{"id":600,"date":"2023-10-19T12:36:51","date_gmt":"2023-10-19T07:06:51","guid":{"rendered":"https:\/\/tnboardsolutions.com\/?p=600"},"modified":"2023-11-10T14:43:54","modified_gmt":"2023-11-10T09:13:54","slug":"samacheer-kalvi-9th-maths-guide-chapter-5-additional-questions","status":"publish","type":"post","link":"https:\/\/tnboardsolutions.com\/samacheer-kalvi-9th-maths-guide-chapter-5-additional-questions\/","title":{"rendered":"Samacheer Kalvi 9th Maths Guide Chapter 5 Coordinate Geometry Additional Questions"},"content":{"rendered":"

Students can download Maths Chapter 5 Coordinate Geometry Additional Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide<\/a> Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.<\/p>\n

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 5 Coordinate Geometry Additional Questions<\/h2>\n

I. Multiple Choice Questions.<\/span><\/p>\n

Question 1.
\nOn which quadrant does the point (- 4, 3) lie?
\n(a) I
\n(b) II
\n(c) III
\n(d) IV
\nSolution:
\n(b) II<\/p>\n

Question 2.
\nThe point whose abscissa is 5 and lies on the x-axis is …….
\n(a) (-5, 0)
\n(b) (5, 5)
\n(c) (0, 5)
\n(d) (5, 0)
\nSolution:
\n(d) (5, 0)<\/p>\n

\"Samacheer<\/p>\n

Question 3.
\nA point which lies in the III quadrant is ……..
\n(a) (5, 4)
\n(b) (5, -4)
\n(c) (-5, -4)
\n(d) (-5, 4)
\nSolution:
\n(c) (-5, -4)<\/p>\n

Question 4.
\nA point on the y-axis is ……..
\n(a) (1, 1)
\n(b) (6, 0)
\n(c) (0, 6)
\n(d) (-1, -1)
\nSolution:
\n(c) (0, 6)<\/p>\n

Question 5.
\nThe distance between the points (4, -1) and the origin is ……..
\n(a) \\(\\sqrt{24}\\)
\n(b) \\(\\sqrt{37}\\)
\n(c) \\(\\sqrt{26}\\)
\n(d) \\(\\sqrt{17}\\)
\nSolution:
\n(d) \\(\\sqrt{17}\\)<\/p>\n

Question 6.
\nThe distance between the points (-1, 2) and (3, 2) is ……..
\n(a) \\(\\sqrt{14}\\)
\n(b) \\(\\sqrt{15}\\)
\n(c) 4
\n(d) 0
\nSolution:
\n(c) 4<\/p>\n

\"Samacheer<\/p>\n

Question 7.
\nThe centre of a circle is (0, 0). One end point of a diameter is (5, -1), then the radius is …….
\n(a) \\(\\sqrt{24}\\)
\n(b) \\(\\sqrt{37}\\)
\n(c) \\(\\sqrt{26}\\)
\n(d) \\(\\sqrt{17}\\)
\nSolution:
\n(c) \\(\\sqrt{26}\\)<\/p>\n

Question 8.
\nThe point (0, -3) lies on
\n(a) + ve x-axis
\n(b) + ve y-axis
\n(c) – ve x-axis
\n(d) – ve y-axis
\nSolution:
\n(d) – ve y-axis<\/p>\n

Question 9.
\nThe point which is on y-axis with ordinate -5 is ……..
\n(a) (0, -5)
\n(b) (-5, 0)
\n(c) (5, 0)
\n(d) (0, 5)
\nSolution:
\n(a) (0, -5)<\/p>\n

Question 10.
\nThe diagonal of a square formed by the points (1, 0), (0, 1), (-1, 0) and (0, -1) is …….
\n(a) 2
\n(b) 4
\n(c) \u221a2
\n(d) 8
\nSolution:
\n(a) 2<\/p>\n

\"Samacheer<\/p>\n

Question 11.
\nThe distance between the points (-2, 2) and (3, 2) is ……..
\n(a) 10 units
\n(b) 5 units
\n(c) 5\u221a3 units
\n(d) 20 units
\nSolution:
\n(b) 5 units<\/p>\n

Question 12.
\nThe midpoint of the line joining the points (1, -1) and (-5, 3) is ……..
\n(a) (2, 1)
\n(b) (2, -1)
\n(c) (-2, -1)
\n(d) (-2, 1)
\nSolution:
\n(d) (-2, 1)<\/p>\n

Question 13.
\nIf the centroid of a triangle is at (1, 3) and two of its vertices are (-7, 6) and (8, 5) then the third vertex is ……..
\n(a) (-2, 2)
\n(b) (2, -2)
\n(c) (-2, -2)
\n(d) (2, 2)
\nSolution:
\n(b) (2, -2)<\/p>\n

Question 14.
\nThe ratio in which the X-axis divides the line segment joining the points (6, 4) and (1, -7) is ……..
\n(a) 1 : 2
\n(b) 2 : 3
\n(c) 4 : 7
\n(d) 7 : 4
\nSolution:
\n(c) 4 : 7<\/p>\n

\"Samacheer<\/p>\n

Question 15.
\nThe centroid of a triangle (3, -5), (-7, 4) and (10, -2) is …….
\n(a) (2, -1)
\n(b) (2, 1)
\n(c) (-2, 1)
\n(d) (1, -2)
\nSolution:
\n(a) (2, -1)<\/p>\n

II. Answer the Following Questions.<\/span><\/p>\n

Question 1.
\nShow that the given points (1, 1), (5, 4), (-2, 5) are the vertices of an isosceles right angled triangle.
\nSolution:
\nLet A (1, 1), B (5, 4) and G (-2, 5)
\n\"Samacheer
\nAB = 5, AC = 5
\n\u2234 ABC is an isosceles triangle …….. (1)
\nBC\u00b2 = AB\u00b2 + AC\u00b2
\n50 = 25 + 25 \u21d2 50 = 50
\n\u2234 \u2220A = 90\u00b0 ……… (2)
\nFrom (1) and (2) we get ABC is an isosceles right angle triangle.<\/p>\n

Question 2.
\nShow that the point (3, -2), (3, 2), (-1, 2) and (-1, -2) taken in order are the vertices of a square.
\nSolution:
\n\"Samacheer
\n= \\(\\sqrt{16}\\)
\n= 4
\nAB = BC = CD = DA = 4. All the four sides are equal.
\n\u2234 ABCD is a Rhombus ……..(1)
\n\"Samacheer
\nDiagonal AC = Diagonal BD = \\(\\sqrt{32}\\) ……..(2)
\nFrom (1) and (2) we get ABCD is a square.<\/p>\n

\"Samacheer<\/p>\n

Question 3.
\nShow that the point A (3, 7) B (6, 5) and C (15, -1) are collinear.
\nSolution:
\n\"Samacheer
\nAB + BC = AC \u21d2 \\(\\sqrt{13}\\) + 3\\(\\sqrt{13}\\) = 4\\(\\sqrt{13}\\)
\n\u2234 The points A, B, C are collinear.<\/p>\n

Question 4.
\nFind the type of triangle formed by (-1, -1), (1, 1) and (-\u221a, \u221a3)
\nSolution:
\nLet the point A (-1, -1), B (1, 1) and C (-\u221a3, \u221a3)
\n\"Samacheer
\nAB = BC = AC = \u221a8
\n\u2234 ABC is an equilateral triangle.<\/p>\n

Question 5.
\nFind x such that PQ = QR where P(6, -1) Q(1, 3) and R(x, 8) respectively.
\nSolution:
\n\"Samacheer
\nBut PQ = QR
\n\\(\\sqrt{(x-1)^{2}+25}\\) = \\(\\sqrt{41}\\)
\nSquaring on both sides
\n(x – 1)\u00b2 + 25 = 41
\n(x – 1)\u00b2 = 41 – 25 = 16
\nx – 1 = \\(\\sqrt{16}\\) = \u00b1 4
\nx – 1 = 4 (or) x – 1 = – 4
\nx = 5 (or) x = -4 + 1 = -3
\nThe value of x = 5 or – 3<\/p>\n

\"Samacheer<\/p>\n

Question 6.
\nFind the coordinate of the point of trisection of the line segment joining (4, -1) and
\nSolution:
\nLet A (4, -1) and B (-2, -3) are the given points
\nLet P (a, b) and Q (c, d) be the points of trisection of AB.
\n\u2234 AP = PQ = QB
\n\"Samacheer
\nThe required coordinate P is (2, –\\(\\frac{5}{3}\\)) and Q is (0, –\\(\\frac{7}{3}\\))<\/p>\n

Question 7.
\nFind the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6).
\nSolution:
\nGiven points are A(-3, 10), B(6, -8) and P(-1, 6)
\ndivides AB internally in the ratio m : n
\nBy section formula.
\nA line divides internally in the ratio m : n the point P =
\n\"Samacheer
\n\u2234 \\(\\frac{6m-3n}{m+n}\\) = -1
\n6m – 3n = -m – n
\n6m + m = 3n – n
\n7m = 2n \u21d2 \\(\\frac{m}{n}\\) = \\(\\frac{2}{7}\\)
\n\u2234 m : n = 2 : 7
\nand
\n\\(\\frac{-8m+10n}{m+n}\\) = 6
\n-8m + 10n = 6m + 6n
\n-8m – 6m = 6n – 10n
\n14m = 4n
\n\u2234 \\(\\frac{m}{n}\\) = \\(\\frac{14}{4}\\) = \\(\\frac{2}{7}\\)
\nHence P divides AB internally in the ratio 2 : 7<\/p>\n

\"Samacheer<\/p>\n

Question 8.
\nIf (1, 2) (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find “x” and “y”.
\nSolution:
\nLet A(1, 2), B(4, y), C(x, 6) and D(3, 5)
\n\"Samacheer
\nSince ABCD is a parallelogram the diagonal bisect each other
\nMid point of AC = Mid point of BD
\n(\\(\\frac{1+x}{2}\\), 4) = (\\(\\frac{7}{2}\\), \\(\\frac{y+5}{2}\\))
\n\\(\\frac{1+x}{2}\\) = \\(\\frac{7}{2}\\)
\n1 + x = 7
\nx = 7 – 1
\n= 6
\nand
\n\\(\\frac{y+5}{2}\\) = 4
\ny + 5 = 8
\ny = 8 – 5
\n= 3
\n\u2234 The value of x = 6 and y = 3<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can download Maths Chapter 5 Coordinate Geometry Additional Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams. Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 5 Coordinate Geometry Additional Questions …<\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[3],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/tnboardsolutions.com\/wp-json\/wp\/v2\/posts\/600"}],"collection":[{"href":"https:\/\/tnboardsolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/tnboardsolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/tnboardsolutions.com\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/tnboardsolutions.com\/wp-json\/wp\/v2\/comments?post=600"}],"version-history":[{"count":1,"href":"https:\/\/tnboardsolutions.com\/wp-json\/wp\/v2\/posts\/600\/revisions"}],"predecessor-version":[{"id":43504,"href":"https:\/\/tnboardsolutions.com\/wp-json\/wp\/v2\/posts\/600\/revisions\/43504"}],"wp:attachment":[{"href":"https:\/\/tnboardsolutions.com\/wp-json\/wp\/v2\/media?parent=600"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tnboardsolutions.com\/wp-json\/wp\/v2\/categories?post=600"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tnboardsolutions.com\/wp-json\/wp\/v2\/tags?post=600"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}