{"id":4599,"date":"2020-09-09T07:12:00","date_gmt":"2020-09-09T01:42:00","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=4599"},"modified":"2021-07-02T15:03:36","modified_gmt":"2021-07-02T09:33:36","slug":"kseeb-solutions-class-8-maths-chapter-14-ex-14-2","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-class-8-maths-chapter-14-ex-14-2\/","title":{"rendered":"KSEEB Solutions for Class 8 Maths Chapter 14 Introduction of Graphs Ex 14.2"},"content":{"rendered":"

Students can Download Maths Chapter 14 Introduction of Graphs Ex 14.2 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 14.2<\/h2>\n

Question 1.
\nDraw the graphs of the following straight lines.
\n(i) y = 3 – x
\n(ii) y = x – 3
\n(iii) y = 3x – 2
\n(iv) y = 5 – 3x
\n(v) 4y = – x + 3
\n(vi) 3y = 4x + 1
\n(vii) x = 4
\n(vii) 3y = 1
\nAnswer:
\n(i) y = 3 – x<\/p>\n\n\n\n\n
x<\/td>\n0<\/td>\n1<\/td>\n2<\/td>\n<\/tr>\n
y<\/td>\n3<\/td>\n2<\/td>\n1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

When x = 0,\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 y = 3 – 0 = (3)
\nWhen x = 1,\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0y = 3 – 1 = (2)
\nWhenx = 2,\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0y = 3 – 2 = (1)
\n\"KSEEB<\/p>\n

(ii) y = x – 3<\/p>\n\n\n\n\n
X<\/td>\n0<\/td>\n1<\/td>\n2<\/td>\n<\/tr>\n
y<\/td>\n-3<\/td>\n-2<\/td>\n-1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

When x = 0,\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 y = 0 – 3 = (- 3)
\nWhen x = 1,\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 y = 1 – 3 = (- 2)
\nWhenx = 2,\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 y = 2 – 3 = (- 1)
\n\"KSEEB<\/p>\n

(iii) y = 3x – 2<\/p>\n\n\n\n\n
X<\/td>\n0<\/td>\n1<\/td>\n2<\/td>\n<\/tr>\n
y<\/td>\n-2<\/td>\n1<\/td>\n4<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

When x = 0,\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 y = 3(0) – 2 = (- 2)
\nWhen x = 1,\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 y = 3(1) – 2 = (1)
\nWhenx = 2,\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 y =3(2) – 2 = (4)
\n\"KSEEB<\/p>\n

(iv) y = 5 – 3x<\/p>\n\n\n\n\n
X<\/td>\n0<\/td>\n1<\/td>\n2<\/td>\n<\/tr>\n
y<\/td>\n5<\/td>\n2<\/td>\n-1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

When x = 0,\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 y = 5 – 3(0) = 5
\nWhen x = 1,\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 y = 5 – 3(1) = 2
\nWhenx = 2,\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 y = 5 – 3(2) = 5 – 6 = -1
\n\"KSEEB<\/p>\n

(v) 4y = – x + 3
\n\"KSEEB
\n\"KSEEB<\/p>\n

(vi) 3y = 4x + 1<\/p>\n\n\n\n\n\n
X<\/td>\n-1<\/td>\n2<\/td>\n5<\/td>\n8<\/td>\n<\/tr>\n
y<\/td>\n-1<\/td>\n3<\/td>\n7<\/td>\n11<\/td>\n<\/tr>\n
(x,y)<\/td>\n(-1,0<\/td>\n(2,3)<\/td>\n(5,7)<\/td>\n(7,11)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\"KSEEB<\/p>\n

(vii) x = 4<\/p>\n\n\n\n\n\n
X<\/td>\n4<\/td>\n4<\/td>\n4<\/td>\n4<\/td>\n<\/tr>\n
y<\/td>\n0<\/td>\n1<\/td>\n2<\/td>\n3<\/td>\n<\/tr>\n
(x,y)<\/td>\n(4,0)<\/td>\n(4,0<\/td>\n(4,2)<\/td>\n(4,3)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\"KSEEB<\/p>\n

(vii) 3y = 1
\n\"KSEEB<\/p>\n

Question 2.
\nDraw the graph of \\(\\frac{y}{x}=\\frac{y+1}{x+2}\\)
\nAnswer:
\nWhen x = 1
\n\"KSEEB<\/p>\n

Question 3.
\nDetermine the equation of the line in each of the following graphs
\n\"KSEEB
\n\"KSEEB
\nAnswer:
\n(i) Let the equation be y = ax + b
\nWe find that the lines passes through (2,0)
\ni. e., when x = 2, y = 0
\n\u2234 y = ax+b
\n0 = ax + b
\n0 = a(2) + b
\n0 = 2a + b
\n\u2234 b = – 2a
\nThe line passes through (4,3) also
\ni. e., when x=4 , y = 3
\n\u2234 y = ax + b
\n3 = a(4) + b
\n3 = 4a + b
\n3 = 4a + (-2a)
\n3 = 4a = 2a
\na = \\(\\frac { 3 }{ 2 }\\) b = – 2a = \\(-2 \\cdot \\frac{3}{2}\\) = -3
\ny = ax + b y
\n\"KSEEB
\nThe equation is 3x – 2y = 6<\/p>\n

(ii) Let the equation be y = ax + b
\nwe find that the line passes through (5,0) i.e., when x = 5, y = 0
\n\u2234 y = ax + b
\n0 = a(5) + b
\n0 = 5a + b
\nb = -5a
\nThe line passes through (0 , 5)also
\ni.e.,when x = 0, y = 5
\ny = ax + b
\n5 = a(0) + b
\n5 = 0 + b
\n5 = 0 + (-5a)
\n5 = -5a
\n\u2234 a =\u00a0\"\\frac
\na = -1
\nb = -5a = -5(-1) = 5 b = 5
\nb = 5
\ny=ax + b
\ny = \u2013 1 x + 5 y = \u2013 x + 5 y + x = 5
\nThe equation of the line of is y + x = 6<\/p>\n

Question 4.
\nA boat is moving in a river, downstream, whose stream has speed 8km\/ hr. the speed of the motor of the boat is 22km\/hr. Draw the graph of the distance covered by the boat versus an hour.
\nAnswer:
\nSpeed of the boat downstream = 22 + 8 = 30 km\/hr
\ny = 30x<\/p>\n\n\n\n\n\n
Time<\/td>\ni<\/td>\n2<\/td>\n3<\/td>\n4<\/td>\n5<\/td>\n<\/tr>\n
Distance<\/td>\n30<\/td>\n60<\/td>\n90<\/td>\n120<\/td>\n150<\/td>\n<\/tr>\n
(1,30)<\/td>\n(2,60)<\/td>\n(3,90)<\/td>\n(4.120)<\/td>\n(5,150)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\"KSEEB<\/p>\n

Question 5.
\nFind the point of intersection of the straight lines 3y + 4x = 7 and 4y + 3x = 7, by drawing their graphs and looking for the point where they meet.
\nAnswer:
\n3y + 4x = 7
\n3y = 7 -4x
\n\\(y-\\frac{7-4 x}{3}\\)<\/p>\n\n\n\n\n\n
X<\/td>\nl<\/td>\n4<\/td>\n<\/tr>\n
y<\/td>\n1<\/td>\n-3<\/td>\n<\/tr>\n
(x,y)<\/td>\n(1 \u00d7 1)<\/td>\n(4-3)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

When x=1, \\(y-\\frac{7-4 x}{3}\\) = \\(\\frac{7-4}{3}=\\frac{3}{3}=1\\)
\nWhen x = 4,
\n\"KSEEB<\/p>\n\n\n\n\n\n
X<\/td>\n-1<\/td>\n5<\/td>\n<\/tr>\n
y<\/td>\n2.5<\/td>\n-2 .<\/td>\n<\/tr>\n
(x,y)<\/td>\n(-1,2.5)<\/td>\n(5-2)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\"KSEEB<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Maths Chapter 14 Introduction of Graphs Ex 14.2 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 14.2 Question 1. Draw the …<\/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\/4599"}],"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=4599"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/4599\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=4599"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=4599"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=4599"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}