{"id":4589,"date":"2020-09-09T06:28:27","date_gmt":"2020-09-09T00:58:27","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=4589"},"modified":"2021-07-02T15:03:36","modified_gmt":"2021-07-02T09:33:36","slug":"kseeb-solutions-class-8-maths-chapter-14-ex-14-1","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-class-8-maths-chapter-14-ex-14-1\/","title":{"rendered":"KSEEB Solutions for Class 8 Maths Chapter 14 Introduction of Graphs Ex 14.1"},"content":{"rendered":"

Students can Download Maths Chapter 14 Introduction of Graphs Ex 14.1 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths<\/a> helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka Board Class 8 Maths Chapter 14 Introduction of Graphs Ex 14.1<\/h2>\n

Question 1.
\nFix up your own coordinate system on a graph paper and locate the following points on the sheet,
\ni. P(-3,5)
\nii. Q(0,-8)
\niii. R(4,0)
\niv. S (-4, -9)
\nAnswer:
\n\"KSEEB<\/p>\n

Question 2.
\nSuppose you are given a coordinate system to determine the quadrant in which the following points lie.
\nAnswer:
\ni. A (4, 5) \u2013 I Quadrant
\nii. B (-4, -5) \u2013 III Quadrant
\niii. C (4, -5) \u2013 IV Quadrant<\/p>\n

Question 3.
\nSuppose p is a point with coordinate (-8, 3) with respect to a coordinate system X1 <\/sup>O X \u2192 Y1 <\/sup>OY. Let X1<\/sup>1<\/sub> O1<\/sub> X1<\/sub> \u2194 Y1<\/sup>OY\u00a0be another system with X1<\/sup>OX\u00a0 ||\u00a0 X1<\/sup>1<\/sub>O1<\/sub>X1<\/sub> and O1<\/sub> has coordinates (9,5) with respect to X1<\/sup>OX \u2194 Y1<\/sup>O Ywhat are the coordinate of p in the system X11<\/sub><\/sup> O1<\/sub>X1<\/sub> Y1<\/sup>1<\/sub>O1<\/sub>Y1<\/sub>?
\nAnswer:
\nLet (x1 <\/sup>,y1<\/sup>) be the coordinates point p in X1<\/sup>1<\/sub>\u00a0O1<\/sub>\u00a0X1<\/sub>\u00a0\u2194 Y1<\/sup>OY.
\nThen x = a + x1\u00a0<\/sup>and y = b + y1<\/sup>
\nbut (a, b) = (9, 5)
\n\u2234 \u2013 8 = 9 +x1<\/sup>\u00a0\u00a0 \u00a0 \u00a0y = b + y1<\/sup>
\n\u2013 8 \u2013 9 = x1<\/sup>\u00a0\u00a0 \u00a0 \u00a0 \u00a0 3 = 5 + y1<\/sup>
\n-17 = x1<\/sup>\u00a0\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 3 \u2013 5 = y1<\/sup>
\n\u2234 (x1<\/sup>\u00a0, y1<\/sup>\u00a0) = (-17,-2)
\n\u2234 Coordinatesof pin X1<\/sup>1<\/sub>\u00a0O1<\/sub>\u00a0X1<\/sub>\u00a0\u2194 Y1<\/sup>OY are (-17,-2)
\n\"KSEEB<\/p>\n

Question 4.
\nSuppose P has coordinates (10,2) in a coordinate system X’OX \u2194 Y’OY and (-3,-6) in an other coordinates system X’1<\/sub>O1 <\/sub>X\u2194Y’1<\/sub>O1<\/sub>X1<\/sub>. Determine the coordinates of O with respect to the system X\u20191<\/sub> O1<\/sub> X1 <\/sub>|| Y\u20191<\/sub> O1<\/sub>Y1<\/sub>\u2019.
\nAnswer:
\nGiven p(x , y) = (10 , 2), (x\u2019, y\u2019) = (-3, -6)
\nLet the coordinates of O, in
\nX\u20191<\/sub>O1<\/sub>\u00a0X1<\/sub>\u00a0\u2194 Y\u20191<\/sub>O1<\/sub>\u00a0Y1<\/sub>\u00a0be (a,b) then
\nx = a + x1<\/sup>\u00a0\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 y = b + y1<\/sup>
\n10 = a + (-3)\u00a0 \u00a0 \u00a02 = b + (-6)
\n10 = a \u2013 3\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 2 = b \u2013 6
\n10 + 3 = a\u00a0 \u00a0 \u00a0 \u00a02 + 6 = b
\na = 13\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 b = 8
\nCoordinates of O1<\/sub>\u00a0in
\nX\u20191<\/sub>\u00a0O\u00a01<\/sub>Y1<\/sub>, \u2194 Y1<\/sub>\u00a0O\u00a01<\/sub>Y\u20191\u00a0<\/sub> is (13 , 8) we should find the coordinates of O
\n(0, 0) in X\u20191<\/sub>O\u00a01<\/sub>X1<\/sub>\u2194 Y\u20191<\/sub>O1<\/sub>Y1<\/sub>
\nLet (x , y) = (0 , 0) , (x\u2019, y\u2019) = (13 , 8) .
\n(a ,b) = ?\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 y = b + y\u2019
\n0 = a + 13\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 0 = b + 8
\n-13 = a\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 -8 = b
\n(a = – 13)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 b= -8
\nThe coordinates of O w.r.t to
\nX\u20191<\/sub>\u00a0O\u00a01<\/sub>Y1<\/sub>, \u2194 Y1<\/sub>\u00a0O\u00a01<\/sub>Y\u20191<\/sub> are (-13 , -8).
\n\"KSEEB<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Maths Chapter 14 Introduction of Graphs Ex 14.1 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka Board Class 8 Maths Chapter 14 Introduction of Graphs Ex 14.1 Question 1. Fix up …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/4589"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=4589"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/4589\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=4589"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=4589"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=4589"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}