Coordinate Geometry (Class 10): Notes और MCQ Practice

आसान भाषा में

Coordinate Geometry की मदद से हम graph paper पर algebra के formulas लगाकर geometry की problems चुटकियों में solve करते हैं। अगर हमें दो points के coordinates पता हों, तो हम scale के बिना उनके बीच की दूरी निकाल सकते हैं। साथ ही, किसी line को divide करने वाले point की exact position भी पता कर सकते हैं!

Key Formulas & Points

• Distance Formula: Points A(x₁, y₁) और B(x₂, y₂) के बीच की दूरी:

AB = √[(x₂ − x₁)² + (y₂ − y₁)²]

• Distance from Origin: Point (x, y) की origin (0, 0) से दूरी = √(x² + y²)
• Section Formula: Point P जो line segment AB को m₁ : m₂ के ratio में divide करता है:

x = (m₁x₂ + m₂x₁) ÷ (m₁ + m₂)

y = (m₁y₂ + m₂y₁) ÷ (m₁ + m₂)

• Mid-point Formula: जब ratio 1 : 1 हो:

Mid-point = ((x₁ + x₂) ÷ 2, (y₁ + y₂) ÷ 2)

• Collinear Points: तीन points A, B, C collinear होते हैं अगर AB + BC = AC.

2 Memory Tricks

1. Section Formula का Cross-Rule: याद रखें कि ratio m₁ हमेशा दूर वाले coordinate x₂ से multiply होता है, और m₂ पास वाले x₁ से (Cross connection!).
2. Distance में Order की टेंशन नहीं: क्योंकि formula में square (²) है, इसलिए (x₂ − x₁)² लिखो या (x₁ − x₂)², result हमेशा same और positive आएगा।

2 Common Board Exam Mistakes

1. Negative Sign का लोचा: अगर coordinates negative हों (जैसे x₁ = −3), तो formula लगाते समय (x₂ − (−3))² = (x₂ + 3)² बनता है। जल्दीबाजी में बच्चे minus लगाना भूल जाते हैं।
2. Axis points में गलती: x-axis पर किसी point के coordinates (x, 0) होते हैं और y-axis पर (0, y)। Students अक्सर 0 की position swap कर देते हैं।

Coordinate Geometry - Class 10 NCERT Study Notes

1. Chapter in Short

Coordinate Geometry bridges algebra and geometry. It allows us to locate points on a flat plane using a pair of numbers called coordinates. In Class 10, this chapter primarily focuses on:

• Finding the straight-line distance between any two points on a plane (Distance Formula).
• Determining if three points lie on the same straight line (Collinearity).
• Finding the coordinates of a point that divides a given line segment in a given ratio (Section Formula and Mid-point Formula).

2. Key Concepts Explained Simply

The Cartesian Plane

Imagine a city where streets run North-South and East-West.

• The horizontal line is the x-axis.
• The vertical line is the y-axis.
• The intersection point is the Origin (0, 0).
• Any position is written as an ordered pair (x, y).
• x is the horizontal distance (Abscissa).
• y is the vertical distance (Ordinate).

Everyday Example: Locating your seat in a cinema hall using Row 4, Seat 5 is just like finding the point (5, 4) on a coordinate grid.

Distance Formula

To find the direct aerial distance between two landmarks on a map, we construct a right-angled triangle between them and apply Pythagoras' Theorem.

• The horizontal leg is (x₂ − x₁).
• The vertical leg is (y₂ − y₁).
• The hypotenuse is the straight-line distance.

Section Formula

If two towns A and B are connected by a straight highway, and a petrol pump needs to be built at a point dividing the distance in the ratio 2 : 3, the Section Formula gives the exact coordinates of that petrol pump.

• If a point divides the line into two equal halves, it is the mid-point.

3. Important Definitions, Formulas, and Facts

1. Coordinates of Axes:

• Any point on the x-axis has coordinates of the form (x, 0).
• Any point on the y-axis has coordinates of the form (0, y).

1. Distance Formula:

The distance d between two points P(x₁, y₁) and Q(x₂, y₂) is:

d = √[(x₂ − x₁)² + (y₂ − y₁)²]

1. Distance from the Origin:

The distance of a point P(x, y) from the origin (0, 0) is:

OP = √(x² + y²)

1. Condition for Collinear Points:

Three points A, B, and C are collinear (lie on the same straight line) if the sum of any two distances equals the third distance (e.g., AB + BC = AC).

1. Section Formula (Internal Division):

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:

x = (m₁x₂ + m₂x₁) ÷ (m₁ + m₂)

y = (m₁y₂ + m₂y₁) ÷ (m₁ + m₂)

1. Mid-point Formula:

The mid-point M of a line segment joining A(x₁, y₁) and B(x₂, y₂) divides it in the ratio 1 : 1:

M = ((x₁ + x₂) ÷ 2, (y₁ + y₂) ÷ 2)

1. Centroid of a Triangle:

The centroid G of a triangle with vertices (x₁, y₁), (x₂, y₂), and (x₃, y₃) is:

G = ((x₁ + x₂ + x₃) ÷ 3, (y₁ + y₂ + y₃) ÷ 3)

4. Solved Board-Exam Style Examples

Example 1: Basic Distance Calculation

Question: Find the distance between the points A(2, 3) and B(4, 1).

Solution:

• Here, x₁ = 2, y₁ = 3 and x₂ = 4, y₂ = 1.
• Using Distance Formula:

AB = √[(4 − 2)² + (1 − 3)²]

AB = √[(2)² + (−2)²]

AB = √(4 + 4) = √8 = 2√2 units

Answer: 2√2 units

Example 2: Equidistant Point on an Axis

Question: Find a point on the x-axis which is equidistant from A(2, −5) and B(−2, 9).

Solution:

1. Since the point lies on the x-axis, its y-coordinate is 0. Let the point be P(x, 0).
2. Since P is equidistant from A and B, PA = PB, which means PA² = PB².
3. Using the distance formula squared:

PA² = (x − 2)² + (0 − (−5))² = (x − 2)² + 25

PB² = (x − (−2))² + (0 − 9)² = (x + 2)² + 81

1. Equating both:

(x − 2)² + 25 = (x + 2)² + 81

x² − 4x + 4 + 25 = x² + 4x + 4 + 81

−4x + 29 = 4x + 85

−8x = 85 − 29

−8x = 56

x = −7

Answer: The required point is (−7, 0).

Example 3: Finding the Ratio

Question: In what ratio does the point P(−4, 6) divide the line segment joining A(−6, 10) and B(3, −8)?

Solution:

1. Let the required ratio be k : 1.
2. Using Section Formula for the x-coordinate:

x = (k × x₂ + 1 × x₁) ÷ (k + 1)

−4 = (k(3) + 1(−6)) ÷ (k + 1)

1. Cross-multiplying:

−4(k + 1) = 3k − 6

−4k − 4 = 3k − 6

−4k − 3k = −6 + 4

−7k = −2

k = 2 ÷ 7

Answer: The point divides the line segment in the ratio 2 : 7.

Example 4: Points of Trisection

Question: Find the coordinates of the points of trisection of the line segment joining A(2, −2) and B(−7, 4).

Solution:

Trisection means dividing a segment into 3 equal parts using two points, say P and Q.

• P divides AB internally in the ratio 1 : 2.
• Q divides AB internally in the ratio 2 : 1 (or Q is the midpoint of PB).

1. For point P (ratio 1 : 2):

x = [1(−7) + 2(2)] ÷ (1 + 2) = (−7 + 4) ÷ 3 = −3 ÷ 3 = −1

y = [1(4) + 2(−2)] ÷ (1 + 2) = (4 − 4) ÷ 3 = 0 ÷ 3 = 0

So, P = (−1, 0).

1. For point Q (midpoint of P and B):

x = (−1 + (−7)) ÷ 2 = −8 ÷ 2 = −4

y = (0 + 4) ÷ 2 = 4 ÷ 2 = 2

So, Q = (−4, 2).

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

5. Common Mistakes and Quick Revision Points

• Sign Errors with Negatives: When substituting negative coordinates into formulas, remember that subtraction of a negative is addition: x₂ − (−x₁) = x₂ + x₁.
• Square Root Property: Remember that √(a² + b²) is not equal to a + b. You must calculate values inside first and then take the square root.
• Distance is Always Positive: Distance is a length; it cannot be negative. If d² = 25, then d = 5, not -5.
• Cross-Multiplication in Section Formula: Be careful with indices. m₁ multiplies with x₂, and m₂ multiplies with x₁. Do not pair m₁ with x₁.
• Ratio Assumption Trick: Whenever you are asked to find an unknown ratio, assume it to be k : 1. This reduces the problem to solving for just one variable k instead of two.