1.The Use of Coordinates Class 9

The Use of Coordinates Class 9

In this article “The Use of Coordinates Class 9”,we will study two dimensional coordinate system uses two lines at right angles to each other to mark points in two-dimensional space (short-form 2-D space).

Also Read This Article:- MCQ Examples of Co-ordinate Geometry

2.Important Points:The Use of Coordinates Class 9

The Use of Coordinates Class 9

• * **Definition of Coordinate System:**
  A structured framework (like a map grid or graph paper) that uses numbers to specify the exact physical location of points or objects in space.
  * **Ancient Indian Origins (Sindhu-Sarasvati Civilisation):**
  The earliest massive urban application of a practical grid system occurred thousands of years ago, where city streets were built with precision in North-South and East-West directions at uniform intervals of roughly 10 meters.
  * **Geometric Foundations by Baudhāyana:**
  Around 800 BCE, the mathematician Baudhāyana utilized North-South and East-West reference lines to construct complex geometric shapes and formulated the Baudhāyana-Pythagoras Theorem.
  * **Prime Meridian of Ujjayinī:**
  As early as the 4th century BCE,ancient Indian *Siddhāntas* established Ujjayinī as the prime longitude meridian (the zero-reference point) for global calculations,which was later recognized by the Greek I’m athematician Ptolemy.
  * **Trigonometric Advancements by Āryabhaṭa:**
  In 499 CE,Āryabhaṭa replaced Greek ‘chords’ with ‘sines’ (jya), vastly simplifying calculations for celestial and terrestrial coordinates based on the ecliptic (the Sun’s path).
  * **Algebraic Groundwork by Brahmagupta:**
  In 628 CE, Brahmagupta formalized zero and negative numbers as active algebraic entities.This foundation directly enabled the development of the modern four-quadrant coordinate plane.
  * **Global Transmission:**
  Indian mathematics reached the Arab world through translations (like the *Sindhind*), establishing Ujjayinī as “Arin” (the zero-longitude line on early Arab maps). Scholars like Al-Bīrūnī and Omar Khayyam later combined these decimal and trigonometric techniques to solve geometric problems algebraically.
  * **Formalization in Europe:**
  In the 17th century, building upon Pierre de Fermat’s work, René Descartes (1637 CE) formalized the modern 2D system where any point is defined by its distance from two perpendicular axes,fusing algebra and geometry into one field.
  * **Structure of the 2D Cartesian Plane:**
     * It uses a horizontal line called the **x-axis** and a vertical line called the **y-axis**.
     * The perpendicular intersection point of these axes is the **origin O**, assigned the coordinates (0, 0).
     * Distances are marked in equal units: positive to the right and upwards, and negative to the left and downwards.

Also Read This Article:- Class 9 Rational and Irrational Number

3.The Use of Coordinates Class 9 Illustrations

• Fig. 1.3 shows Reiaan’s room with points OABC marking its corners.The x- and y-axes are marked in the figure.Point O is the origin.

The Use of Coordinates Class 9

• Referring to Fig. 1.3, answer the following questions:
  Illustration:(i).If D_{1}R_{1}, represents the door to Reiaan’s room,how far is the door from the left wall (the y-axis) of the room? How far is the door from the x-axis?
  Solution: D_{1}=(8,0) , R_1=(11.5,0)
  The distance of door from y-axis (left wall) is 8 units and distance of door from x-axis is 0 units.
  Illustration:(ii). What are the coordinates of D_{1}?
  Solution:Coordinates of D_{1}=(8,0)
  Illustration:(iii). If R_{1} is the point (11.5,0), how wide is the door? Do you think this is a comfortable width for the room door? If a person in a wheelchair wants to enter the room,will he/she be able to do so easily?
  Solution: R_1=(11.5,0) and D_{1}=(8,0)
  Width of door=11.5-8=3.5
  Which is sufficient to enter the room for a person.
  Illustration:(iv).If B_{1} (0, 1.5) and B_{2} (0, 4) represent the ends of the bathroom door, is the bathroom door narrower or wider than the room door?
  Solution:width of bathroom door=4-1.5=2.5
  Yes the bathroom door narrower than the room door.
  With the above illustrations,one can understand the “The Use of Coordinates Class 9”.

4..Practice Problems for Students

The Use of Coordinates Class 9

• Problem 1:Grid Navigation
  In an ancient Sindhu-Sarasvati city, a merchant starts from the city center (the origin, (0,0)).Each block is exactly 10 meters wide.If the merchant walks 4 blocks directly East and then 3 blocks directly North to reach a warehouse:
  1. What are the coordinates of the warehouse?
  2. What is the total physical distance walked by the merchant in meters?
  Problem 2:Understanding Dimensions
  Shalini creates a floor map of Reiaan’s bedroom using a flat,2D rectangular grid. Explain in your own words why Shalini can map the location of the wardrobe and the bathroom walls on this floor map, but cannot mark the exact position and height of the bedroom windows.
  Problem 3:The Four Quadrants
  Brahmagupta’s formalization of negative numbers made the modern 4-quadrant Cartesian plane possible. If you start at the origin (0,0) on a standard coordinate plane:
  1. In which direction (Left/Right and Up/Down) must you move to plot a point with coordinates (-4, -5)?
  2. Which coordinate (x or y) stays at 0 if you only move straight up along the vertical axis?
  By solving the above questions,you can “The Use of Coordinates Class 9” well because the concept is well understood when you solve it practically.
• ### 📢 If you liked this math article:
  * 👥 **Share with Friends:** Knowledge grows by sharing,so be sure to share it with your friends.
  * 🔔 **Follow the Website:** If this is your first time here,follow our **email subscription** so that you get instant notifications for every new article.
  * 💬 **Give Your Suggestions:** If you have any issues or would like to make any suggestions,do let us know by **commenting** below.
  *Welcome to read the full article!*

Also Read This Article:- Coordinate system

5.Frequently Asked Questions Related to The Use of Coordinates Class 9

\begin{array}{|c|} \hline  \text{**छात्र-छात्राओं से आज का प्रश्न**} \\ \text{*"किसी आयताकार भूखण्ड की लम्बाई } \\  \text{उसकी चौड़ाई से 40\% अधिक है।} \\ \text{अगर लम्बाई और चौड़ाई के बीच अन्तर } \\ \text{ 20 मीटर है तो उस आयत का क्षेत्रफल क्या होगा?"*} \\ \text{**Today's Question to Students**} \\ \text{*"The length of a rectangular plot is} \\ \text{ 40\% more than  its breadth.} \\ \text{ If the difference between length  } \\ \text{and breadth is 20 m, then what} \\   \text{will be the area of the rectangle?"*} \\ \text{दिनांक 09.07.2026 के प्रश्न का उत्तर:} \\ \text{माना उसे रिटायर्मेंट पर धनराशि मिली=x} \\ \text{उसने पारिवारिक दायित्वों पर खर्च की=y} \\ \text{ शेष राशि=x-y} \\ \text{1 वर्ष बाद धनराशि=2(x-y)} \\ \text{y खर्च करने के बाद शेष राशि=2(x-y)-y \\ =2x-3y} \\ \text{2 वर्ष बाद धनराशि=2(2x-3y)} \\ \text{ y खर्च करने के बाद शेष राशि=2(2x-3y)-y \\ =4x-7y} \\ \text{3 वर्ष बाद धनराशि=2(4x-7y)}  \\ \text{y खर्च करने के बाद शेष राशि} =2(4x-7y)-y \\ =8x-15y \\ \text{4 वर्ष बाद धनराशि} =2(8x-15y) \\ \text{y खर्च करने के बाद शेष राशि}=16x-30y-y \\ =16x-31y \\ \text{5 वर्ष बाद धनराशि} =2(16x-31y) \\ \text{y खर्च करने के बाद शेष राशि}=32x-63y \\ \text{ 6 वर्ष बाद धनराशि} =2(32x-63y) \\ 64x-126y=y \\ \Rightarrow 64x=127 y  \\ \text{यदि y=64 तो } x=\frac{127 \times 64}{64}=127 \\ \text{अतः प्रारम्भ में उसके पास 127 थे,}  \\ \text{हर बार उसने 64 खर्च किए।} \\  \text{Answer to Question Dated 09.07.2026 } \\ \text{Suppose he received the money } \\ \text{at retirement = x } \\ \text{He spent on family obligations=y} \\ \text{Balance=x-y} \\ \text{Amount after 1 year=2(x-y)} \\ \text{Balance after spending y}=2(x-y)-y \\ =2x-3y \\ \text{Amount after 2 years}=2(2x-3y) \\ \text{Balance after spending y} =2(2x-3y)-y \\=4x-7y \\ \text{ Amount after 3 years} =2(4x-7y) \\ \text{Balance after spending y}=2(4x-7y)-y \\ =8x-15y \\ \text{Amount after 4 years}=2(8x-15y) \\ \text{Balance after spending y}=16x-30y-y \\ =16x-31y \\ \text{Amount after 5 years}=2(16x-31y) \\ \text{Balance after spending y}=32x-63y \\ \text{Amount after 6 years} =2(32x-63y) \\ 64x-126y=y \\ \Rightarrow 64x=127 y \\ \text{ If y=64 then} \\  x=\frac{127 \times 64}{64}=127 \\ \text{So initially he had 127, } \\ \text{each time he spent 64.}  \\  \\ \hline \end{array}
This article has been prepared by **Satyam Coaching Centre** on the **Satyam Mathematics** blog.”*