{"id":128338,"date":"2021-09-16T14:51:46","date_gmt":"2021-09-16T09:21:46","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=128338"},"modified":"2021-09-16T14:51:50","modified_gmt":"2021-09-16T09:21:50","slug":"rd-sharma-class-10-solutions-chapter-14-exercise-14-5","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-14-exercise-14-5\/","title":{"rendered":"RD Sharma Class 10 Solutions Chapter 14 Co-ordinate Geometry Exercise 14.5 (Updated for 2021-22)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-128402\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-14-Exercise-14.5.jpg\" alt=\"RD Sharma Class 10 Solutions Chapter 14 Exercise 14.5\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-14-Exercise-14.5.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-14-Exercise-14.5-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 14 Exercise 14.5:\u00a0<\/strong>In this exercise, students will solve problems involving the area of a triangle and quadrilateral when the vertices&#8217; coordinates are given, as well as the collinearity of three points. The <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-for-maths\/\"><strong>RD Sharma Class 10 Solutions<\/strong><\/a> is created to assist students in grasping the notion of problem-solving. In this article, you will get 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<\/strong><\/a> Co-ordinate Geometry Exercise 14.5 PDF.<\/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-69d4c3f5de39b\" 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-145-free-pdf\"><\/span>Download RD Sharma Class 10 Solutions Chapter 14 Exercise 14.5 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-14.5.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-14.5.pdf\">RD Sharma Class 10 Solutions Chapter 14 Exercise 14.5<\/a><\/p>\n<h3><span class=\"ez-toc-section\" id=\"access-answers-to-rd-sharma-solutions-class-10-maths-chapter-14-exercise-145-important-question-with-answers\"><\/span>Access answers to RD Sharma Solutions Class 10 Maths Chapter 14 Exercise 14.5- Important Question with Answers<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>RD Sharma Class 10 Solutions Chapter 14 Exercise 14.5 Question 1.<br \/>Find the area of a triangle whose vertices are :<br \/>(i) (6, 3), (-3, 5) and (4, -2)<br \/>(ii) (at21, 2at<sub>1<\/sub>), (at22, 2at<sub>2<\/sub>) and (at23, 2at<sub>3<\/sub>)<br \/>(iii) (a, c + a), (a, c) and (-a, c \u2013 a)<br \/>Solution:<br \/>(i) Co-ordinates of \u2206ABC are A (6, 3), B (-3, 5) and C (4, -2)<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1759\/42014283834_3eab835e68_o.png\" alt=\"RD Sharma Class 10 Chapter 14 Co-Ordinate Geometry\" width=\"362\" height=\"555\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1746\/42014283584_be35beefd7_o.png\" alt=\"Co-Ordinate Geometry Class 10 RD Sharma\" width=\"367\" height=\"414\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1728\/28858298048_3a64410801_o.png\" alt=\"RD Sharma Class 10 Chapter 14 Co-Ordinate Geometry\" width=\"334\" height=\"162\" \/><\/p>\n<p>RD Sharma Class 10 Solutions Chapter 14 Exercise 14.5 Question 2.<br \/>Find the area of the quadrilaterals, the coordinates of whose vertices are<br \/>(i) (-3, 2), (5, 4), (7, -6) and (-5, -4)<br \/>(ii) (1, 2), (6, 2), (5, 3) and (3, 4)<br \/>(iii) (-4, -2), (-3, -5), (3, -2), (2, 3) (C.B.S.E. 2009)<br \/>Solution:<br \/>(i) Let vertices of quadrilateral ABCD be A (-3, 2), B (5, 4), C (7, -6) and D (-5, -4)<br \/>Join AC<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1758\/41831916095_6018bef6ec_o.png\" alt=\"Co-Ordinate Geometry Class 10 RD Sharma\" width=\"330\" height=\"560\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1727\/28858298438_2e7bc55f69_o.png\" alt=\"RD Sharma Class 10 Solutions Co-Ordinate Geometry\" width=\"357\" height=\"553\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1738\/41831914955_e97f40fe94_o.png\" alt=\"RD Sharma Class 10 Solutions Co-Ordinate Geometry Exercise 14.5\" width=\"340\" height=\"418\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1736\/41831915275_879a0a8d2d_o.png\" alt=\"RD Sharma Class 10 Solutions Co-Ordinate Geometry\" width=\"367\" height=\"557\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1749\/41831915455_0188557d30_o.png\" alt=\"RD Sharma Class 10 Solutions Co-Ordinate Geometry Exercise 14.5\" width=\"334\" height=\"425\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1742\/41831915785_09609336f7_o.png\" alt=\"RD Sharma Class 10 Solutions Chapter 14 Co-Ordinate Geometry\" width=\"350\" height=\"111\" \/><\/p>\n<p>RD Sharma Class 10 Solutions Chapter 14 Exercise 14.5 Question 3.<br \/>The four vertices of a quadrilaterals are (1, 2), (-5, 6), (7, -4) and (k, -2) taken in order. If the area of the quadrilateral is zero, find the value of k ?<br \/>Solution:<br \/>Let the vertices of quadrilateral ABCD be<br \/>A (1, 2), B (-5, 6), C (7, -4) and D (k, -2)<br \/>Join AC<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1733\/28858300008_551a6027b3_o.png\" alt=\"RD Sharma Class 10 Solutions Chapter 14 Co-Ordinate Geometry\" width=\"335\" height=\"466\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1731\/41831916345_e3922a4f53_o.png\" alt=\"RD Sharma Class 10 Pdf Chapter 14 Co-Ordinate Geometry\" width=\"335\" height=\"200\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1758\/41831916645_05c5680e93_o.png\" alt=\"RD Sharma Solutions Class 10 Chapter 14 Co-Ordinate Geometry\" width=\"350\" height=\"312\" \/><\/p>\n<p>RD Sharma Class 10 Solutions Chapter 14 Exercise 14.5 Question 4.<br \/>The vertices of \u2206ABC are (-2, 1), (5, 4) and (2, -3) respectively. Find the area of the triangle and the length of the altitude through A.<br \/>Solution:<br \/>Vertices of \u2206ABC are A (-2, 1), B (5, 4) and C (2, -3) and AD \u22a5 BC, let AD = h<br \/>Now area of \u2206ABC<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1733\/28858300198_df4ecc96c4_o.png\" alt=\"Learncbse.In Class 10 Chapter 14 Co-Ordinate Geometry\" width=\"357\" height=\"463\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1725\/28858300088_0e58b1bc33_o.png\" alt=\"Class 10 RD Sharma Solutions Chapter 14 Co-Ordinate Geometry\" width=\"352\" height=\"376\" \/><\/p>\n<p>RD Sharma Class 10 Solutions Chapter 14 Exercise 14.5 Question 5.<br \/>Show that the following sets of points are collinear<br \/>(a) (2, 5), (4, 6) and (8, 8)<br \/>(b) (1, -1), (2, 1) and (4, 5)<br \/>Solution:<br \/>We know that points are collinear if the area of the triangle formed by them is zero<br \/>(a) Vertices of \u2206ABC are (2, 5), (4, 6) and (8, 8)<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1742\/28858300598_26588f0c21_o.png\" alt=\"RD Sharma Class 10 Pdf Free Download Full Book Chapter 14 Co-Ordinate Geometry \" width=\"360\" height=\"412\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1733\/28858300328_9306abab7a_o.png\" alt=\"RD Sharma Class 10 Solution Chapter 14 Co-Ordinate Geometry\" width=\"345\" height=\"277\" \/><\/p>\n<p>Question 6.<br \/>Find the area of a quadrilateral ABCD, the coordinates of whose varities are A (-3, 2), B (5, 4), C (7, -6) and D (-5, -4). [CBSE 2016]<br \/>Solution:<br \/>Area of quadrilateral ABCD<br \/>= area of \u2206ABC + area of \u2206ACD<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1756\/28858300708_589c4c8490_o.png\" alt=\"RD Sharma Class 10 Pdf Free Download Full Book Chapter 14 Co-Ordinate Geometry \" width=\"343\" height=\"544\" \/><\/p>\n<p>Question 7.<br \/>In \u2206ABC, the coordinates of vertex A are (0, -1) and D (1, 0) and E (0, 1) respectively the mid-points of the sides AB and AC. If F is the mid-point of side C, find the area of \u2206DEF. [CBSE 2016]<br \/>Solution:<br \/>Let B (p, q), C (r, s) and F (x, y)<br \/>Mid-point of AB = Coordinates of D<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1755\/28858301548_5d5a9b8e40_o.png\" alt=\"RD Sharma Class 10 Pdf Ebook Chapter 14 Co-Ordinate Geometry\" width=\"300\" height=\"424\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1746\/28858301228_a3d193def7_o.png\" alt=\"RD Sharma Class 10 Book Pdf Free Download Chapter 14 Co-Ordinate Geometry\" width=\"314\" height=\"356\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1726\/28858301438_c845d338df_o.png\" alt=\"RD Sharma Maths Class 10 Solutions Pdf Free Download Chapter 14 Co-Ordinate Geometry\" width=\"355\" height=\"360\" \/><\/p>\n<p>Question 8.<br \/>Find the area of the triangle PQR with Q (3, 2) and the mid-points of the sides through Q being (2, -1) and (1, 2). [CBSE 2015]<br \/>Solution:<br \/>In \u2206PQR, L and N are mid points of QR and QP respectively coordinates of Q are (3, 2) of L are (2, -1) and of N are (1, 2)<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1729\/28858302058_5f1bca47db_o.png\" alt=\"Class 10 RD Sharma Chapter 14 Co-Ordinate Geometry\" width=\"347\" height=\"416\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1738\/28858301778_aca3dc360c_o.png\" alt=\"RD Sharma Maths Class 10 Solutions Chapter 14 Co-Ordinate Geometry\" width=\"354\" height=\"407\" \/><\/p>\n<p>Question 9.<br \/>If P (-5, -3), Q (-4, -6), R (2, -3) and S (1, 2) are the vertices of a quadrilateral PQRS, find its area. [CBSE 2015]<br \/>Solution:<br \/>P (-5, -3), Q (-4, -6), R (2, -3) and S (1,2) are the vertices of a quadrilateral PQRS<br \/>Join PR which forms two triangles PQR and PSR<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1732\/41831918745_0a4af11ff6_o.png\" alt=\"RD Sharma 10 Class Solutions Chapter 14 Co-Ordinate Geometry\" width=\"333\" height=\"415\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1758\/41831918675_c31760a7ac_o.png\" alt=\"RD Sharma Maths Class 10 Solutions Chapter 14 Co-Ordinate Geometry\" width=\"357\" height=\"282\" \/><\/p>\n<p>Question 10.<br \/>If A (-3, 5), B (-2, -7), C (1, -8) and D (6, 3) are the vertices of a quadrilateral ABCD, find its area. [CBSE 2014]<br \/>Solution:<br \/>A (-3, 5), B (-2, -7), C (1,-8) and D (6, 3) are the vertices of a quadrilateral ABCD<br \/>Join AC<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1745\/28858302578_97f7b295ab_o.png\" alt=\"RD Sharma 10 Class Solutions Chapter 14 Co-Ordinate Geometry\" width=\"339\" height=\"518\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1736\/41831918875_b41936bb0d_o.png\" alt=\"RD Sharma Class 10 Textbook PDF Chapter 14 Co-Ordinate Geometry\" width=\"346\" height=\"366\" \/><\/p>\n<p>Question 11.<br \/>For what value of \u2018a\u2019 the points (a, 1), (1, -1) and (11, 4) are collinear ?<br \/>Solution:<br \/>Let the vertices of \u2206ABC are A (a, 1), B (1, -1) and C (11, 4)<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1739\/28858302668_27f25593e2_o.png\" alt=\"Maths RD Sharma Class 10 Solutions Chapter 14 Co-Ordinate Geometry\" width=\"352\" height=\"434\" \/><\/p>\n<p>Question 12.<br \/>Prove that the points (a, b), (a<sub>1<\/sub>, b<sub>1<\/sub>) and (a \u2013 a<sub>1<\/sub>, b \u2013 b<sub>1<\/sub>) are collinear if ab<sub>1<\/sub>\u00a0= a<sub>1<\/sub>b.<br \/>Solution:<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1759\/28858302858_dee0261f95_o.png\" alt=\"RD Sharma Class 10 Textbook PDF Chapter 14 Co-Ordinate Geometry\" width=\"355\" height=\"559\" \/><\/p>\n<p>Question 13.<br \/>If the vertices of a triangle are (1, -3), (4, p) and (-9, 7) and its area is 15 sq. units, find the value(s) of p. [CBSE 2012]<br \/>Solution:<br \/>The vertices of a triangle are (1, -3), (4, p) and (-9, 7) and area of triangle = 15 sq. units<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1757\/28858303218_1259680194_o.png\" alt=\"Maths RD Sharma Class 10 Solutions Chapter 14 Co-Ordinate Geometry\" width=\"335\" height=\"257\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1727\/28858303028_73085006bd_o.png\" alt=\"10th Maths Solution Book Pdf Chapter 14 Co-Ordinate Geometry\" width=\"180\" height=\"134\" \/><\/p>\n<p>Question 14.<br \/>If (x, y) be on the line joining the two points (1, -3) and (-4, 2), prove that x + y + 2 = 0.<br \/>Solution:<br \/>Point (x, y) be on the line joining the two points (1, -3) and (-4, 2)<br \/>Points (x, y), (1, -3) and (-4, 2) are collinear<br \/>Let the points (x, y) (1, -3) and (-4, 2) are the vertices of a triangle, then<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1744\/28858303348_7238944624_o.png\" alt=\"RD Sharma 10 Chapter 14 Co-Ordinate Geometry\" width=\"355\" height=\"424\" \/><\/p>\n<p>Question 15.<br \/>Find the value of k if points (k, 3), (6, -2) and (-3, 4) are collinear. [CBSE 2008]<br \/>Solution:<br \/>Let the points (k, 3), (6, -2) and (-3, 4) be the vertices of a triangle, then<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1744\/28858303718_8be751d000_o.png\" alt=\"10th Maths Solution Book Pdf Chapter 14 Co-Ordinate Geometry\" width=\"341\" height=\"93\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1752\/28858303528_0b3cd1b334_o.png\" alt=\"RD Sharma 10 Chapter 14 Co-Ordinate Geometry\" width=\"351\" height=\"455\" \/><\/p>\n<p>Question 16.<br \/>Find the value of k, if the points A (7, -2), B (5, 1) and C (3, 2k) are collinear. [CBSE 2010]<br \/>Solution:<br \/>Points A (7, -2), B (5, 1) and C (3, 2k) are collinear<br \/>area of \u2206ABC = 0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1750\/28858303808_fb731e8e6b_o.png\" alt=\"RD Sharma Class 10 Book Pdf Chapter 14 Co-Ordinate Geometry\" width=\"346\" height=\"356\" \/><\/p>\n<p>Question 17.<br \/>If the point P (m, 3) lies on the line segment joining the points A (\u221225\u00a0, 6) and B (2, 8), find the value of m.<br \/>Solution:<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1731\/28858303968_ca65eef75b_o.png\" alt=\"Solution Of RD Sharma Class 10 Chapter 14 Co-Ordinate Geometry\" width=\"348\" height=\"509\" \/><\/p>\n<p>Question 18.<br \/>If R (x, y) is a point on the line segment joining the points P (a, b) and Q (b, a), then prove that x + y = a + b. [CBSE 2010]<br \/>Solution:<br \/>Point R (x, y) lies on the line segment joining the points P (a, b) and Q (b, a)<br \/>Area of \u2206PRQ = 0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1752\/28858304158_487c0314e2_o.png\" alt=\"RD Sharma Class 10 Book Pdf Chapter 14 Co-Ordinate Geometry\" width=\"302\" height=\"141\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1721\/28858304068_c32b9d32cb_o.png\" alt=\"Solution Of RD Sharma Class 10 Chapter 14 Co-Ordinate Geometry\" width=\"348\" height=\"384\" \/><\/p>\n<p>Question 19.<br \/>Find the value of k, if the points A (8, 1), B (3, -4) and C (2, k) are collinear. [CBSE 2010]<br \/>Solution:<br \/>The points A (8, 1), B (3, -4) and C (2, k) are collinear<br \/>Area of \u2206ABC = 0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1747\/28858304288_1bac441cea_o.png\" alt=\"RD Sharma 10 Solutions Chapter 14 Co-Ordinate Geometry\" width=\"352\" height=\"421\" \/><\/p>\n<p>Question 20.<br \/>Find the value of a for which the area of the triangle formed by the points A (a, 2a), B (-2, 6) and C (3, 1) is 10 square units.<br \/>Solution:<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1749\/28858304418_66eaf5d100_o.png\" alt=\"RD Sharma Maths Book For Class 10 Solution Chapter 14 Co-Ordinate Geometry\" width=\"348\" height=\"550\" \/><\/p>\n<p>Question 21.<br \/>If a \u2260 b \u2260 0, prove that the points (a, a\u00b2), (b, b\u00b2), (0, 0) are never collinear. [CBSE 2017]<br \/>Solution:<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1729\/28858304658_0d1e002bb1_o.png\" alt=\"RD Sharma 10 Solutions Chapter 14 Co-Ordinate Geometry\" width=\"339\" height=\"211\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1737\/28858304538_02b568aa97_o.png\" alt=\"RD Sharma Maths Book For Class 10 Solution Chapter 14 Co-Ordinate Geometry\" width=\"335\" height=\"278\" \/><\/p>\n<p>Question 22.<br \/>The area of a triangle is 5 sq. units. Two of its vertices are at (2, 1) and (3, -2). If the third vertex is (72\u00a0, y), find y. [CBSE 2017]<br \/>Solution:<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1744\/28858304778_6001f962aa_o.png\" alt=\"RD Sharma Class 10 Maths Chapter 14 Co-Ordinate Geometry\" width=\"354\" height=\"534\" \/><\/p>\n<p>Question 23.<br \/>Prove that the points (a, 0), (0, b) and (1, 1) are collinear if,\u00a01a\u00a0+\u00a01b\u00a0= 1.<br \/>Solution:<br \/>Let the points are A (a, 0), B (0, b) and C (1, 1) which form a triangle<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1749\/28858304818_632293312b_o.png\" alt=\"RD Sharma Mathematics Class 10 Pdf Download Free Chapter 14 Co-Ordinate Geometry\" width=\"732\" height=\"381\" \/><\/p>\n<p>Question 24.<br \/>The point A divides the join of P (-5, 1) and Q (3, 5) in the ratio k : 1. Find the two values of k for which the area of \u2206ABC where B is (1, 5) and C (7, -2) is equal to 2 units.<br \/>Solution:<br \/>Let the coordinates of A be (x, y) which divides the join of P (-5, 1) and Q (3, 5) in the ratio. Then<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1749\/28858305288_a88d83cbb2_o.png\" alt=\"RD Sharma Class 10 Maths Chapter 14 Co-Ordinate Geometry\" width=\"550\" height=\"360\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1724\/28858304898_1eb1a01c2d_o.png\" alt=\"RD Sharma Mathematics Class 10 Pdf Download Free Chapter 14 Co-Ordinate Geometry\" width=\"363\" height=\"333\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1734\/28858305158_5d31b04334_o.png\" alt=\"Answers Of RD Sharma Class 10 Chapter 14 Co-Ordinate Geometry\" width=\"358\" height=\"369\" \/><\/p>\n<p>Question 25.<br \/>The area of a triangle is 5. Two of its vertices are (2, 1) and (3, -2). The third vertex lies on y = x + 3. Find the third vertex.<br \/>Solution:<br \/>Let the coordinates of third vertex of the triangle be (x, y) and other two vertices are (2, 1) and (3, 2)<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1759\/28858305668_5e18dd6420_o.png\" alt=\"Answers Of RD Sharma Class 10 Chapter 14 Co-Ordinate Geometry\" width=\"322\" height=\"86\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1734\/28858305498_c20fb05825_o.png\" alt=\"Class 10 RD Sharma Pdf Chapter 14 Co-Ordinate Geometry\" width=\"342\" height=\"350\" \/><\/p>\n<p>Question 26.<br \/>If a \u2260 b \u2260 c, prove that the points (a, a\u00b2), (b, b\u00b2), (c, c\u00b2) can never be collinear.<br \/>Solution:<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1722\/28858305898_80761378df_o.png\" alt=\"Class 10 RD Sharma Pdf Chapter 14 Co-Ordinate Geometry\" width=\"360\" height=\"545\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1742\/28858305768_9edeb711e7_o.png\" alt=\"RD Sharma Class 10 Solutions Pdf Free Download Chapter 14 Co-Ordinate Geometry\" width=\"356\" height=\"323\" \/><\/p>\n<p>Question 27.<br \/>Four points A (6, 3), B (-3, 5), C (4, -2) and D (x, 3x) are given in such a way that\u00a0\u25b3DBC\u25b3ABC=12\u00a0, find x?<br \/>Solution:<br \/>Let A (6, 3), B (-3, 5), C (4, -2) and D (x, 3x) are the vertices of quadrilateral ABCD<br \/>AC and BD are joined<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1755\/28858306278_51aa1cfd28_o.png\" alt=\"RD Sharma Class 10 Solutions Pdf Free Download Chapter 14 Co-Ordinate Geometry\" width=\"344\" height=\"450\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1738\/42014301404_585de22f0c_o.png\" alt=\"Answers Of RD Sharma Class 10 Chapter 14 Co-Ordinate Geometry\" width=\"344\" height=\"456\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1729\/28858306148_9c51e26c63_o.png\" alt=\"RD Sharma Class 10 Maths Chapter 14 Co-Ordinate Geometry\" width=\"333\" height=\"327\" \/><\/p>\n<p>Question 28.<br \/>If three points (x<sub>1<\/sub>, y<sub>1<\/sub>), (x<sub>2<\/sub>, y<sub>2<\/sub>), (x<sub>3<\/sub>, y<sub>3<\/sub>) lie on the same line, prove that<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1724\/28858306648_32f5c32391_o.png\" alt=\"Answers Of RD Sharma Class 10 Chapter 14 Co-Ordinate Geometry\" width=\"289\" height=\"61\" \/><br \/>Solution:<br \/>Let the points (x<sub>1<\/sub>, y<sub>1<\/sub>), (x<sub>2<\/sub>, y<sub>2<\/sub>), (x<sub>3<\/sub>, y<sub>3<\/sub>) are the vertices of a triangle<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1739\/28858306508_1aabcc3aff_o.png\" alt=\"RD Sharma Class 10 Maths Chapter 14 Co-Ordinate Geometry\" width=\"350\" height=\"360\" \/><\/p>\n<p>Question 29.<br \/>Find the area of a parallelogram ABCD if three of its vertices are A (2, 4), B (2 + \u221a3, 5) and C (2, 6). [CBSE 2013]<br \/>Solution:<br \/>Three vertices of a ||gm ABCD are A (2, 4), B (2 + \u221a3 , 5) and C (2, 6).<br \/>Draw one diagonal AC of ||gm ABCD<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1751\/41831922995_9ccd4a9a57_o.png\" alt=\"RD Sharma 10 Solutions Chapter 14 Co-Ordinate Geometry\" width=\"352\" height=\"421\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1746\/28858306808_9dab6a2dcc_o.png\" alt=\"RD Sharma Class 10 Book Pdf Chapter 14 Co-Ordinate Geometry\" width=\"352\" height=\"136\" \/><\/p>\n<p>Question 30.<br \/>Find the value (s) of k for which the points (3k \u2013 1, k \u2013 2), (k, k \u2013 7) and (k \u2013 1, -k \u2013 2) are collinear. [CBSE 2014]<br \/>Solution:<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1737\/28858307038_43e76c5f68_o.png\" alt=\"RD Sharma 10 Solutions Chapter 14 Co-Ordinate Geometry\" width=\"348\" height=\"526\" \/><\/p>\n<p>Question 31.<br \/>If the points A (-1, -4), B (b, c) and C (5, -1) are collinear and 2b + c = 4, find the values of b and c. [CBSE 2014]<br \/>Solution:<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1723\/41831923505_8dd487f634_o.png\" alt=\"RD Sharma Class 10 Book Pdf Chapter 14 Co-Ordinate Geometry\" width=\"348\" height=\"133\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1735\/41831923325_166e109bf2_o.png\" alt=\"10th Maths Solution Book Pdf Chapter 14 Co-Ordinate Geometry\" width=\"356\" height=\"546\" \/><\/p>\n<p>Question 32.<br \/>If the points A (-2, 1), B (a, b) and C (4, -1) are collinear and a \u2013 b = 1, find the values of a and 6. [CBSE 2014]<br \/>Solution:<br \/>Points A (-2, 1), B (a, b) and C (4, -1) are<br \/>collinear if area \u2206ABC = 0<br \/>Now area of \u2206ABC<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1759\/28858307458_f09ab0a4b2_o.png\" alt=\"RD Sharma Class 10 Textbook PDF Chapter 14 Co-Ordinate Geometry\" width=\"334\" height=\"272\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1758\/28858307328_79e5ece000_o.png\" alt=\"RD Sharma Maths Class 10 Solutions Chapter 14 Co-Ordinate Geometry\" width=\"345\" height=\"200\" \/><\/p>\n<p>Question 33.<br \/>If the points A (1, -2), B (2, 3), C (a, 2) and D (-4, -3) form a parallelogram, find the value of a and height of the parallelogram taking AB as base. [NCERT Exemplar]<br \/>Solution:<br \/>In parallelogram, we know that, diagonals bisects each other<br \/>i.e., mid-point of AC = mid-point of BD<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1746\/28858307988_ac4ffc4c6a_o.png\" alt=\"RD Sharma Class 10 Book Pdf Free Download Chapter 14 Co-Ordinate Geometry\" width=\"360\" height=\"568\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1721\/28858307568_0b8ed3e380_o.png\" alt=\"RD Sharma Class 10 Pdf Ebook Chapter 14 Co-Ordinate Geometry\" width=\"371\" height=\"547\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1747\/40921463940_506161763f_o.png\" alt=\"Learncbse.In Class 10 Chapter 14 Co-Ordinate Geometry\" width=\"363\" height=\"437\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1756\/41831924015_fb7b5fda73_o.png\" alt=\"RD Sharma Class 10 Pdf Free Download Full Book Chapter 14 Co-Ordinate Geometry \" width=\"364\" height=\"425\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1722\/40921464240_aeb7b24f43_o.png\" alt=\"Learncbse.In Class 10 Chapter 14 Co-Ordinate Geometry\" width=\"361\" height=\"295\" \/><\/p>\n<p>Question 34.<br \/>A (6, 1), B (8, 2) and C (9, 4) are three vertices of a parallelogram ABCD. If E is the mid-point of DC, find the area of \u2206ADE. [NCERT Exemplar]<br \/>Solution:<br \/>Given that, A (6,1), B (8,2) and C (9,4) are three vertices of a parallelogram ABCD.<br \/>Let the fourth vertex of parallelogram be (x, y).<br \/>We know that, the diagonal of a parallelogram bisect each other.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1748\/28858308208_b0b4fdfd13_o.png\" alt=\"RD Sharma Class 10 Pdf Chapter 14 Co-Ordinate Geometry\" width=\"551\" height=\"440\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1744\/28858308108_6f6b62cc16_o.png\" alt=\"RD Sharma Class 10 Solutions Co-Ordinate Geometry Exercise 14.5\" width=\"586\" height=\"354\" \/><\/p>\n<p>Question 35.<br \/>If D (\u221212,\u00a052) E (7, 3) and F (72,\u00a072) are the mid-points of sides of \u2206ABC, find the area of \u2206ABC. [NCERT Exemplar]<br \/>Solution:<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1754\/28858308668_1c296da17f_o.png\" alt=\"RD Sharma Class 10 Pdf Chapter 14 Co-Ordinate Geometry\" width=\"585\" height=\"410\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1725\/40921464680_4b60c66351_o.png\" alt=\"RD Sharma Class 10 Solutions Co-Ordinate Geometry Exercise 14.5\" width=\"457\" height=\"328\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1731\/28858308398_ddf10307b8_o.png\" alt=\"RD Sharma Class 10 Solutions Chapter 14 Co-Ordinate Geometry \" width=\"288\" height=\"352\" \/><\/p>\n<p>We have provided complete details of RD Sharma Class 10 Solutions Chapter 14 Exercise 14.5. 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-145\"><\/span>FAQs on RD Sharma Class 10 Solutions Chapter 14 Exercise 14.5<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-1631771895857\" 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-145-available-on-the-kopykitab-website\"><\/span>Is the RD Sharma Class 10 Solutions Chapter 14 Exercise 14.5 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 14.5 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-1631772022667\" 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-145-free-pdf\"><\/span>Where can I get RD Sharma Class 10 Solutions Chapter 14 Exercise 14.5 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 14.5 Free PDF from the above article.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631772094240\" 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-145-for-their-exam-preparations-from-kopykitab\"><\/span>Can students depend on the RD Sharma Class 10 Solutions Chapter 14 Exercise 14.5 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 14.5 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 14.5:\u00a0In this exercise, students will solve problems involving the area of a triangle and quadrilateral when the vertices&#8217; coordinates are given, as well as the collinearity of three points. The RD Sharma Class 10 Solutions is created to assist students in grasping the notion of problem-solving. In &#8230; <a title=\"RD Sharma Class 10 Solutions Chapter 14 Co-ordinate Geometry Exercise 14.5 (Updated for 2021-22)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-14-exercise-14-5\/\" aria-label=\"More on RD Sharma Class 10 Solutions Chapter 14 Co-ordinate Geometry Exercise 14.5 (Updated for 2021-22)\">Read more<\/a><\/p>\n","protected":false},"author":238,"featured_media":128402,"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\/128338"}],"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=128338"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/128338\/revisions"}],"predecessor-version":[{"id":128428,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/128338\/revisions\/128428"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/128402"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=128338"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=128338"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=128338"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}