{"id":128341,"date":"2021-09-16T14:51:08","date_gmt":"2021-09-16T09:21:08","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=128341"},"modified":"2021-09-16T14:51:12","modified_gmt":"2021-09-16T09:21:12","slug":"rd-sharma-class-10-solutions-chapter-14-exercise-14-2","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-14-exercise-14-2\/","title":{"rendered":"RD Sharma Class 10 Solutions Chapter 14 Co-ordinate Geometry Exercise 14.2 (Updated for 2021-22)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-128396\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-14-Exercise-14.2.jpg\" alt=\"RD Sharma Class 10 Solutions Chapter 14 Exercise 14.2\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-14-Exercise-14.2.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-14-Exercise-14.2-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 14 Exercise 14.2:\u00a0<\/strong>Students will solve problems involving finding the distance between two points in this exercise. The <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-for-maths\/\"><strong>RD Sharma Class 10 Solutions<\/strong><\/a> is an excellent resource for students who need to address problems in class. Also download the <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-14-co-ordinate-geometry\/\"><strong>RD Sharma Solutions for Class 10 Maths Chapter 14 Co-ordinate Geometry<\/strong><\/a> Exercise 14.2 in PDF format.<\/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-69d4f3bed49db\" 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\" 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class=\"ez-toc-section\" id=\"download-rd-sharma-class-10-solutions-chapter-14-exercise-142-free-pdf\"><\/span>Download RD Sharma Class 10 Solutions Chapter 14 Exercise 14.2 Free PDF<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<div id=\"example1\" style=\"text-align: justify;\">\u00a0<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 800px;}<br \/>\n.pdfobject { border: 1px solid #666; }<br \/>\n<\/style><\/p>\n<p style=\"text-align: justify;\"><script src=\"https:\/\/www.kopykitab.com\/_utility\/js\/pdfobject.min.js\"><\/script><br \/><script>PDFObject.embed(\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solution-Class-10-Chapter-14-Exercise-14.2.pdf\", \"#example1\");<\/script><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solution-Class-10-Chapter-14-Exercise-14.2.pdf\">RD Sharma Class 10 Solutions Chapter 14 Exercise 14.2<\/a><\/p>\n<h3><span class=\"ez-toc-section\" id=\"access-answers-to-rd-sharma-solutions-class-10-maths-chapter-14-exercise-142-important-question-with-answers\"><\/span>Access answers to RD Sharma Solutions Class 10 Maths Chapter 14 Exercise 14.2- Important Question with Answers<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>RD Sharma Class 10 Solutions Chapter 14 Exercise 14.2 Question 1.<br \/>Find the distance between the following pair of points :<br \/>(i) (-6, 7) and (-1, -5)<br \/>(ii) (a + b, b + c) and (a \u2013 b, c \u2013 b)<br \/>(iii) (a sin \u03b1, -b cos \u03b1) and (-a cos \u03b1, -b sin \u03b1)<br \/>(iv) (a, 0) and (0, b)<br \/>Solution:<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1743\/42726526381_bfa9aba8d3_o.png\" alt=\"RD Sharma Class 10 Chapter 14 Co-Ordinate Geometry \" width=\"350\" height=\"150\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1760\/42726525711_0ccd240fab_o.png\" alt=\"Co-Ordinate Geometry Class 10 RD Sharma \" width=\"362\" height=\"551\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1732\/42726525841_d1bb2f979f_o.png\" alt=\"RD Sharma Class 10 Chapter 14 Co-Ordinate Geometry \" width=\"302\" height=\"161\" \/><\/p>\n<p>RD Sharma Class 10 Solutions Chapter 14 Exercise 14.2 Question 2.<br \/>Find the value of a when the distance between the points (3, a) and (4, 1) is \u221a10<br \/>Solution:<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1760\/42726526851_005d746095_o.png\" alt=\"RD Sharma Class 10 Solutions Co-Ordinate Geometry \" width=\"303\" height=\"165\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1734\/42726526571_239df2721c_o.png\" alt=\"RD Sharma Class 10 Solutions Chapter 14 Co-Ordinate Geometry \" width=\"349\" height=\"287\" \/><\/p>\n<p>RD Sharma Class 10 Solutions Chapter 14 Exercise 14.2 Question 3.<br \/>If the points (2, 1) and (1, -2) are equidistant from the point (x, y), show that x + 3y = 0.<br \/>Solution:<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1753\/42726527301_62b89be581_o.png\" alt=\"RD Sharma Class 10 Solutions Co-Ordinate Geometry \" width=\"359\" height=\"483\" \/><\/p>\n<p>RD Sharma Class 10 Solutions Chapter 14 Exercise 14.2 Question 4.<br \/>Find the values of x, y if the distances of the point (x, y) from (-3, 0) as well as from (3, 0) are 4.<br \/>Solution:<br \/>Distance between (x, y) and (-3, 0) is<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1738\/42726527721_f057fde6a9_o.png\" alt=\"RD Sharma Class 10 Solutions Co-Ordinate Geometry Exercise 14.2 \" width=\"336\" height=\"435\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1728\/42726527491_dee6a572b2_o.png\" alt=\"RD Sharma Class 10 Solutions Chapter 14 Co-Ordinate Geometry \" width=\"343\" height=\"215\" \/><\/p>\n<p>RD Sharma Class 10 Solutions Chapter 14 Exercise 14.2 Question 5.<br \/>The length of a line segment is of 10 units and the coordinates of one end-point are (2, -3). If the abscissa of the other end is 10, find the ordinate of the other end.<br \/>Solution:<br \/>Let the ordinate of other end by y, then The distance between (2, -3) and (10, y) is<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1743\/42726527981_08381b2339_o.png\" alt=\"RD Sharma Class 10 Pdf Chapter 14 Co-Ordinate Geometry \" width=\"355\" height=\"142\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1742\/40916233780_5a8ab7d030_o.png\" alt=\"RD Sharma Solutions Class 10 Chapter 14 Co-Ordinate Geometry \" width=\"349\" height=\"246\" \/><\/p>\n<p>RD Sharma Class 10 Solutions Chapter 14 Exercise 14.2 Question 6.<br \/>Show that the points (-4, -1), (-2, -4), (4, 0) and (2, 3) are the vertices points of a rectangle. (C.B.S.E. 2006C)<br \/>Solution:<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1743\/42726528291_a278688cc0_o.png\" alt=\"RD Sharma Class 10 Pdf Chapter 14 Co-Ordinate Geometry \" width=\"331\" height=\"645\" \/><br \/>AB = CD and AD = BC<br \/>and diagonal AC = BD<br \/>ABCD is a rectangle<\/p>\n<p>RD Sharma Class 10 Solutions Chapter 14 Exercise 14.2 Question 7.<br \/>Show that the points A (1, -2), B (3, 6), C (5, 10) and D (3, 2) are the vertices of a parallelogram.<br \/>Solution:<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1724\/42726528671_d4b420926e_o.png\" alt=\"Learncbse.In Class 10 Chapter 14 Co-Ordinate Geometry \" width=\"344\" height=\"498\" \/><\/p>\n<p>RD Sharma Class 10 Solutions Chapter 14 Exercise 14.2 Question 8.<br \/>Prove that the points A (1, 7), B (4, 2), C (-1, -1) and D (-4, 4) are the vertices of a square. [NCERT]<br \/>Solution:<br \/>Vertices A (1, 7), B (4, 2), C (-1,-1), D (-4, 4)<br \/>If these are the vertices of a square, then its diagonals and sides are equal<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1758\/42726529101_0dbb68da28_o.png\" alt=\"Class 10 RD Sharma Solutions Chapter 14 Co-Ordinate Geometry \" width=\"355\" height=\"134\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1750\/42726528951_76f5fb4cb8_o.png\" alt=\"RD Sharma Class 10 Pdf Free Download Full Book Chapter 14 Co-Ordinate Geometry \" width=\"332\" height=\"469\" \/><\/p>\n<p>RD Sharma Class 10 Solutions Chapter 14 Exercise 14.2 Question 9.<br \/>Prove that the points (3, 0), (6, 4) and (-1, 3) are the vertices of a right-angled isosceles triangle. (C.B.S.E. 2006C)<br \/>Solution:<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1736\/42726529551_4d945c03b6_o.png\" alt=\"Class 10 RD Sharma Solutions Chapter 14 Co-Ordinate Geometry \" width=\"342\" height=\"415\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1730\/27857739367_b7ee57dd38_o.png\" alt=\"RD Sharma Class 10 Pdf Free Download Full Book Chapter 14 Co-Ordinate Geometry \" width=\"347\" height=\"136\" \/><\/p>\n<p>RD Sharma Class 10 Solutions Chapter 14 Exercise 14.2 Question 10.<br \/>Prove that (2, -2), (-2, 1) and (5, 2) are the vertices of a right angled triangle. Find the area of the triangle and the length of the hypotenuse.<br \/>Solution:<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1754\/42726530071_bc9a6ba911_o.png\" alt=\"RD Sharma Class 10 Solution Chapter 14 Co-Ordinate Geometry \" width=\"356\" height=\"449\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1733\/42726529751_861c30684c_o.png\" alt=\"RD Sharma Class 10 Pdf Ebook Chapter 14 Co-Ordinate Geometry \" width=\"349\" height=\"184\" \/><\/p>\n<p>RD Sharma Class 10 Solutions Chapter 14 Exercise 14.2 Question 11.<br \/>Prove that the points (2a, 4a), (2a, 6a) and (2a + \u221a3 a , 5a) are the vertices of an equilateral triangle.<br \/>Solution:<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1725\/42726530221_6d2941c8d1_o.png\" alt=\"RD Sharma Class 10 Solution Chapter 14 Co-Ordinate Geometry \" width=\"354\" height=\"571\" \/><\/p>\n<p>RD Sharma Class 10 Solutions Chapter 14 Exercise 14.2 Question 12.<br \/>Prove that the points (2, 3), (-4, -6) and (1,\u00a032\u00a0)do not form a triangle.<br \/>Solution:<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1756\/42726530551_f46eff4f3a_o.png\" alt=\"RD Sharma Class 10 Pdf Ebook Chapter 14 Co-Ordinate Geometry \" width=\"329\" height=\"276\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1729\/42726530371_71b801cf2b_o.png\" alt=\"RD Sharma Maths Class 10 Solutions Pdf Free Download Chapter 14 Co-Ordinate Geometry \" width=\"292\" height=\"228\" \/><\/p>\n<p>RD Sharma Class 10 Solutions Chapter 14 Exercise 14.2 Question 13.<br \/>The points A (2, 9), B (a, 5) and C (5, 5) are the vertices of a triangle ABC right angled at B. Find the values of a and hence the area of \u2206ABC. [NCERT Exemplar]<br \/>Solution:<br \/>Given that, the points A (2, 9), B (a, 5) and C (5, 5) are the vertices of a \u2206ABC right angled at B.<br \/>By Pythagoras theorem, AC\u00b2 = AB\u00b2 + BC\u00b2 \u2026\u2026\u2026(i)<br \/>Now, by distance formula,<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1750\/42726531441_b206f78435_o.png\" alt=\"RD Sharma Maths Class 10 Solutions Pdf Free Download Chapter 14 Co-Ordinate Geometry \" width=\"347\" height=\"475\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1753\/42726530981_da9692524c_o.png\" alt=\"RD Sharma Class 10 Book Pdf Free Download Chapter 14 Co-Ordinate Geometry \" width=\"722\" height=\"479\" \/><\/p>\n<p>Question 14.<br \/>Show that the quadrilateral whose vertices are (2, -1), (3, 4), (-2, 3) and (-3, -2) is a rhombus.<br \/>Solution:<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1740\/27857741577_b17eb7795f_o.png\" alt=\"RD Sharma Class 10 Book Pdf Free Download Chapter 14 Co-Ordinate Geometry \" width=\"674\" height=\"438\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1746\/28853143458_474923886f_o.png\" alt=\"Class 10 RD Sharma Chapter 14 Co-Ordinate Geometry \" width=\"319\" height=\"176\" \/><\/p>\n<p>Question 15.<br \/>Two vertices of an isosceles triangle are (2, 0) and (2, 5). Find the third vertex if the length of the equal sides is 3.<br \/>Solution:<br \/>Two vertices of an isosceles \u2206ABC are A (2, 0) and B (2, 5). Let co-ordinates of third vertex C be (x, y)<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1755\/42726533061_d7a3a6527d_o.png\" alt=\"RD Sharma Maths Class 10 Solutions Chapter 14 Co-Ordinate Geometry \" width=\"280\" height=\"217\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1750\/27857741737_2c614af9e1_o.png\" alt=\"RD Sharma 10 Class Solutions Chapter 14 Co-Ordinate Geometry \" width=\"353\" height=\"425\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1730\/27857742007_4d87044502_o.png\" alt=\"RD Sharma Maths Class 10 Solutions Chapter 14 Co-Ordinate Geometry \" width=\"364\" height=\"535\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1757\/27857742257_6215a50ed6_o.png\" alt=\"RD Sharma 10 Class Solutions Chapter 14 Co-Ordinate Geometry \" width=\"350\" height=\"188\" \/><\/p>\n<p>Question 16.<br \/>Which point on x-axis is equidistant from (5, 9) and (-4, 6) ?<br \/>Solution:<br \/>Let co-ordinates of two points are A (5, 9), B (-4, 6)<br \/>The required point is on x-axis<br \/>Its ordinates or y-co-ordinates will be 0<br \/>Let the co-ordinates of the point C be (x, 0)<br \/>AC = CB<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1739\/42726533871_3091eda0ee_o.png\" alt=\"RD Sharma Class 10 Textbook PDF Chapter 14 Co-Ordinate Geometry \" width=\"238\" height=\"48\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1737\/42726533561_99b8b82b5a_o.png\" alt=\"RD Sharma Class 10 Textbook PDF Chapter 14 Co-Ordinate Geometry \" width=\"318\" height=\"420\" \/><\/p>\n<p>Question 17.<br \/>Prove that the point (-2, 5), (0, 1) and (2, -3) are collinear.<br \/>Solution:<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1730\/28853145048_022a5560a4_o.png\" alt=\"Maths RD Sharma Class 10 Solutions Chapter 14 Co-Ordinate Geometry \" width=\"332\" height=\"493\" \/><br \/>Now AB + BC = 2\u221a5 +2\u221a5<br \/>and CA = 4\u221a5<br \/>AB + BC = CA<br \/>A, B and C are collinear<\/p>\n<p>Question 18.<br \/>The co-ordinates of the point P are (-3,2). Find the co-ordinates of the point Q which lies on the line joining P and origin such that OP = OQ.<br \/>Solution:<br \/>Co-ordinates of P are (-3, 2) and origin O are (0, 0)<br \/>Let co-ordinates of Q be (x, y)<br \/>O is the mid point of PQ<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1722\/28853145368_767c2c867f_o.png\" alt=\"Maths RD Sharma Class 10 Solutions Chapter 14 Co-Ordinate Geometry \" width=\"338\" height=\"322\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1723\/28853145248_d3f4db9c92_o.png\" alt=\"10th Maths Solution Book Pdf Chapter 14 Co-Ordinate Geometry \" width=\"353\" height=\"288\" \/><br \/>= 9 + 4 = (\u00b13)\u00b2 + (\u00b12)\u00b2<br \/>The point will be in fourth quadrant<br \/>Its y-coordinates will be negative<br \/>and x-coordinates will be positive<br \/>Now comparing the equation<br \/>x\u00b2 = (\u00b13)\u00b2 =&gt; x = \u00b13<br \/>y\u00b2 = (\u00b12)\u00b2 =&gt; y = \u00b12<br \/>x = 3, y = -2<br \/>Co-ordinates of the point Q are (3, -2)<\/p>\n<p>Question 19.<br \/>Which point on y-axis is equidistant from (2, 3) and (-4, 1) ?<br \/>Solution:<br \/>The required point lies on y-axis<br \/>Its abscissa will be zero<br \/>Let the point be C (0, y) and A (2, 3), B (-4, 1)<br \/>Now,<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1744\/42726535341_87684e2f44_o.png\" alt=\"10th Maths Solution Book Pdf Chapter 14 Co-Ordinate Geometry \" width=\"349\" height=\"501\" \/><\/p>\n<p>Question 20.<br \/>The three vertices of a parallelogram are (3, 4), (3, 8) and (9, 8). Find the fourth vertex.<br \/>Solution:<br \/>Let ABCD be a parallelogram and vertices will be A (3, 4), B (3, 8), C (9, 8)<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1753\/42726536771_e1eb247cab_o.png\" alt=\"RD Sharma 10 Chapter 14 Co-Ordinate Geometry \" width=\"295\" height=\"557\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1736\/42726535711_8e9777d92d_o.png\" alt=\"RD Sharma Class 10 Book Pdf Chapter 14 Co-Ordinate Geometry \" width=\"352\" height=\"333\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1722\/42726536061_331f78637e_o.png\" alt=\"RD Sharma 10 Chapter 14 Co-Ordinate Geometry \" width=\"351\" height=\"399\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1747\/42726536501_b397faf7cb_o.png\" alt=\"RD Sharma Class 10 Book Pdf Chapter 14 Co-Ordinate Geometry \" width=\"345\" height=\"232\" \/><\/p>\n<p>Question 21.<br \/>Find a point which is equidistant from the point A (-5, 4) and B (-1, 6). How many such points are there? [NCERT Exemplar]<br \/>Solution:<br \/>Let P (h, k) be the point which is equidistant from the points A (-5, 4) and B (-1, 6).<br \/>PA = PB<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1749\/42008828744_5be64f75cd_o.png\" alt=\"Solution Of RD Sharma Class 10 Chapter 14 Co-Ordinate Geometry \" width=\"356\" height=\"166\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1754\/42726536971_d3f2e8b731_o.png\" alt=\"Solution Of RD Sharma Class 10 Chapter 14 Co-Ordinate Geometry \" width=\"361\" height=\"330\" \/><br \/>So, the mid-point of AB satisfy the Eq. (i).<br \/>Hence, infinite number of points, in fact all points which are solution of the equation 2h + k + 1 = 0, are equidistant from the point A and B.<br \/>Replacing h, k, by x, y in above equation, we have 2x + y + 1 = 0<\/p>\n<p>Question 22.<br \/>The centre of a circle is (2a, a \u2013 7). Find the values of a if the circle passes through the point (11, -9) and has diameter 10\u221a2 units. [NCERT Exemplar]<br \/>Solution:<br \/>By given condition,<br \/>Distance between the centre C (2a, a-1) and the point P (11, -9), which lie on the circle = Radius of circle<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1750\/42726537701_170628a6bd_o.png\" alt=\"RD Sharma 10 Solutions Chapter 14 Co-Ordinate Geometry \" width=\"348\" height=\"274\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1746\/42726537451_8cbd392c6a_o.png\" alt=\"RD Sharma 10 Solutions Chapter 14 Co-Ordinate Geometry \" width=\"366\" height=\"561\" \/><\/p>\n<p>Question 23.<br \/>Ayush starts walking from his house to office. Instead of going to the office directly, he goes to a bank first, from there to his daughter\u2019s school and then reaches the office. What is the extra distance travelled by Ayush in reaching the office? (Assume that all distance covered are in straight lines). If the house is situated at (2, 4), bank at (5, 8), school at (13, 14) and office at (13, 26) and coordinates are in kilometers. [NCERT Exemplar]<br \/>Solution:<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1744\/42726538361_f298b47932_o.png\" alt=\"RD Sharma Maths Book For Class 10 Solution Chapter 14 Co-Ordinate Geometry \" width=\"348\" height=\"146\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1741\/42726537931_8aaabd70f2_o.png\" alt=\"RD Sharma Class 10 Maths Chapter 14 Co-Ordinate Geometry \" width=\"339\" height=\"442\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1754\/42726538161_7db46bc23b_o.png\" alt=\"RD Sharma Maths Book For Class 10 Solution Chapter 14 Co-Ordinate Geometry \" width=\"333\" height=\"400\" \/><\/p>\n<p>Question 24.<br \/>Find the value of k, if the point P (0, 2) is equidistant from (3, k) and (k, 5).<br \/>Solution:<br \/>Let P (0, 2) is equidistant from A (3, k) and B (k, 5)<br \/>PA = PB<br \/>=&gt; PA\u00b2 = PB\u00b2<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1743\/42726538681_40760431bc_o.png\" alt=\"RD Sharma Mathematics Class 10 Pdf Download Free Chapter 14 Co-Ordinate Geometry \" width=\"295\" height=\"418\" \/><\/p>\n<p>Question 25.<br \/>If (-4, 3) and (4, 3) are two vertices of an equilateral triangle, find the coordinates of the third vertex, given that the origin lies in the<br \/>(i) interior,<br \/>(ii) exterior of the triangle. [NCERT Exemplar]<br \/>Solution:<br \/>Let the third vertex of an equilateral triangle be (x, y).<br \/>Let A (-4, 3), B (4,3) and C (x, y).<br \/>We know that, in equilateral triangle the angle between two adjacent side is 60 and all three sides are equal.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1735\/42726539411_580eac899e_o.png\" alt=\"RD Sharma Mathematics Class 10 Pdf Download Free Chapter 14 Co-Ordinate Geometry \" width=\"354\" height=\"316\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1757\/42677482742_85e35b43d7_o.png\" alt=\"Answers Of RD Sharma Class 10 Chapter 14 Co-Ordinate Geometry \" width=\"349\" height=\"577\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1736\/42726539261_9da3d8d426_o.png\" alt=\"RD Sharma Mathematics Class 10 Pdf Download Free Chapter 14 Co-Ordinate Geometry \" width=\"311\" height=\"395\" \/><br \/>But given that, the origin lies in the interior of the \u2206ABC and the x-coordinate of third vertex is zero.<br \/>Then, y-coordinate of third vertex should be negative.<br \/>Hence, the require coordinate of third vertex,<br \/>C = (0, 3 \u2013 4\u221a3). [C \u2260 (0, 3 + 4\u221a3)]<\/p>\n<p>Question 26.<br \/>Show that the points (-3, 2), (-5, -5), (2, -3) and (4, 4) are the vertices of a rhombus. Find the area of this rhombus.<br \/>Solution:<br \/>Let the co-ordinates of the vertices A, B, C and D of a rhombus are A (-3, 2), B (-5, -5), C (2, -3) and D (4, 4)<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1734\/42726539911_10bda9934d_o.png\" alt=\"Answers Of RD Sharma Class 10 Chapter 14 Co-Ordinate Geometry\" width=\"344\" height=\"624\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1734\/42726539511_3174a3d923_o.png\" alt=\"Class 10 RD Sharma Pdf Chapter 14 Co-Ordinate Geometry\" width=\"186\" height=\"183\" \/><\/p>\n<p>Question 27.<br \/>Find the coordinates of the circumcentre of the triangle whose vertices are (3, 0), (-1, -6) and (4, -1). Also, find its circumradius.<br \/>Solution:<br \/>Let ABC is a triangle whose vertices are A (3, 0), B (-1, -6) and C (4, -1)<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1722\/42726540041_08a4663908_o.png\" alt=\"Class 10 RD Sharma Pdf Chapter 14 Co-Ordinate Geometry\" width=\"359\" height=\"617\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1747\/42726539981_d996b9d6b2_o.png\" alt=\"RD Sharma Class 10 Solutions Pdf Free Download Chapter 14 Co-Ordinate Geometry\" width=\"352\" height=\"472\" \/><\/p>\n<p>Question 28.<br \/>Find a point on the x-axis which is equidistant from the points (7, 6) and (-3, 4). [CBSE 2005]<br \/>Solution:<br \/>The required point is on x-axis<br \/>Its ordinate will be O<br \/>Let the co-ordinates of the required point P (x, 0)<br \/>Let the point P is equidistant from the points A (7, 6) and B (-3, 4)<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1739\/42726540261_a9217e4557_o.png\" alt=\"RD Sharma Class 10 Solutions Pdf Free Download Chapter 14 Co-Ordinate Geometry\" width=\"298\" height=\"294\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1733\/42726540101_6eb2bb935a_o.png\" alt=\"RD Sharma Class 10 Solutions Pdf Free Download Chapter 14 Co-Ordinate Geometry\" width=\"323\" height=\"183\" \/><\/p>\n<p>Question 29.<br \/>(i) Show that the points A (5, 6), B (1, 5), C (2, 1) and D (6, 2) are the vertices of a square. [CBSE 2004]<br \/>(ii) Prove that the points A (2, 3), B (-2, 2), C (-1, -2) and D (3, -1) are the vertices of a square ABCD. [CBSE 2013]<br \/>(iii) Name the type of triangle PQR formed by the point P(\u221a2 , \u221a2), Q(- \u221a2, \u2013 \u221a2) and R (-\u221a6 , \u221a6 ). [NCERT Exemplar]<br \/>Solution:<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1728\/42726540561_3ecd7532d7_o.png\" alt=\"Answers Of RD Sharma Class 10 Chapter 14 Co-Ordinate Geometry \" width=\"346\" height=\"560\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1721\/42677484442_2eb4c7788c_o.png\" alt=\"RD Sharma Class 10 Maths Chapter 14 Co-Ordinate Geometry \" width=\"353\" height=\"605\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1730\/42677484532_0ceb93a002_o.png\" alt=\"Answers Of RD Sharma Class 10 Chapter 14 Co-Ordinate Geometry \" width=\"246\" height=\"385\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1721\/42726540501_8427c975c9_o.png\" alt=\"RD Sharma Class 10 Maths Chapter 14 Co-Ordinate Geometry \" width=\"365\" height=\"540\" \/><\/p>\n<p>Question 30.<br \/>Find the point on x-axis which is equidistant from the points (-2, 5) and (2, -3). [CBSE 2004]<br \/>Solution:<br \/>The point P lies on x-axis<br \/>The ordinates of P will be 0 Let the point P be (x, 0)<br \/>Let P is equidistant from A (-2, 5) and B (2, -3)<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1738\/42726540851_ee9cc65e5b_o.png\" alt=\"RD Sharma 10 Solutions Chapter 14 Co-Ordinate Geometry\" width=\"259\" height=\"216\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1731\/42726540681_e5f9117d4e_o.png\" alt=\"RD Sharma Class 10 Book Pdf Chapter 14 Co-Ordinate Geometry\" width=\"293\" height=\"236\" \/><\/p>\n<p>Question 31.<br \/>Find the value of x such that PQ = QR where the co-ordinates of P, Q and R are (6, -1) (1, 3) and (x, 8) respectively. [CBSE 2005]<br \/>Solution:<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1724\/42726540971_3e9301fd3e_o.png\" alt=\"RD Sharma 10 Solutions Chapter 14 Co-Ordinate Geometry\" width=\"305\" height=\"453\" \/><\/p>\n<p>Question 32.<br \/>Prove that the points (0, 0), (5, 5) and (-5, 5) are the vertices of a right isosceles triangle. [CBSE 2005]<br \/>Solution:<br \/>Let the vertices of a triangle be A (0, 0), B (5, 5) and C (-5, 5)<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1751\/42677485772_f38ca7bccb_o.png\" alt=\"10th Maths Solution Book Pdf Chapter 14 Co-Ordinate Geometry \" width=\"270\" height=\"42\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1760\/42726541201_a6ef4f48b8_o.png\" alt=\"RD Sharma Class 10 Textbook PDF Chapter 14 Co-Ordinate Geometry\" width=\"340\" height=\"380\" \/><\/p>\n<p>Question 33.<br \/>If the points P (x, y) is equidistant from the points A (5, 1) and B (1,5), prove that x = y. [CBSE 2005]<br \/>Solution:<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1732\/42677485982_42ba59b8aa_o.png\" alt=\"10th Maths Solution Book Pdf Chapter 14 Co-Ordinate Geometry\" width=\"354\" height=\"504\" \/><\/p>\n<p>Question 34.<br \/>If Q (0, 1) is equidistant from P (5, -3) and R (x, 6) find the values of x. Also find the distances QR and PR. [NCERT]<br \/>Solution:<br \/>Q (0, 1) is equidistant from P (5, -3) and R (x, 6)<br \/>PQ = RQ<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1732\/42677486382_9be711b1d8_o.png\" alt=\"RD Sharma Maths Class 10 Solutions Chapter 14 Co-Ordinate Geometry\" width=\"338\" height=\"344\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1743\/42677486162_696db54394_o.png\" alt=\"RD Sharma Maths Class 10 Solutions Chapter 14 Co-Ordinate Geometry\" width=\"291\" height=\"318\" \/><\/p>\n<p>Question 35.<br \/>Find the values ofy for which the distance between the points P (2, -3) and Q (10, y) is 10 units. [NCERT]<br \/>Solution:<br \/>Distance between P (2, -3) and Q (10, y) = 10<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1729\/42726799191_83c8e730da_o.png\" alt=\"RD Sharma Class 10 Book Pdf Free Download Chapter 14 Co-Ordinate Geometry\" width=\"303\" height=\"375\" \/><\/p>\n<p>Question 36.<br \/>If the point P (k \u2013 1, 2) is equidistant from the points A (3, k) and B (k, 5), find the values of k. [CBSE 2014]<br \/>Solution:<br \/>Point P (k \u2013 1, 2) is equidistant from A (3, k) and B (k, 5)<br \/>PA= PB<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1753\/42726541771_f0b4fa338f_o.png\" alt=\"RD Sharma Class 10 Book Pdf Free Download Chapter 14 Co-Ordinate Geometry\" width=\"360\" height=\"289\" \/><\/p>\n<p>Question 37.<br \/>If the point A (0, 2) is equidistant from the point B (3, p) and C (p, 5), find p. Also, find the length of AB. [CBSE 2014]<br \/>Solution:<br \/>Point A (0, 2) is equidistant from B (3, p) and C (p, 5)<br \/>AB = AC<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1751\/42726541821_6b6cfc47cd_o.png\" alt=\"RD Sharma Class 10 Pdf Ebook Chapter 14 Co-Ordinate Geometry\" width=\"322\" height=\"293\" \/><\/p>\n<p>Question 38.<br \/>Name the quadrilateral formed, if any, by the following points, and give reasons for your answers :<br \/>(i) A (-1, -2), B (1, 0), C (-1, 2), D (-3, 0)<br \/>(ii) A (-3, 5), B (3, 1), C (0, 3), D (-1, -4)<br \/>(iii) A (4, 5), B (7, 6), C (4, 3), D (1, 2)<br \/>Solution:<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1738\/42677487682_086b886c52_o.png\" alt=\"RD Sharma Class 10 Pdf Ebook Chapter 14 Co-Ordinate Geometry\" width=\"216\" height=\"191\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1722\/42726541871_963ac2d12f_o.png\" alt=\"RD Sharma Class 10 Pdf Free Download Full Book Chapter 14 Co-Ordinate Geometry \" width=\"406\" height=\"551\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1759\/42726541981_1a44ea0a17_o.png\" alt=\"RD Sharma Class 10 Pdf Ebook Chapter 14 Co-Ordinate Geometry\" width=\"349\" height=\"447\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1722\/42726542091_eaf028e68a_o.png\" alt=\"RD Sharma Class 10 Pdf Free Download Full Book Chapter 14 Co-Ordinate Geometry \" width=\"360\" height=\"470\" \/><\/p>\n<p>Question 39.<br \/>Find the equation of the perpendicular bisector of the line segment joining points (7, 1) and (3, 5).<br \/>Solution:<br \/>Let the given points are A (7, 1) and B (3, 5) and mid point be M<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1747\/42677488312_0303be757d_o.png\" alt=\"RD Sharma Class 10 Solution Chapter 14 Co-Ordinate Geometry\" width=\"314\" height=\"252\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1724\/42677488092_26736f5560_o.png\" alt=\"RD Sharma Class 10 Solution Chapter 14 Co-Ordinate Geometry\" width=\"352\" height=\"386\" \/><\/p>\n<p>Question 40.<br \/>Prove that the points (3, 0), (4, 5), (-1, 4) and (-2, -1) taken in order, form a rhombus. Also find its area.<br \/>Solution:<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1748\/42677489292_654566275f_o.png\" alt=\"RD Sharma Maths Class 10 Solutions Pdf Free Download Chapter 14 Co-Ordinate Geometry\" width=\"329\" height=\"166\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1750\/42677488952_692276e274_o.png\" alt=\"RD Sharma Maths Class 10 Solutions Pdf Free Download Chapter 14 Co-Ordinate Geometry\" width=\"334\" height=\"460\" \/><\/p>\n<p>Question 41.<br \/>In the seating arrangement of desks in a classroom three students Rohini, Sandhya and Bina are seated at A (3, 1), B (6, 4) and C (8, 6). Do you think they are seated in a line ?<br \/>Solution:<br \/>A (3, 1), B (6, 4) and C (8, 6)<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1730\/42677489652_80572ff6c6_o.png\" alt=\"Class 10 RD Sharma Chapter 14 Co-Ordinate Geometry \" width=\"318\" height=\"393\" \/><\/p>\n<p>Question 42.<br \/>Find a point ony-axis which is equidistant from the points (5, -2) and (-3, 2).<br \/>Solution:<br \/>The point lies on y-axis<br \/>Its x = 0<br \/>Let the required point be (0, y) and let A (5, -2) and B (-3, 2)<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1746\/42726543321_595952a3ec_o.png\" alt=\"RD Sharma 10 Class Solutions Chapter 14 Co-Ordinate Geometry\" width=\"348\" height=\"174\" \/><\/p>\n<p>Question 43.<br \/>Find a relation between x and y such that the point (x, y) is equidistant from the points (3, 6) and (-3, 4). [NCERT]<br \/>Solution:<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1755\/42726543551_eddc5630ab_o.png\" alt=\"Maths RD Sharma Class 10 Solutions Chapter 14 Co-Ordinate Geometry \" width=\"367\" height=\"253\" \/><\/p>\n<p>Question 44.<br \/>If a point A (0, 2) is equidistant from the points B (3, p) and C (p, 5), then find the value of p. [CBSE 2012]<br \/>Solution:<br \/>Point A (0, 2) is equidistant from the points B (3, p) and C (p, 5)<br \/>AB = AC<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1756\/27857753777_944638a070_o.png\" alt=\"RD Sharma 10 Chapter 14 Co-Ordinate Geometry \" width=\"344\" height=\"217\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1759\/42726543741_dd1c3476c1_o.png\" alt=\"RD Sharma 10 Chapter 14 Co-Ordinate Geometry\" width=\"356\" height=\"152\" \/><\/p>\n<p>Question 45.<br \/>Prove that the points (7, 10), (-2, 5) and (3, -4) are the vertices of an isosceles right triangle. [CBSE 2013]<br \/>Solution:<br \/>Let points are A (7, 10), B (-2, 5) and C (3, -4)<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1724\/28853153538_49f2161d01_o.png\" alt=\"RD Sharma Class 10 Book Pdf Chapter 14 Co-Ordinate Geometry\" width=\"349\" height=\"490\" \/><\/p>\n<p>Question 46.<br \/>If the point P (x, 3) is equidistant from the points A (7, -1) and B (6, 8), find the value of x and And the distance AP. [CBSE 2014]<br \/>Solution:<br \/>Point P (x, 3) is equidistant from the points A (7, -1) and B (6, 8)<br \/>PA = PB<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1738\/28853153848_3aa0270917_o.png\" alt=\"RD Sharma Class 10 Book Pdf Chapter 14 Co-Ordinate Geometry\" width=\"328\" height=\"96\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1746\/28853153628_b3803c50e4_o.png\" alt=\"RD Sharma 10 Solutions Chapter 14 Co-Ordinate Geometry\" width=\"338\" height=\"223\" \/><\/p>\n<p>Question 47.<br \/>If A (3, y) is equidistant from points P (8, -3) and Q (7, 6), find the value of y and find the distance AQ. [CBSE 2014]<br \/>Solution:<br \/>Point A (3, y) is equidistant from P (8, -3) and Q (7, 6)<br \/>i.e., AP = AQ<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1759\/27857754857_4b34645e65_o.png\" alt=\"RD Sharma 10 Solutions Chapter 14 Co-Ordinate Geometry\" width=\"343\" height=\"309\" \/><\/p>\n<p>Question 48.<br \/>If (0, -3) and (0, 3) are the two vertices of an equilateral triangle, find the coordinates of its third vertex. [CBSE 2014]<br \/>Solution:<br \/>Let A (0, -3) and B (0, 3) are vertices of an equilateral triangle<br \/>Let the coordinates of the third vertex be C (x, y)<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1747\/27857755577_88dfd030b9_o.png\" alt=\"RD Sharma Class 10 Maths Chapter 14 Co-Ordinate Geometry\" width=\"192\" height=\"195\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1752\/27857755197_1c050c974f_o.png\" alt=\"RD Sharma Class 10 Maths Chapter 14 Co-Ordinate Geometry\" width=\"361\" height=\"580\" \/><\/p>\n<p>Question 49.<br \/>If the point P (2, 2) is equidistant from the points A (-2, k) and B (-2k, -3), find k. Also, find the length of AP.<br \/>Solution:<br \/>Point P (2, 2) is equidistant from the points A (-2, k) and B (-2k, -3)<br \/>AP = BP<br \/>Now,<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1754\/41827002465_9c9a0687cd_o.png\" alt=\"Answers Of RD Sharma Class 10 Chapter 14 Co-Ordinate Geometry\" width=\"327\" height=\"236\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1758\/41827002075_a4cb8c5dec_o.png\" alt=\"Answers Of RD Sharma Class 10 Chapter 14 Co-Ordinate Geometry\" width=\"246\" height=\"215\" \/><\/p>\n<p>Question 50.<br \/>Show that \u2206ABC, where A (-2, 0), B (2, 0), C (0, 2) and \u2206PQR, where P (-4, 0), Q (4, 0), R (0, 4) are similar.<br \/>Solution:<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1741\/40916260080_40ce0d272f_o.png\" alt=\"RD Sharma Class 10 Solutions Pdf Free Download Chapter 14 Co-Ordinate Geometry \" width=\"270\" height=\"343\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1739\/41827002795_03d35a017a_o.png\" alt=\"RD Sharma Class 10 Solutions Pdf Free Download Chapter 14 Co-Ordinate Geometry\" width=\"259\" height=\"304\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1721\/41827003355_8bed0802ff_o.png\" alt=\"RD Sharma Class 10 Solutions Pdf Free Download Chapter 14 Co-Ordinate Geometry\" width=\"227\" height=\"158\" \/><\/p>\n<p>Question 51.<br \/>An equilateral triangle has two vertices at the points (3, 4), and (-2, 3). Find the co-ordinates of the third vertex.<br \/>Solution:<br \/>Let two vertices of an equilateral triangle are A (3,4), and B (-2,3) and let the third vertex be C (x, y)<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1737\/41827005605_a42f83d4c4_o.png\" alt=\"Class 10 RD Sharma Pdf Chapter 14 Co-Ordinate Geometry\" width=\"347\" height=\"633\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1743\/40916260280_8e2cd15177_o.png\" alt=\"Answers Of RD Sharma Class 10 Chapter 14 Co-Ordinate Geometry \" width=\"375\" height=\"518\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1750\/41827004865_fdc06459be_o.png\" alt=\"RD Sharma Mathematics Class 10 Pdf Download Free Chapter 14 Co-Ordinate Geometry\" width=\"259\" height=\"460\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1730\/41827004935_04dc85bd5c_o.png\" alt=\"RD Sharma Class 10 Maths Chapter 14 Co-Ordinate Geometry \" width=\"329\" height=\"472\" \/><\/p>\n<p>Question 52.<br \/>Find the circumcentre of the triangle whose vertices are (-2, -3), (-1, 0), (7, -6).<br \/>Solution:<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1725\/41827006115_31076649c6_o.png\" alt=\"RD Sharma Maths Book For Class 10 Solution Chapter 14 Co-Ordinate Geometry \" width=\"344\" height=\"417\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1753\/41827005525_366094f0d7_o.png\" alt=\"RD Sharma 10 Solutions Chapter 14 Co-Ordinate Geometry \" width=\"356\" height=\"402\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1721\/41827005825_5e53cf4d19_o.png\" alt=\"Solution Of RD Sharma Class 10 Chapter 14 Co-Ordinate Geometry\" width=\"356\" height=\"428\" \/><\/p>\n<p>Question 53.<br \/>Find the angle subtended at the origin by the line segment whose end points are (0, 100) and (10, 0).<br \/>Solution:<br \/>Let co-ordinates of the end points of a line segment are A (0, 100), B (10, 0) and origin is O (0, 0)<br \/>Abscissa of A is 0<br \/>It lies on y-axis<br \/>Similarly, ordinates of B is 0<br \/>It lies on x-axis<br \/>But axes intersect each other at right angle<br \/>AB will subtended 90\u00b0 at the origin<br \/>Angle is 90\u00b0 or\u00a0\u03c02<\/p>\n<p>Question 54.<br \/>Find the centre of the circle passing through (5, -8), (2, -9) and (2, 1).<br \/>Solution:<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1743\/28853157508_3b1c95df53_o.png\" alt=\"RD Sharma Class 10 Book Pdf Chapter 14 Co-Ordinate Geometry \" width=\"352\" height=\"559\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1760\/28853157108_4214d351b5_o.png\" alt=\"RD Sharma 10 Chapter 14 Co-Ordinate Geometry \" width=\"358\" height=\"159\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1729\/28853157318_cfe207d946_o.png\" alt=\"10th Maths Solution Book Pdf Chapter 14 Co-Ordinate Geometry\" width=\"274\" height=\"228\" \/><\/p>\n<p>Question 55.<br \/>If two opposite vertices of a square are (5, 4) and (1, -6), find the coordinates of its remaining two vertices.<br \/>Solution:<br \/>Two opposite points of a square are (5, 4) and (1, -6)<br \/>Let ABCD be a square and A (5, 4) and C (1, -6) are the opposite points<br \/>Let the co-ordinates of B be (x, y). Join AC<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1739\/28853158058_5b6a34a620_o.png\" alt=\"Maths RD Sharma Class 10 Solutions Chapter 14 Co-Ordinate Geometry\" width=\"368\" height=\"570\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1734\/28853157748_e2426aa7f0_o.png\" alt=\"Maths RD Sharma Class 10 Solutions Chapter 14 Co-Ordinate Geometry\" width=\"360\" height=\"428\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1726\/28853157868_da4fdbff43_o.png\" alt=\"Maths RD Sharma Class 10 Solutions Chapter 14 Co-Ordinate Geometry\" width=\"341\" height=\"207\" \/><\/p>\n<p>Question 56.<br \/>Find the centre of the circle passing through (6, -6), (3, -7) and (3, 3).<br \/>Solution:<br \/>Let O is the centre of the circle is (x, 7) Join OA, OB and OC<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1747\/40916263390_6a754a0e4d_o.png\" alt=\"RD Sharma Class 10 Textbook PDF Chapter 14 Co-Ordinate Geometry\" width=\"341\" height=\"247\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1738\/41827008585_23052322d8_o.png\" alt=\"RD Sharma Class 10 Textbook PDF Chapter 14 Co-Ordinate Geometry\" width=\"367\" height=\"597\" \/><\/p>\n<p>Question 57.<br \/>Two opposite vertices of a square are (-1, 2) and (3, 2). Find the co-ordinates of other two vertices.<br \/>Solution:<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1730\/27857761237_fbb11c1b4c_o.png\" alt=\"RD Sharma 10 Class Solutions Chapter 14 Co-Ordinate Geometry\" width=\"346\" height=\"303\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1752\/27857760757_364d2106b9_o.png\" alt=\"RD Sharma Maths Class 10 Solutions Chapter 14 Co-Ordinate Geometry\" width=\"355\" height=\"461\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1743\/27857760997_be0582f72f_o.png\" alt=\"RD Sharma 10 Class Solutions Chapter 14 Co-Ordinate Geometry\" width=\"293\" height=\"212\" \/><\/p>\n<p>We have provided complete details of RD Sharma Class 10 Solutions Chapter 14 Exercise 14.2. If you have any queries related to <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\"><strong>CBSE<\/strong><\/a>\u00a0Class 10, feel free to ask us in the comment section below.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-class-10-solutions-chapter-14-exercise-142\"><\/span>FAQs on RD Sharma Class 10 Solutions Chapter 14 Exercise 14.2<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-1631771873688\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"is-the-rd-sharma-class-10-solutions-chapter-14-exercise-142-available-on-the-kopykitab-website\"><\/span>Is the RD Sharma Class 10 Solutions Chapter 14 Exercise 14.2 available on the Kopykitab website?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, the PDFs of RD Sharma Class 10 Solutions Chapter 14 Exercise 14.2 are available. These solutions are created in a unique method by Kopykitab\u2019s expert faculty.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631772003939\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"where-can-i-get-rd-sharma-class-10-solutions-chapter-14-exercise-142-free-pdf\"><\/span>Where can I get RD Sharma Class 10 Solutions Chapter 14 Exercise 14.2 Free PDF?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can get RD Sharma Class 10 Solutions Chapter 14 Exercise 14.2 Free PDF from the above article.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631772072519\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"can-students-depend-on-the-rd-sharma-class-10-solutions-chapter-14-exercise-142-for-their-exam-preparations-from-kopykitab\"><\/span>Can students depend on the RD Sharma Class 10 Solutions Chapter 14 Exercise 14.2 for their exam preparations from Kopykitab?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, students can trust the RD Sharma Class 10 Solutions Chapter 14 Exercise 14.2 on the Kopykitab website since subject matter experts have created the solutions in accordance with the most recent CBSE guidelines and exam patterns. Furthermore, students can swiftly study the main concepts of the chapters in both online and offline formats at any time and from anywhere.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Class 10 Solutions Chapter 14 Exercise 14.2:\u00a0Students will solve problems involving finding the distance between two points in this exercise. The RD Sharma Class 10 Solutions is an excellent resource for students who need to address problems in class. Also download the RD Sharma Solutions for Class 10 Maths Chapter 14 Co-ordinate Geometry &#8230; <a title=\"RD Sharma Class 10 Solutions Chapter 14 Co-ordinate Geometry Exercise 14.2 (Updated for 2021-22)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-14-exercise-14-2\/\" aria-label=\"More on RD Sharma Class 10 Solutions Chapter 14 Co-ordinate Geometry Exercise 14.2 (Updated for 2021-22)\">Read more<\/a><\/p>\n","protected":false},"author":238,"featured_media":128396,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411,2985,73410],"tags":[3243,9206,73520,4388],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/128341"}],"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\/238"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=128341"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/128341\/revisions"}],"predecessor-version":[{"id":128425,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/128341\/revisions\/128425"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/128396"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=128341"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=128341"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=128341"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}