{"id":67268,"date":"2021-01-04T18:35:03","date_gmt":"2021-01-04T13:05:03","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=67268"},"modified":"2023-12-27T11:01:22","modified_gmt":"2023-12-27T05:31:22","slug":"rs-aggarwal-chapter-6-class-9-maths-exercise-6-2-solutions","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rs-aggarwal-chapter-6-class-9-maths-exercise-6-2-solutions\/","title":{"rendered":"RS Aggarwal Chapter 6 Class 9 Maths Exercises 6.2 (ex 6b) Solutions (Updated For 2024)"},"content":{"rendered":"\n<p>RS Aggarwal Chapter 6 Class 9 Maths Exercise 6.2 Solutions: The word \u2018geometry\u2019 comes from the Greek words \u2018geo\u2019, meaning the \u2018earth\u2019, and \u2018metrein\u2019, meaning \u2018to measure\u2019. Geometry appears to have originated from the need for measuring land. This branch of mathematics was studied in various forms in every ancient civilization, be it in Egypt, Babylonia, China, India, Greece, the Incas, etc. The people of these civilizations faced several practical problems which required the development of geometry in various ways. Know more on <a href=\"https:\/\/www.kopykitab.com\/blog\/rs-aggarwal-solutions-class-9-maths-chapter-6-introduction-to-euclids-geometry\/\">Euclids Geometry.<\/a><\/p>\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_47_1 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"ez-toc-toggle-icon-1\"><label for=\"item-69d02cce87b47\" aria-label=\"Table of Content\"><span style=\"display: flex;align-items: center;width: 35px;height: 30px;justify-content: center;direction:ltr;\"><svg style=\"fill: #000000;color:#000000\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #000000;color:#000000\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/label><input  type=\"checkbox\" id=\"item-69d02cce87b47\"><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 eztoc-visibility-hide-by-default' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/www.kopykitab.com\/blog\/rs-aggarwal-chapter-6-class-9-maths-exercise-6-2-solutions\/#download-rs-aggarwal-chapter-6-class-9-maths-exercise-62-solutions\" title=\"Download RS Aggarwal Chapter 6 Class 9 Maths Exercise 6.2 Solutions\">Download RS Aggarwal Chapter 6 Class 9 Maths Exercise 6.2 Solutions<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/www.kopykitab.com\/blog\/rs-aggarwal-chapter-6-class-9-maths-exercise-6-2-solutions\/#access-the-rs-aggarwal-chapter-6-class-9-maths-exercise-62-solutions\" title=\"Access The RS Aggarwal Chapter 6 Class 9 Maths Exercise 6.2 Solutions\">Access The RS Aggarwal Chapter 6 Class 9 Maths Exercise 6.2 Solutions<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.kopykitab.com\/blog\/rs-aggarwal-chapter-6-class-9-maths-exercise-6-2-solutions\/#important-definition-for-rs-aggarwal-chapter-6-class-9-maths-ex-6b-solutions\" title=\"Important Definition for RS Aggarwal Chapter 6\u00a0Class 9\u00a0Maths Ex 6b\u00a0Solutions\">Important Definition for RS Aggarwal Chapter 6\u00a0Class 9\u00a0Maths Ex 6b\u00a0Solutions<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/www.kopykitab.com\/blog\/rs-aggarwal-chapter-6-class-9-maths-exercise-6-2-solutions\/#faqs-on-rs-aggarwal-chapter-6-class-9-maths-exercise-62-solutions\" title=\"FAQs on RS Aggarwal Chapter 6 Class 9 Maths Exercise 6.2 Solutions\">FAQs on RS Aggarwal Chapter 6 Class 9 Maths Exercise 6.2 Solutions<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/www.kopykitab.com\/blog\/rs-aggarwal-chapter-6-class-9-maths-exercise-6-2-solutions\/#from-where-can-i-find-the-download-link-for-the-rs-aggarwal-chapter-6-class-9-maths-exercise-62-solutions-pdf\" title=\"From where can I find the download link for the RS Aggarwal Chapter 6 Class 9 Maths Exercise 6.2 Solutions PDF?\">From where can I find the download link for the RS Aggarwal Chapter 6 Class 9 Maths Exercise 6.2 Solutions PDF?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/www.kopykitab.com\/blog\/rs-aggarwal-chapter-6-class-9-maths-exercise-6-2-solutions\/#how-much-does-it-cost-to-download-the-rs-aggarwal-chapter-6-class-9-maths-exercise-62-solutions-pdf\" title=\"How much does it cost to download the RS Aggarwal Chapter 6 Class 9 Maths Exercise 6.2 Solutions PDF?\">How much does it cost to download the RS Aggarwal Chapter 6 Class 9 Maths Exercise 6.2 Solutions PDF?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-7\" href=\"https:\/\/www.kopykitab.com\/blog\/rs-aggarwal-chapter-6-class-9-maths-exercise-6-2-solutions\/#can-i-access-the-rs-aggarwal-chapter-6-class-9-maths-exercise-62-solutions-pdf-offline\" title=\"Can I access the RS Aggarwal Chapter 6 Class 9 Maths Exercise 6.2 Solutions PDF Offline?\">Can I access the RS Aggarwal Chapter 6 Class 9 Maths Exercise 6.2 Solutions PDF Offline?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rs-aggarwal-chapter-6-class-9-maths-exercise-62-solutions\"><\/span>Download RS Aggarwal Chapter 6 Class 9 Maths Exercise 6.2 Solutions<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<div tabindex=\"0\" title=\"Advertisement\" aria-label=\"Advertisement\">\u00a0<\/div>\n<div tabindex=\"0\" title=\"Advertisement\" aria-label=\"Advertisement\">\u00a0<\/div>\n<div id=\"example1\" class=\" pdfobject-container\"><embed class=\"pdfobject\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Maths-Chapter-6.pdf\" type=\"application\/pdf\" width=\"622\" height=\"622\" data-mce-fragment=\"1\"><\/embed><\/div>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Maths-Chapter-6.pdf\" target=\"_blank\" rel=\"noopener\">RS Aggarwal Solutions Class 9 Maths Chapter 6 Introduction To Euclids Geometry<\/a><\/p>\n<h2><span class=\"ez-toc-section\" id=\"access-the-rs-aggarwal-chapter-6-class-9-maths-exercise-62-solutions\"><\/span>Access The RS Aggarwal Chapter 6 Class 9 Maths Exercise 6.2 Solutions<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<div class=\"entry-content\">\n<p><strong>Question 1:<\/strong><br \/>Draw the perpendiculars from the AF, BG, CH, DI and EJ on the x-axis.<br \/>(1) The distance of A from the y-axis = OF = -6 units<br \/>The distance of A from the x-axis = AF = 5 units<br \/>Hence, the coordinates of A are (-6, 5)<\/p>\n<p>(2)The distance of B from the y-axis = OG = 5 units<br \/>The distance of B from the x-axis = BG = 4 units<br \/>Hence, the coordinate of B are (5, 4)<\/p>\n<p>(3)The distance of C from the y-axis = OH = -3 units<br \/>The distance of C from the x-axis = HC = 2 units<br \/>Hence, the coordinate of C are (-3, 2)<\/p>\n<p>(4)The distance of D from the y-axis = OI = 2 units<br \/>The distance of D from the x-axis = ID = -2 units<br \/>Hence, the coordinate of D are (2, -2)<\/p>\n<p>(5)The distance of E from the y-axis = OJ = -1 unit<br \/>The distance of E from the x-axis = JE = -4 units<br \/>Hence, the coordinate of E are (-1, -4)<\/p>\n<p>Thus, the coordinates of A, B, C, D and E are respectively, A(-6,5), B(5,4), C(-3,2), D(2,-2) and E(-1,-4)<\/p>\n<p><strong>Question 2:<\/strong><br \/>Let X\u2019OX and Y\u2019OY be the coordinate axes.<br \/>Fix the side of the small squares as one units.<br \/>(i) Starting from O, take +7 units on the x-axis and then +4 units on the\u00a0y-axis to obtain the point P(7, 4)<br \/>(ii) Starting from O, take -5 units on the x-axis and then +3 units on the\u00a0y-axis to obtain the point Q(-5, 3)<br \/>(iii) Starting from O, take -6 units on the x-axis and then -3 units on the\u00a0y-axis to obtain the point R(-6, -3)<br \/>(iv) Starting from O, take +3 units on the x-axis and then -7 units on the\u00a0y-axis to obtain the point S(3, -7)<br \/>(v) Starting from O, take 6 units on the x-axis to obtain the point A(6, 0)<br \/>(vi) Starting from O, take 9 units on the y-axis to obtain the point B(0,9)<br \/>(vii) Mark the point O as O(0, 0)<br \/>(viii) Starting from O, take -3 units on the x-axis and then -3 units on the\u00a0y-axis to obtain the point C(-3, -3)<br \/>These points are shown in the following graph:<br \/><img class=\"alignnone size-full wp-image-114543\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-9-Chapter-6-Coordinate-Geometry-6a-2.1.jpg\" sizes=\"(max-width: 516px) 100vw, 516px\" srcset=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-9-Chapter-6-Coordinate-Geometry-6a-2.1.jpg 516w, https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-9-Chapter-6-Coordinate-Geometry-6a-2.1-239x300.jpg 239w\" alt=\"RS Aggarwal Solutions Class 9 Chapter 6 Coordinate Geometry 6a 2.1\" width=\"516\" height=\"647\" \/><\/p>\n<p><strong>Question 3:<\/strong><br \/>(i) In (7, 0), we have the ordinate = 0.<br \/>Therefore, (7,0) lies on the x-axis<\/p>\n<p>(ii) In (0, -5), we have the abscissa = 0.<br \/>Therefore, (0,-5) lies on the y-axis<\/p>\n<p>(iii) In (0,1), we have the abscissa = 0.<br \/>Therefore, (0,1) lies on the y-axis<\/p>\n<p>(iv) In (-4,0), we have the ordinate = 0.<br \/>Therefore, (-4,0) lies on the x-axis<\/p>\n<p><strong>Question 4:<\/strong><br \/>(i) Points of the type (-, +) lie in the second quadrant. Therefore, the point (-6,5) lies in the II quadrant.<\/p>\n<p>(ii) Points of the type (-, -) lie in the third quadrant. Therefore, the point\u00a0 (-3,-2) lies in the III quadrant.<\/p>\n<p>(iii) Points of the type (+, -) lie in the fourth quadrant. Therefore, the point\u00a0 (2,-9) lies in the IV quadrant.<\/p>\n<p><strong>Question 5:<\/strong><br \/>The given equation is y = x + 1<br \/>Putting x = 1, we get y = 1 + 1 = 2<br \/>Putting x = 2, we get y = 2 + 1 = 3<\/p>\n<p>Thus, we have the following table:<br \/><img class=\"alignnone size-full wp-image-114544\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-9-Chapter-6-Coordinate-Geometry-6a-5.1.png\" alt=\"RS Aggarwal Solutions Class 9 Chapter 6 Coordinate Geometry 6a 5.1\" width=\"111\" height=\"66\" \/><br \/>On a graph paper, draw the lines X\u2019OX and YOY\u2019 as the x-axis and y-axis respectively.<br \/>Then, plot points P (1, 2) and Q (2, 3) on the graph paper. Join PQ and extend it to both sides.<br \/>Then, line PQ is the graph of the equation y = x + 1.<br \/><img class=\"alignnone size-full wp-image-114545\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-9-Chapter-6-Coordinate-Geometry-6a-5.2.png\" sizes=\"(max-width: 451px) 100vw, 451px\" srcset=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-9-Chapter-6-Coordinate-Geometry-6a-5.2.png 451w, https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-9-Chapter-6-Coordinate-Geometry-6a-5.2-284x300.png 284w\" alt=\"RS Aggarwal Solutions Class 9 Chapter 6 Coordinate Geometry 6a 5.2\" width=\"451\" height=\"477\" \/><\/p>\n<p><strong>Question 6:<\/strong><br \/>The give equation is y = 3x + 2<br \/>Putting x = 1, we get y = (3 1) + 2 = 5<br \/>Putting x = 2, we get y = (3 2) + 2 = 8<\/p>\n<p>Thus, we have the following table:<br \/><img class=\"alignnone size-full wp-image-114546\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-9-Chapter-6-Coordinate-Geometry-6a-6.1.png\" sizes=\"(max-width: 102px) 100vw, 102px\" srcset=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-9-Chapter-6-Coordinate-Geometry-6a-6.1.png 102w, https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-9-Chapter-6-Coordinate-Geometry-6a-6.1-100x73.png 100w\" alt=\"RS Aggarwal Solutions Class 9 Chapter 6 Coordinate Geometry 6a 6.1\" width=\"102\" height=\"73\" \/><br \/>On the graph paper, draw the lines X\u2019OX and YOY\u2019 as the x-axis and y-axis respectively.<br \/>Now, plot points P(1,5) and Q(2,8) on the graph paper.<br \/>Join PQ and extend it to both sides.<br \/>Then, line PQ is the graph of the equation y = 3x + 2.<br \/><img class=\"alignnone size-full wp-image-114547\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-9-Chapter-6-Coordinate-Geometry-6a-6.2.png\" sizes=\"(max-width: 450px) 100vw, 450px\" srcset=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-9-Chapter-6-Coordinate-Geometry-6a-6.2.png 450w, https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-9-Chapter-6-Coordinate-Geometry-6a-6.2-278x300.png 278w\" alt=\"RS Aggarwal Solutions Class 9 Chapter 6 Coordinate Geometry 6a 6.2\" width=\"450\" height=\"486\" \/><\/p>\n<p><strong>Question 7:<\/strong><br \/>The given equation is y = 5x \u2013 3<br \/>Putting x = 0, we get y = (5 \u00d7 0) \u2013 3 = -3<br \/>Putting x = 1, we get y = (5 \u00d7 1) \u2013 3 = 2<\/p>\n<p>Thus, we have following table:<br \/><img class=\"alignnone size-full wp-image-114548\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-9-Chapter-6-Coordinate-Geometry-6a-7.1.png\" alt=\"RS Aggarwal Solutions Class 9 Chapter 6 Coordinate Geometry 6a 7.1\" width=\"106\" height=\"71\" \/><br \/>On a graph paper, draw the lines X\u2019OX and YOY\u2019 as the x-axis and y-axis respectively.<br \/>Now plot the points P(0,-3) and Q(1,2).<br \/>Join PQ and extend it in both the directions.<br \/>Then, line PQ is the graph of the equation, y = 5x \u2013 3.<br \/><img class=\"alignnone size-full wp-image-114549\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-9-Chapter-6-Coordinate-Geometry-6a-7.2.png\" sizes=\"(max-width: 452px) 100vw, 452px\" srcset=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-9-Chapter-6-Coordinate-Geometry-6a-7.2.png 452w, https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-9-Chapter-6-Coordinate-Geometry-6a-7.2-150x150.png 150w, https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-9-Chapter-6-Coordinate-Geometry-6a-7.2-300x300.png 300w, https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-9-Chapter-6-Coordinate-Geometry-6a-7.2-100x100.png 100w\" alt=\"RS Aggarwal Solutions Class 9 Chapter 6 Coordinate Geometry 6a 7.2\" width=\"452\" height=\"450\" \/><\/p>\n<p><strong>Question 8:<\/strong><br \/>The given equation is y = 3x<br \/>Putting x = 1, we get y = (3 1) = 3<br \/>Putting x = 2, we get y = (3 2) = 6<\/p>\n<p>Thus, we have the following table:<br \/><img class=\"alignnone size-full wp-image-114550\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-9-Chapter-6-Coordinate-Geometry-6a-8.1.png\" alt=\"RS Aggarwal Solutions Class 9 Chapter 6 Coordinate Geometry 6a 8.1\" width=\"103\" height=\"71\" \/><br \/>On a graph paper draw the lines X\u2019OX and YOY\u2019 as the x-axis and y-axis respectively.<br \/>Now, plot points P(1,3) and Q(2,6).<br \/>Join PQ and extend it in both the directions.<br \/>Then, line PQ is the graph of the equation y = 3x.<br \/><img class=\"alignnone size-full wp-image-114551\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-9-Chapter-6-Coordinate-Geometry-6a-8.2.png\" sizes=\"(max-width: 453px) 100vw, 453px\" srcset=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-9-Chapter-6-Coordinate-Geometry-6a-8.2.png 453w, https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-9-Chapter-6-Coordinate-Geometry-6a-8.2-295x300.png 295w\" alt=\"RS Aggarwal Solutions Class 9 Chapter 6 Coordinate Geometry 6a 8.2\" width=\"453\" height=\"460\" \/><\/p>\n<p><strong>Question 9:<\/strong><br \/>The given equation is y = -x<br \/>Putting x = 1, we get y = -1<br \/>Putting x = 2, we get y = -2<\/p>\n<p>Thus, we have the following table:<br \/><img class=\"alignnone size-full wp-image-114552\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-9-Chapter-6-Coordinate-Geometry-6a-9.1.png\" alt=\"RS Aggarwal Solutions Class 9 Chapter 6 Coordinate Geometry 6a 9.1\" width=\"123\" height=\"71\" \/><br \/>On a graph paper, draw the lines X\u2019OX and YOY\u2019 as the x-axis and y-axis respectively.<br \/>Now, plot the points P(1,-1) and Q(2,-2).<br \/>Join PQ and extend it in both directions.<br \/>Then, line PQ is the graph of the equation y = -x.<br \/><img class=\"alignnone size-full wp-image-114553\" src=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-9-Chapter-6-Coordinate-Geometry-6a-9.2.png\" sizes=\"(max-width: 452px) 100vw, 452px\" srcset=\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-9-Chapter-6-Coordinate-Geometry-6a-9.2.png 452w, https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/07\/RS-Aggarwal-Solutions-Class-9-Chapter-6-Coordinate-Geometry-6a-9.2-281x300.png 281w\" alt=\"RS Aggarwal Solutions Class 9 Chapter 6 Coordinate Geometry 6a 9.2\" width=\"452\" height=\"483\" \/><\/p>\n<\/div>\n<footer class=\"entry-meta\" aria-label=\"Entry meta\"><span class=\"cat-links\"><span class=\"screen-reader-text\">Categories<\/span><\/span><\/footer>\n<h2><span class=\"ez-toc-section\" id=\"important-definition-for-rs-aggarwal-chapter-6-class-9-maths-ex-6b-solutions\"><\/span>Important Definition for RS Aggarwal Chapter 6\u00a0Class 9\u00a0Maths Ex 6b\u00a0Solutions<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<ul>\n<li>A line is an endless length.<\/li>\n<li>A point has no dimension (length, breadth, and width).<\/li>\n<li>A line that lies evenly with the points on itself is straight.<\/li>\n<li>Points are the ends of a line.<\/li>\n<li>A surface is that which has breadth and length only.<\/li>\n<li>A plane surface is a surface that lies evenly with straight lines on itself.<\/li>\n<li>Lines are the edges of a surface.<\/li>\n<\/ul>\n<p>Know More at the <a href=\"https:\/\/cbse.nic.in\/\" target=\"_blank\" rel=\"noopener\">official website<\/a>.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rs-aggarwal-chapter-6-class-9-maths-exercise-62-solutions\"><\/span>FAQs on RS Aggarwal Chapter 6 Class 9 Maths Exercise 6.2 Solutions<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1703654943431\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"from-where-can-i-find-the-download-link-for-the-rs-aggarwal-chapter-6-class-9-maths-exercise-62-solutions-pdf\"><\/span>From where can I find the download link for the RS Aggarwal Chapter 6 Class 9 Maths Exercise 6.2 Solutions PDF?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can find the download link in the above blog.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1703654969802\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"how-much-does-it-cost-to-download-the-rs-aggarwal-chapter-6-class-9-maths-exercise-62-solutions-pdf\"><\/span>How much does it cost to download the RS Aggarwal Chapter 6 Class 9 Maths Exercise 6.2 Solutions PDF?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can download it for free.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1703654985499\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"can-i-access-the-rs-aggarwal-chapter-6-class-9-maths-exercise-62-solutions-pdf-offline\"><\/span>Can I access the RS Aggarwal Chapter 6 Class 9 Maths Exercise 6.2 Solutions PDF Offline?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Once you have downloaded the PDF online, you can access it offline whenever you want.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RS Aggarwal Chapter 6 Class 9 Maths Exercise 6.2 Solutions: The word \u2018geometry\u2019 comes from the Greek words \u2018geo\u2019, meaning the \u2018earth\u2019, and \u2018metrein\u2019, meaning \u2018to measure\u2019. Geometry appears to have originated from the need for measuring land. This branch of mathematics was studied in various forms in every ancient civilization, be it in Egypt, &#8230; <a title=\"RS Aggarwal Chapter 6 Class 9 Maths Exercises 6.2 (ex 6b) Solutions (Updated For 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rs-aggarwal-chapter-6-class-9-maths-exercise-6-2-solutions\/\" aria-label=\"More on RS Aggarwal Chapter 6 Class 9 Maths Exercises 6.2 (ex 6b) Solutions (Updated For 2024)\">Read more<\/a><\/p>\n","protected":false},"author":236,"featured_media":67546,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[1,73395,73412,2985,73410],"tags":[3081,3086,4711],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/67268"}],"collection":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/users\/236"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=67268"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/67268\/revisions"}],"predecessor-version":[{"id":527256,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/67268\/revisions\/527256"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/67546"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=67268"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=67268"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=67268"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}