{"id":1086,"date":"2020-09-15T10:21:18","date_gmt":"2020-09-15T04:51:18","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=1086"},"modified":"2021-07-02T15:01:50","modified_gmt":"2021-07-02T09:31:50","slug":"kseeb-sslc-class-10-maths-solutions-chapter-7-ex-7-2","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-sslc-class-10-maths-solutions-chapter-7-ex-7-2\/","title":{"rendered":"KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.2"},"content":{"rendered":"

KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.2<\/strong> are part of KSEEB SSLC Class 10 Maths Solutions<\/a>. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.2.<\/p>\n

Karnataka SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.2<\/h2>\n

Question 1.
\nFind the coordinates of the point which divides the join of (-1, 7) and (4, -3) in the ratio 2 : 3.
\nSolution:
\nLet P(x, y) is the required point.
\nP(x, y) divides the points A(-1, 7),
\nB (4, -3) in the ratio m1<\/sub> : m2<\/sub> = 2 : 3.
\nAs per Section Formula, coordinates of P(x, y) are
\n\"KSEEB<\/p>\n

Question 2.
\nFind the coordinates of the points of trisection of the line segment joining (4, -1) and (-2, -3)
\n\"KSEEB
\nSolution:
\nLet A(x1<\/sub>, y1<\/sub>) = A(4,-1)
\nB(x2<\/sub>, y2<\/sub>) = B(-2, -3).
\nP and Q are the points of trisection of AB.
\nAP = PQ = QB
\ni) P divides AB in the ratio m1<\/sub> : m2<\/sub> :: 1 : 2.
\n\u2234 As per Section Formula, Coordinates of P are :
\n\"KSEEB<\/p>\n

ii) Q divides AB in the ratio 2 : 1.
\n\u2234 As per Section formula,
\n\"KSEEB<\/p>\n

Question 3.
\nTo conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1m from each other along AD, as shown in the following figure.
\n\"KSEEB
\nNiharika runs \\(\\frac{1}{4}\\) th the distance of AD on then 2nd line and posts a green flag
\nPreet runs \\(\\frac{1}{5}\\) th the distance AD on the 5 eighthline and posts a red flag. What is the distance between both the flags ? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?
\nSolution:
\nOrigin, A(0, 0)
\nAB along x – axis
\nAD along y – axis
\nIn the parallel line AD in 2nd<\/sup> row, Rashmi sees green flag at the distance of \\(\\frac{1}{4}\\) .
\n\u2234 AD = \\(\\frac{1}{4}\\) \u00d7 100 = 25 m.
\n\u2234 Coordinates of green flag are G(2, 25)
\nSimilarly we can find out coordinates of Red flag as R(8, 20).
\n\u2234 Distance between G and R is
\n\"KSEEB
\nIt means blue flag is in 5th line and parallel to AD at the distance of 22.5 m.<\/p>\n

Question 4.
\nFind the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6).
\nSolution:
\nWe have to check whether the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6).
\nLet P(- 1, 6) divides AB in the ratio m1<\/sub> : m2<\/sub>.
\nBy using Section formula,
\n\"KSEEB
\n\u2234 P(-1, 6) point divides the line segment in the ratio 2 : 7.<\/p>\n

Question 5.
\nFind the ratio in whcih the line segment joining A(1, -5) and B(-4, 5) is divided by the x-axis. Also find the coordinates of the point of division.
\n\"KSEEB
\nSolution:
\nLet the line segment joining A and B points divides x-axis in the ratio k : 1.
\nIf coordinates of y, P(x, 0) compared,
\n\\(\\frac{k \\times(5)+1 \\times(-5)}{k+1}=0\\)
\n5k – 5 = 0
\n\u2234 k = 1
\n\u2234 Ratio will be k : 1 = 1 : 1
\n\u2234 Point P(x, 0) is the mid-point of line segment AB.
\n\u2234 x = \\(\\frac{1-4}{2}=\\frac{-3}{2}\\)
\n\u2234 Dividing point, P(\\(-\\frac{3}{2}\\), 1)<\/p>\n

Question 6.
\nIf (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
\n\"KSEEB
\nSolution:
\nAC and BD are the diagonals of a parallelogram ABCD and diagonals bisect at O’.
\n\u2234 Coordinate of the midpoint of AC = Coordinate of midpoint BD.
\nAs per Mid-point formula,
\n\"KSEEB
\n\u2234 y = 3
\n\u2234 x = 6, y = 3.<\/p>\n

Question 7.
\nFind the coordinates of point A, where AB is the diameter of a circle whose centre is (2, -3) and B is (1, 4).
\n\"KSEEB
\nSolution:
\nAB is the diameter.
\n‘O’ is centre of the circle, bisects AB
\nLet coordinates of A are (x, y)
\nAs per the Mid-point formula,
\n\"KSEEB
\n\u2234 The coordinates of point A are (3, -10)<\/p>\n

Question 8.
\nIf A and B are (-2, -2) and (2, -4), respectively, find the coordinates of P such that AP = \\(=\\frac{3}{7}\\) AB and P lies on the line segment AB.
\n\"KSEEB
\nSolution:
\nLet coordinates of P are (x, y).
\nAP = \\(\\frac{3}{7}\\) AB AP : AB = 3 : 7
\nAs per Section formula,
\nAP : AB = 3 : 7.
\n\u2234 AP + PB = AB
\n3 + PB = 7
\n\u2234 PB = = 7 – 3 = 4
\n\u2234 Let AP : PB = 3 : 4 = m1 <\/sub>: m2<\/sub>.
\nCoordinates of P are
\n\"KSEEB
\n\"KSEEB<\/p>\n

Question 9.
\nFind the coordinates of the points which divide the line segment joining A(-2, 2) and B(2, 8) into four equal parts.
\n\"KSEEB
\nSolution:
\nLet A (-2, 2) = (x1<\/sub>, y1<\/sub>).
\nB (2, 8) = (x4<\/sub>, y4<\/sub>).
\nP, Q, R points divide equally the side AB.
\nHere, we have AP = PQ = QR = RB.
\n\u2234 AP : PB = 1 : 3
\nAQ : QB = 2 : 2 = 1 : 1
\n\u2234 AR : RB = 3 : 1.
\nAs per Section formula
\ni) P divides AB in the ratio 1 : 3
\n\"KSEEB<\/p>\n

ii) Q divides AB in the ratio 2 : 2 means it is mid-point of AB.
\n\u2234 As per the Mid-point formula,
\n\"KSEEB
\nQ(x2<\/sub>, y2<\/sub>) = (0, 5)<\/p>\n

iii) R divides AB in the ratio 3 : 1
\nm1<\/sub> = 3, m2<\/sub> = 1
\n\"KSEEB<\/p>\n

Question 10.
\nFind the area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2, -1) taken in order. [Hint: Area of a rhombus = \\(\\frac{1}{2}\\) (product of its diagonals]
\n\"KSEEB
\nSolution:
\nABCD is a rhombus.
\nAC and BD are the diagonals and intersects at ‘O’ and bisects perpendicularly.
\n\u2234 Let Diagonal BD = d1<\/sub> Diagonal AC = d2<\/sub>,
\n\"KSEEB<\/p>\n

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.2 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.2, drop a comment below and we will get back to you at the earliest<\/p>\n","protected":false},"excerpt":{"rendered":"

KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.2 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.2. Karnataka SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.2 Question 1. Find the coordinates of …<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/1086"}],"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\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=1086"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/1086\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=1086"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=1086"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=1086"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}