{"id":6063,"date":"2023-08-20T07:12:19","date_gmt":"2023-08-20T07:12:19","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=6063"},"modified":"2023-08-20T07:12:19","modified_gmt":"2023-08-20T07:12:19","slug":"ncert-solutions-class-10-maths-chapter-7-coordinate-geometry-ex-7-3","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-7-coordinate-geometry-ex-7-3\/","title":{"rendered":"NCERT Solutions Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.3"},"content":{"rendered":"<h2 style=\"text-align: center;\"><strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">NCERT Solutions Class 10 Maths\u00a0<\/span><\/strong><br \/>\n<strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Chapter &#8211; 7 (Coordinate Geometry)\u00a0<\/span><\/strong><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">The NCERT Solutions in English Language for Class 10 Mathematics <strong>Chapter &#8211; 7 Coordinate Geometry <\/strong>Exercise 7.3 has been provided here to help the students in solving the questions from this exercise.\u00a0<\/span><\/p>\n<p><span style=\"color: #ff0000;\"><strong>Chapter : 7 Coordinate Geometry<\/strong> <\/span><\/p>\n<ul>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-7-coordinate-geometry-ex-7-1\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 7.1<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-7-coordinate-geometry-ex-7-2\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 7.2<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-7-coordinate-geometry-ex-7-4\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 7.4<\/a><\/span><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 7.3<\/span><\/strong><\/h2>\n<p style=\"text-align: justify;\"><span style=\"font-family: Georgia, Palatino; color: #000000;\"><strong>1. Find the area of the triangle whose vertices are:<br \/>\n<\/strong>(i) (2, 3), (-1, 0), (2, -4)<br \/>\n(ii) (-5, -1), (3, -5), (5, 2)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: Georgia, Palatino; color: #000000;\"><strong>Solution &#8211;<br \/>\n<\/strong>Area of a triangle formula = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u00d7 [x<sub>1<\/sub>(y<sub>2<\/sub>\u00a0\u2013 y<sub>3<\/sub>) + x<sub>2<\/sub>(y<sub>3<\/sub>\u00a0\u2013 y<sub>1<\/sub>) + x<sub>3<\/sub>(y<sub>1<\/sub>\u00a0\u2013 y<sub>2<\/sub>)]<br \/>\n<strong>(i) Here,<br \/>\n<\/strong>x<sub>1<\/sub>\u00a0= 2, x<sub>2<\/sub>\u00a0= -1, x<sub>3<\/sub>\u00a0= 2, y<sub>1<\/sub>\u00a0= 3, y<sub>2<\/sub>\u00a0= 0 and y<sub>3<\/sub> = -4<br \/>\nArea of triangle = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> [2 {0- (-4)} + (-1) {(-4) \u2013 (3)} + 2 (3 \u2013 0)]<br \/>\n= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> {8 + 7 + 6}<br \/>\n= 21\/2<br \/>\nSo, the area of the triangle is 21\/2 square units.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: Georgia, Palatino; color: #000000;\"><strong>(ii) Here,<br \/>\n<\/strong>x<sub>1<\/sub>\u00a0= -5, x<sub>2<\/sub>\u00a0= 3, x<sub>3<\/sub>\u00a0= 5, y<sub>1<\/sub>\u00a0= -1, y<sub>2<\/sub>\u00a0= -5 and y<sub>3<\/sub> = 2<br \/>\nArea of the triangle = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/>\u00a0 [-5 { (-5)- (2)} + 3(2-(-1)) + 5{-1 \u2013 (-5)}]<br \/>\n= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> {35 + 9 + 20} = 32<br \/>\nTherefore, the area of the triangle is 32 square units.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: Georgia, Palatino; color: #000000;\"><strong>2. In each of the following, find the value of \u2018k\u2019, for which the points are collinear.<br \/>\n<\/strong>(i) (7, -2), (5, 1), (3, -k)<br \/>\n(ii) (8, 1), (k, -4), (2, -5) <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: Georgia, Palatino; color: #000000;\"><strong>Solution &#8211;<br \/>\n<\/strong><strong>(i)<\/strong> For collinear points, the area of triangle formed by them is always zero.<br \/>\nLet points (7, -2), (5, 1), and (3, k) are vertices of a triangle.<br \/>\nArea of triangle = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> [7 { 1- k} + 5(k-(-2)) + 3{(-2) \u2013 1}] = 0<br \/>\n7 \u2013 7k + 5k + 10 &#8211; 9 = 0<br \/>\n-2k + 8 = 0<br \/>\nk = 4<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"font-family: Georgia, Palatino; color: #000000;\"><strong>(ii)<\/strong> For collinear points, the area of triangle formed by them is zero.<br \/>\nTherefore, for points (8, 1), (k, \u2013 4), and (2, \u2013 5), area = 0<br \/>\n<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> [8 { -4- (-5)} + k{(-5)-(1)} + 2{1 -(-4)}] = 0<br \/>\n8 \u2013 6k + 10 = 0<br \/>\n6k = 18<br \/>\nk = 3<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>3. Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, -1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.<br \/>\n<\/strong><strong>Solution &#8211; <\/strong>Let the vertices of the triangle be A (0, -1), B (2, 1), C (0, 3).<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6457\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex7.3-Que3.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"254\" height=\"216\" \/><br \/>\n<\/span><\/p>\n<p><span style=\"color: #000000; font-family: Georgia, Palatino;\">Let D, E, F be the mid-points of the sides of this triangle. Coordinates of D, E, and F are given by<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">D = ((0 + 2)\/2 , (-1 + 1)\/2) = (1, 0)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">E = ((0 + 0)\/2 , (-1 + 3)\/2) = (0, 1)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">F = ((2 + 0)\/2 , (1 + 3)\/2) = (1, 2)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Area of a triangle = 1\/2 {<i>x<\/i><sub>1<\/sub> (<i>y<\/i><sub>2<\/sub> &#8211; <i>y<\/i><sub>3<\/sub>)\u00a0+ <i>x<\/i><sub>2<\/sub> (<i>y<\/i><sub>3<\/sub> &#8211; <i>y<\/i><sub>1<\/sub>)\u00a0+ <i>x<\/i><sub>3<\/sub> (<i>y<\/i><sub>1<\/sub> &#8211; <i>y<\/i><sub>2<\/sub>)}<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Area of \u0394DEF = 1\/2 {1(2 &#8211; 1) + 1(1 &#8211; 0) + 0(0 &#8211; 2)}<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 1\/2 (1 + 1)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"> = 1 square units<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Area of \u0394ABC = 1\/2 [0(1 &#8211; 3) + 2{3 &#8211; (-1)} + 0(-1 &#8211; 1)]<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 1\/2 {8}<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"> = 4 square units<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Therefore, the required ratio is 1: 4.<\/span><\/p>\n<p><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>4. Find the area of the quadrilateral whose vertices, taken in order, are <\/strong><strong>(-4, -2), (-3, -5), (3, -2) and (2, 3).<br \/>\n<\/strong><strong>Solution &#8211; <\/strong>Let the vertices of the quadrilateral be A ( &#8211; 4, &#8211; 2), B ( &#8211; 3, &#8211; 5), C (3, &#8211; 2), and D (2, 3). Join AC to form two triangles \u0394ABC and \u0394ACD.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6458\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex7.3-Ans4.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"383\" height=\"222\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex7.3-Ans4.png 383w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex7.3-Ans4-300x174.png 300w\" sizes=\"auto, (max-width: 383px) 100vw, 383px\" \/><br \/>\nArea of a triangle = 1\/2 {<i>x<\/i><sub>1<\/sub>\u00a0(<i>y<\/i><sub>2<\/sub>\u00a0&#8211;\u00a0<i>y<\/i><sub>3<\/sub>)\u00a0+\u00a0<i>x<\/i><sub>2<\/sub>\u00a0(<i>y<\/i><sub>3<\/sub>\u00a0&#8211;\u00a0<i>y<\/i><sub>1<\/sub>)\u00a0+\u00a0<i>x<\/i><sub>3<\/sub>\u00a0(<i>y<\/i><sub>1<\/sub>\u00a0&#8211;\u00a0<i>y<\/i><sub>2<\/sub>)}<br \/>\nArea of \u0394ABC = 1\/2 [(-4) {(-5) &#8211; (-2)} + (-3) {(-2) &#8211; (-2)} + 3 {(-2) &#8211; (-5)}]<br \/>\n= \u00a01\/2 (12 + 0 + 9)<br \/>\n= 21\/2 square units<\/span><\/p>\n<p><span style=\"color: #000000; font-family: Georgia, Palatino;\">Area of \u0394ACD = 1\/2 [(-4) {(-2) &#8211; (3)} + 3{(3) &#8211; (-2)} + 2 {(-2) &#8211; (-2)}]<br \/>\n= 1\/2 (20 + 15 + 0)<br \/>\n= 35\/2 square units<\/span><\/p>\n<p><span style=\"color: #000000; font-family: Georgia, Palatino;\">Area of \u2610ABCD \u00a0= Area of \u0394ABC + Area of \u0394ACD<br \/>\n= (21\/2 + 35\/2) square units<br \/>\n= 28 square units<\/span><\/p>\n<p><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>5. You have studied in Class IX that a median of a triangle divides it into two triangles of equal areas. Verify this result for \u0394ABC, whose vertices are A (4, \u2013 6), B (3, \u2013 2) and C (5, 2).<br \/>\n<\/strong><strong>Solution &#8211; <\/strong>Let the vertices of the triangle be A (4, -6), B (3, -2), and C (5, 2).<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6459\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/08\/NCERT-Class-10-Maths-Ex7.3-Ans5.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"296\" height=\"264\" \/><\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Let D be the mid-point of side BC of \u0394ABC. Therefore, AD is the median in \u0394ABC. <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Coordinates of point D = ((3 + 5)\/2, (-2 + 2)\/2) = (4, 0)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Area of a triangle = 1\/2 {<i>x<\/i><sub>1<\/sub>\u00a0(<i>y<\/i><sub>2<\/sub>\u00a0&#8211;\u00a0<i>y<\/i><sub>3<\/sub>)\u00a0+\u00a0<i>x<\/i><sub>2<\/sub>\u00a0(<i>y<\/i><sub>3<\/sub>\u00a0&#8211;\u00a0<i>y<\/i><sub>1<\/sub>)\u00a0+\u00a0<i>x<\/i><sub>3<\/sub>\u00a0(<i>y<\/i><sub>1<\/sub>\u00a0&#8211;\u00a0<i>y<\/i><sub>2<\/sub>)}<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Area of\u00a0\u0394ABD =<\/strong> 1\/2 [(4) {(-2) &#8211; (0)}\u00a0+ 3{(0) &#8211; (-6)}\u00a0+ (4) {(-6) &#8211; (-2)}]<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 1\/2 (-8 + 18 &#8211; 16)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= -3 square units<\/span><\/p>\n<p><span style=\"color: #000000; font-family: Georgia, Palatino;\">However, area cannot be negative. Therefore, area of \u0394ABD is 3 square units. <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Area of<\/strong> \u0394ABD = 1\/2 [(4) {0 &#8211; (2)} + 4{(2) &#8211; (-6)} + (5) {(-6) &#8211; (0)}]<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 1\/2 (-8 + 32 &#8211; 30)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= -3 square units <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">However, area cannot be negative. Therefore, area of \u0394ABD is 3 square units. <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">The area of both sides is same. Thus, median AD has divided \u0394ABC in two triangles of equal areas.<\/span><\/p>\n<p><span style=\"color: #0000ff; font-family: Georgia, Palatino;\"><b><i><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths\/\">Go Back To Chapters<\/a><\/i><\/b><\/span><\/p>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 10 Maths\u00a0 Chapter &#8211; 7 (Coordinate Geometry)\u00a0 The NCERT Solutions in English Language for Class 10 Mathematics Chapter &#8211; 7 Coordinate Geometry Exercise 7.3 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter : 7 Coordinate Geometry NCERT Class 10 Maths Solution Ex &#8211; 7.1 [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1043],"tags":[1045,1086,1089,1044,1049,1048],"class_list":["post-6063","post","type-post","status-publish","format-standard","hentry","category-class-10-maths","tag-class-10-ncert-mathematics-solutions","tag-ncert-class-10-mathematics-chapter-7-coordinate-geometry-solutions","tag-ncert-class-10-mathematics-exercise-7-3-solutions","tag-ncert-class-10-mathematics-solutions-class-10-ncert-solutions","tag-ncert-solutions-class-10-mathematics","tag-ncert-solutions-class-10-maths"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.9 (Yoast SEO v27.3) - 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