-axis.
  • Quadrants: The axes divide the plane into four quadrants (I, II, III, IV) with specific sign conventions for (x, y).
    • Quadrant I: (+, +)
    • Quadrant II: (-, +)
    • Quadrant III: (-, -)
    • Quadrant IV: (+, -)
  • Point on X-axis: (x, 0)
  • Point on Y-axis: (0, y)

2. The Distance Formula

• Purpose: To find the distance between any two points in the Cartesian plane whose coordinates are known.
• Derivation: Based on the Pythagorean theorem applied to a right-angled triangle formed by the two points and lines parallel to the axes.
• Formula: Let P(x₁, y₁) and Q(x₂, y₂) be two points. The distance PQ is given by:
  PQ = √[(x₂ - x₁)² + (y₂ - y₁)²]
• Key Points:
  • Distance is always non-negative.
  • The order of points doesn't matter for distance (PQ = QP) because squaring the differences eliminates negative signs.
  • Distance from Origin: The distance of a point P(x, y) from the origin O(0, 0) is OP = √[(x - 0)² + (y - 0)²] = √(x² + y²).
• Applications:
  • Finding the length of a line segment.
  • Determining if points are collinear (If A, B, C are collinear, then AB + BC = AC, or AC + CB = AB, or BA + AC = BC).
  • Identifying types of triangles (Equilateral, Isosceles, Scalene, Right-angled) by comparing side lengths.
  • Identifying types of quadrilaterals (Square, Rhombus, Rectangle, Parallelogram) by comparing side lengths and diagonal lengths.
    • Parallelogram: Opposite sides equal.
    • Rectangle: Opposite sides equal AND diagonals equal.
    • Rhombus: All four sides equal.
    • Square: All four sides equal AND diagonals equal.

3. Section Formula

• Purpose: To find the coordinates of a point that divides the line segment joining two given points internally in a specific ratio.
• Formula: Let R(x, y) be the point that divides the line segment joining P(x₁, y₁) and Q(x₂, y₂) internally in the ratio m₁ : m₂. The coordinates of R are:
  x = (m₁x₂ + m₂x₁)/(m₁ + m₂)
y = (m₁y₂ + m₂y₁)/(m₁ + m₂)
• Key Points:
  • The ratio m₁ : m₂ corresponds to the division PR : RQ.
  • This formula is for internal division (the point R lies between P and Q).
• Special Case: Mid-point Formula
  • If R is the mid-point of PQ, it divides the segment in the ratio 1 : 1 (m₁ = 1, m₂ = 1).
  • Substituting m₁ = 1 and m₂ = 1 in the section formula gives the coordinates of the mid-point:
    x = (1*x₂ + 1*x₁)/(1 + 1) = (x₁ + x₂)/2
y = (1*y₂ + 1*y₁)/(1 + 1) = (y₁ + y₂)/2
  • Mid-point M(x, y) = ( (x₁ + x₂)/2 , (y₁ + y₂)/2 )
• Finding the Ratio: If the coordinates of P, Q, and the dividing point R are given, you can use the section formula (either the x-coordinate or y-coordinate formula) to find the ratio m₁ : m₂. It's often easier to assume the ratio is k : 1 and solve for k.
  x = (kx₂ + 1x₁)/(k + 1) or y = (ky₂ + 1y₁)/(k + 1)
  If k is positive, the division is internal. If k is negative, the division is external (though external division is generally outside the Class 10 scope).
• Applications:
  • Finding coordinates of points of trisection (ratio 1:2 or 2:1).
  • Finding the ratio in which a line segment is divided by the X-axis (y-coordinate of the dividing point is 0) or Y-axis (x-coordinate of the dividing point is 0).
  • Problems involving centroids (point of intersection of medians) of triangles, although the direct centroid formula ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3) might not be explicitly derived in NCERT, it's based on the section formula (median is divided in 2:1 ratio by centroid). Useful for competitive exams.
  • Finding vertices of parallelograms when mid-points are involved (diagonals of a parallelogram bisect each other).

4. Area of a Triangle

• Purpose: To find the area of a triangle when the coordinates of its three vertices are given.
• Formula: Let the vertices of a triangle be A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃). The area of triangle ABC is given by:
  Area = ½ |x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|
• Key Points:
  • The absolute value (modulus, denoted by |...|) is taken because area must be positive.
  • Remember the cyclic order of coordinates: 1 -> 2 -> 3 -> 1. (x₁ is multiplied by the difference of y₂ and y₃, x₂ by y₃ and y₁, x₃ by y₁ and y₂).
• Condition for Collinearity:
  • Three points A, B, and C are collinear (lie on the same straight line) if the area of the triangle formed by them is zero.
  • So, A(x₁, y₁), B(x₂, y₂), C(x₃, y₃) are collinear if:
    x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂) = 0
    (No need for the ½ or the absolute value here, as we are equating it to zero).
• Applications:
  • Calculating the area of any triangle given vertices.
  • Checking if three points are collinear.
  • Finding the area of quadrilaterals (by dividing them into two triangles using a diagonal).

Summary of Formulas:

1. Distance: √[(x₂ - x₁)² + (y₂ - y₁)²]
2. Section (Internal): ( (m₁x₂ + m₂x₁)/(m₁ + m₂), (m₁y₂ + m₂y₁)/(m₁ + m₂) )
3. Mid-point: ( (x₁ + x₂)/2 , (y₁ + y₂)/2 )
4. Area of Triangle: ½ |x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|
5. Collinearity: x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂) = 0

Remember to practice problems involving each of these concepts to gain proficiency. Pay attention to the specific requirements of each question – whether it asks for distance, coordinates, ratio, or area.

Multiple Choice Questions (MCQs)

Here are 10 MCQs based on Coordinate Geometry for practice:

1. The distance of the point P(–6, 8) from the origin is:
   (A) 8
   (B) 2√7
   (C) 10
   (D) 6

2. If the distance between the points (2, –2) and (–1, x) is 5, one of the values of x is:
   (A) –2
   (B) 2
   (C) –1
   (D) 1

3. The mid-point of the line segment joining the points A(–2, 8) and B(–6, –4) is:
   (A) (–4, –6)
   (B) (2, 6)
   (C) (–4, 2)
   (D) (4, 2)

4. 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

5. The coordinates of the point which divides the line segment joining the points (5, –2) and (9, 6) internally in the ratio 3 : 1 are:
   (A) (8, 4)
   (B) (7, 3)
   (C) (6, 2)
   (D) (4, 8)

6. 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

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 area of the triangle whose vertices are A(5, 2), B(4, 7) and C(7, –4) is (in square units):
   (A) 2
   (B) 5
   (C) 1
   (D) 0

9. The coordinates of a point A, where AB is the diameter of a circle whose centre is (2, –3) and B is (1, 4), are:
   (A) (3, –10)
   (B) (10, –3)
   (C) (–3, 10)
   (D) (–10, 3)

10. The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is:
    (A) 5
    (B) 12
    (C) 11
    (D) 7 + √5

Answers to MCQs:

1. (C)
2. (B)
3. (C)
4. (B)
5. (A)
6. (B)
7. (C)
8. (A)
9. (A)
10. (B)

Study these notes thoroughly and practice solving various problems. Good luck with your preparation!