{"id":25460,"date":"2020-03-10T11:50:00","date_gmt":"2020-03-10T06:20:00","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=25460"},"modified":"2021-07-02T15:20:25","modified_gmt":"2021-07-02T09:50:25","slug":"kseeb-solutions-for-class-10-maths-chapter-7-additional-questions","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-for-class-10-maths-chapter-7-additional-questions\/","title":{"rendered":"KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Additional Questions"},"content":{"rendered":"

Students can Download Class 10 Maths Chapter 7 Coordinate Geometry Additional Questions, Notes Pdf, KSEEB Solutions for Class 10 Maths<\/a>\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka State Syllabus Class 10 Maths Chapter 7 Coordinate Geometry Additional Questions<\/h2>\n

I. Multiple-choice Questions:<\/span><\/p>\n

Question 1.
\nThe co-ordinates of origin are
\na. 0,0.
\nb. 0, 1.
\nc. 1, 0.
\nd. 1, 1.
\nAnswer:
\na. 0,0.<\/p>\n

Question 2.
\nThe distance of the point (3, 4) x – co-ordinate is equal to
\na. 3
\nb. 4
\nc. 1
\nd. 7
\nAnswer:
\nb. 4<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\nThe distance of the point (3, 4) from y-axis is
\na. 3
\nb. 4
\nc. 5
\nd. 1
\nAnswer:
\na. 3<\/p>\n

Question 4.
\nThe angle between x-axis and y-axis is
\na. 0\u00b0
\nb. 45\u00b0
\nc. 60\u00b0
\nd. 90\u00b0
\nAnswer:
\nd. 90\u00b0<\/p>\n

Question 5.
\nThe distance of the point (\u03b1, \u03b2) from origin is
\na. \u03b1.\u03b2
\nb. \u03b12<\/sup>. \u03b22<\/sup>
\nc. \\(\\sqrt{\\alpha^{2}-\\beta^{2}}\\)
\nd. \\(\\sqrt{\\alpha^{2}+\\beta^{2}}\\)
\nAnswer:
\nd. \\(\\sqrt{\\alpha^{2}+\\beta^{2}}\\)<\/p>\n

Question 6.
\nThe distance between the point (x1<\/sub>,y1<\/sub>) and (x2<\/sub>, y2<\/sub>) is
\n\"KSEEB
\nAnswer:
\n\"KSEEB<\/p>\n

Question 7.
\nWhich of the following is distance formula?
\nAnswer:
\na. The distance between 2 points (x1<\/sub>,y1<\/sub>) and (x2<\/sub>, y2<\/sub>) is
\n\"KSEEB<\/p>\n

b. The mid point of the lines segment join the points (x, y) and (x2<\/sub>, y2<\/sub>)
\n\"KSEEB<\/p>\n

c. The co-ordinates of the centroid of the \u2206le<\/sup> ABC where A, B, C are (x1<\/sub>,y1<\/sub>)
\n\"KSEEB<\/p>\n

d.
\n\"KSEEB<\/p>\n

\"KSEEB<\/p>\n

Question 8.
\nThe distance b\/w the point (0, 3) and (- 2, 0) is
\na. \\(\\sqrt{14}\\)
\nb. \\(\\sqrt{15}\\)
\nc. \\(\\sqrt{13}\\)
\nd. \\(\\sqrt{5}\\)
\nAnswer:
\nc. \\(\\sqrt{13}\\)<\/p>\n

Question 9.
\nThe \u2206le<\/sup> with vertices (- 2, 1) (2, – 2) and (5, 2) is
\na. Scalene
\nb. Equilateral
\nc. Isosceles
\nd. right angled isosceles
\nAnswer:
\nd. right angled isosceles<\/p>\n

Question 10.
\nThe Co-ordinates of the point which divides the join of (x, y) and (x2<\/sub>, y2<\/sub>) in the ratio m1<\/sub> : m2<\/sub> internally are
\n\"KSEEB
\nAnswer:
\n\"KSEEB<\/p>\n

Question 11.
\nIf the distance b\/w the point (3, a) and (4,1) is \\(\\sqrt{10}\\), then, find the values of a
\na. 3,- 1
\nb. 4, – 2
\nc. 2, – 2
\nd. 5, – 3
\nAnswer:
\nc. 2, – 2<\/p>\n

Question 12.
\nThe distance of the mid – point of the line -segment joining the points 6, 8. and 2, 4. from the point (1, 2) is
\na. 3
\nb. 4
\nc. 5
\nd. 6
\nAnswer:
\nc. 5<\/p>\n

Question 13.
\nThe coordinates of the mid – point of the line – segment joining (- 8, 13) and (x, – 7)
\na. 16
\nb. 10
\nc. 4
\nd. 8
\nAnswer:
\na. 16<\/p>\n

\"KSEEB<\/p>\n

Question 14.
\nFind the value of k if the distance b\/w (k, 3) and (2, 3) is 5
\na. 5
\nb. 6
\nc. 7
\nd. 8
\nAnswer:
\nc. 7<\/p>\n

Question 15.
\nThe distance b\/w the point (Sin \u03b8, Sin \u03b8) & (cos \u03b8, – cos \u03b8) is
\na. \\(\\sqrt{2}\\)
\nb. 2
\nc. \\(\\sqrt{3}\\)
\nd. 1
\nAnswer:
\na. \\(\\sqrt{2}\\)<\/p>\n

II. Short Answers Questions:<\/span><\/p>\n

Question 1.
\nWrite the coordinate of the origin.
\nAnswer:
\n(0, 0)<\/p>\n

Question 2.
\nWhat is the ordinate of all the point on x-axis
\nAnswer:
\n0(zero)<\/p>\n

Question 3.
\nFind the perpendicular distance of the point (- 3, 5) form the x – axis.
\nAnswer:
\n5<\/p>\n

Question 4.
\nFind the distance of B(3 + \\(\\sqrt{3}\\), 3 – \\(\\sqrt{3}\\)) form origin.
\nAnswer:
\n\"KSEEB
\nd = \\(\\sqrt{24}\\)
\nd = \\(\\sqrt{4 \\times 6}\\)
\nd = 2\\(\\sqrt{6}\\)<\/p>\n

Question 5.
\nWrite the formula to find distance b\/w two points
\nAnswer:
\n\"KSEEB<\/p>\n

\"KSEEB<\/p>\n

Question 6.
\nWrite the formula to find distance b\/W point and origin.
\nAnswer:
\nd = \\(\\sqrt{x^{2}+y^{2}}\\)<\/p>\n

Question 7.
\nWrite the formula to find area of triangle.
\nAnswer:
\narea of triangle
\n\"KSEEB<\/p>\n

Question 8.
\nFind the value of a, so that the point (3, a) lie on the line 2x – 3y = 5.
\nAnswer:
\nSince (3, a) lies on the line 2x – 3y = 5
\nThen 2(3) – 3(a) = 5
\n6 – 3a = 5
\n– 3a = 5 – 6
\n– 3a = – 1
\na = 1\/3<\/p>\n

Question 9.
\nWrite the section formula when it divide in ration p : q
\nAnswer:
\nCo-ordinates
\n\"KSEEB<\/p>\n

III. Long Answer Questions:<\/span><\/p>\n

Question 1.
\nFind the distance between the following pairs of points.
\ni) (6, 5) and (4, 4)
\nAnswer:
\n\"KSEEB
\n\"KSEEB<\/p>\n

\"KSEEB<\/p>\n

Question 2.
\nThe distance between the point (3, 1) and (0, x) in 5 unit. Find x.
\nAnswer:
\n(x1<\/sub>, y1<\/sub>) = (3, 1); (x2<\/sub>, y2<\/sub>) = (o, x)
\nd = 5
\n\"KSEEB
\n(\u2235 squaring on both the sides)
\n25 = (3)2<\/sub> + (x – 1)2<\/sub>
\nx2<\/sub> + 1 – 2x + 9 = 25
\nx2<\/sub> + 1 – 2x + 9 – 25 = 0
\nx2<\/sub> + 1 – 2x + 9 – 25 = 0
\nx2<\/sub> – 2x – 15 = 0
\nx2<\/sub> – 5x + 3x – 15 = 0
\nx(x – 5) + 3(x – 5) = 0
\n(x – 5)(x + 3) = 0
\nx – 5 = 0, x + 3 = 0
\nx = 5, x = – 3<\/p>\n

Question 3.
\nFind a point on y-axis which is equidistant from the points (5, 2) and (- 4, 3)
\nAnswer:
\nA Point lie on y – axis = (0, y) the points are equidistant from the point on y – axis.
\n\"KSEEB
\n6y – 4y = 25 – 29
\n2y = – 4
\ny = \\(\\frac{-4}{2}\\) \u21d2 y = – 2
\n\u2234 point on y – axis is y = – 2<\/p>\n

Question 4.
\nFind the radius of a circle whose centre is (- 5, 4) and which passes through the point (- 7,1)
\n\"KSEEB<\/p>\n

Question 5.
\nProve that each of the set of co\u00acordinates are the vertices of parallelograms.
\ni. (- 5, – 3), (1, – 11), (7, – 6), (1, 2).
\nAnswer:
\n\"KSEEB
\n\"KSEEB
\n\u2234 AD = BC & AB = CD
\n\u2234 Given vertices from a parallelogram.<\/p>\n

\"KSEEB<\/p>\n

Question 6.
\nIn what ratio does the point (- 2, 3) divide the line segment joining the point (- 3, 5) and (4, – a)?
\nAnswer:
\n(x1<\/sub>, y1<\/sub>) = (3, 5)
\n(x2<\/sub>, y2<\/sub>) = (4, – a)
\np(x, y) = (- 2, 3)
\n\"KSEEB
\n\"KSEEB
\n4m – 3n = – 2(m + n)
\n4m – 3n = – 2m – 2n
\n4m + 2m = – 2n + 3n
\n6m = n
\n\\(\\frac{m}{n}=\\frac{1}{6}\\)
\nm : n = 1 : 6 (or)
\n\\(\\frac{-9 m+5 n}{m+n}\\) = 3
\n– 9m + 5n = 3(m + n)
\n– 9m + 5n = 3m + 3n
\n– 9m – 3m = 3n – 5n
\n– 12m = – 2n
\n\\(\\frac{m}{n}=\\frac{-2}{-12}=\\frac{1}{6}\\)
\nm : n = 1 : 6<\/p>\n

Question 7.
\nIn the point C(1, 1) divides the line segment joining A(- 2, 7) and B in the ration 3 : 2 find the co-ordinates of B.
\nAnswer:
\nPoint C(1, 1)
\n(x1<\/sub>, y1<\/sub>) = A(- 2, 7) m : n = 3 : 2
\nB(x2<\/sub>, y2<\/sub>) =?
\n\"\"
\n\"KSEEB
\n\u2234 Co-ordinate of B = (3, – 3)<\/p>\n

Question 8.
\nFind the ratio in which the point (- 1, k) divides the line joining the points (- 3, 10) and (6, – 8) and also find ‘k’.
\nAnswer:
\nP(x, y) = (- 1, k); (x1<\/sub>, y1<\/sub>) = (- 3,10)
\n(x2<\/sub>, y2<\/sub>) = (6,- 8)
\n\"KSEEB
\n6m – 3n = – (m + n)
\n6m – 3n = – m – n
\n6m + m = 3n – n, \\(\\frac{-8 m+10 n}{m+n}\\) = k
\n\"KSEEB<\/p>\n

\"KSEEB<\/p>\n

Question 9.
\nThree Consecutive vertices of a parallelogram are A(1, 2), B(2, 3) and (8, 5). Find the fourth vertex.
\nAnswer:
\n\"KSEEB
\n‘O’ is the mid point of AC and by using mid point formulas AC is a line and ‘O’ is the point
\nA(x1<\/sub>, y1<\/sub>) =(1, 2);
\nc(x2<\/sub>, y2<\/sub>) = (8, 5)
\nCo-ordinates of
\n\"KSEEB
\nCo-ordinates of ‘O’ = \\(\\frac{9}{2}, \\frac{7}{2}\\)
\nBD is a line and ‘O’ is midpoint
\nB(2, 3) = (x1<\/sub>, y1<\/sub>); D(x, y) = (x2<\/sub>, y2<\/sub>)
\n\\(\\left(\\frac{9}{2} \\frac{7}{2}\\right)\\) = (x, y)
\nBy midpoint formula
\nCo-ordinates of
\n\"KSEEB
\n2 + x = 9 3 + y = 7
\nx = 9 – 2 y = 7 – 3
\nx = 7 y = 4
\nthe fourth vertex is (7, 4)<\/p>\n

Question 10.
\nFind the relation between x and y, if the points (x, y) (1, 2) and (7, 0) are colliner.
\nAnswer:
\nGiven point are A(x, y), B(1, 2) and C(7, 0)
\nThese points will be collinear if the area of the triangle formed by them is zero.
\nNow, ar(ABC)
\n\"KSEEB
\n\u21d2 2x + 6y – 14 = 0
\n\u21d2 x + 3y = 7
\nwhich is the required relation between x & y<\/p>\n

Question 11.
\nFind the ratio in which the line segment joining the points A(3, – 3) and B(- 2,7) is divides by x-axis. Also find the Co-ordinates of the point of division.
\nAnswer:
\n\"KSEEB
\nHere, point Q is an x-axis its ordinate & 0
\nLet ratio be k : 1 and Co-ordinate of point Q be (x, 0)
\n\"KSEEB<\/p>\n

\"KSEEB<\/p>\n

Question 12.
\nIf A(4, 2), B = (7, 6) and C(1, 4) are the vertices of \u2206 ABC and AD is its median, P.T that median AD divides \u2206 ABC into two triangle of equal areas.D is the median on BC
\n\u21d2 The Co-ordinates of midpoint D are given by
\n\"KSEEB
\nCo-ordinates of D are (4, 5)
\nNow, Area of triangle
\nABD = \\(\\frac{1}{2}\\)[x1<\/sub> (y2<\/sub> – y3<\/sub>) + x2<\/sub> (y3<\/sub> – y1<\/sub>) + x3<\/sub> (y1<\/sub> – y2<\/sub>)
\n= \\(\\frac{1}{2}\\)[4(6 – 5) + 7(52) + 4(2 – 6)
\n= \\(\\frac{1}{2}\\)[4 + 21 – 16] = \\(\\frac{1}{2}\\) sq.units
\nArea of
\n\u2206 ACD = \\(\\frac{1}{2}\\)[4[4 – 5] + 1[5 – 2] + 4[2 – 4]]
\n= \\(\\frac{1}{2}\\)[- 4 + 3 – 8] = \\(\\frac{-9}{2}=\\frac{9}{2}\\) sq.units<\/p>\n

Question 13.
\nIn what ratio does the y – axis divide the line segment joining the point p(- 4, 5) and Q(3, 7). Also, find the co-ordinates of the intersection.
\nAnswer:
\n\"KSEEB
\nSuppose y – axis divides PQ in the ratio k : 1. Then the co-ordinates of the point of division are
\n\"KSEEB
\nSince, R lies on y-axis and X Co-ordinate of every point on y-axis is zero.
\n\u2234 \\(\\frac{3 \\mathrm{K}-4}{\\mathrm{K}+1}\\) = 0
\n\u21d2 3k – 4 = 0 \u21d2 K = \\(\\frac{4}{3}\\)
\nHence, the required ratio is 4\/3 : 1 i.e, 4 : 3
\nPutting K = 4\/3 in the Co-ordinates of R, we find that its Co-ordinates are (0, – 13\/7)<\/p>\n

Question 14.
\nShow that \u2206 ABC with vertices A(- 2, 0), B(2, 0) and C(0, 2) is similar to \u2206 DEF with vertices D(- 4, 0), E(4, 0) and F(0, 4)
\nAnswer:
\nGiven vertices of \u2206 ABC and \u2206 DEF are
\n\"KSEEB
\nA(- 2, 0), B(2, 0), C(0, 2), D(- 4, 0), E(4, 0) and F(0, 4)
\n\"KSEEB
\nHere, we see that sides of \u2206 DEF are twice the sides of a \u2206 ABC.
\nHence, both triangle are similar.<\/p>\n

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

Students can Download Class 10 Maths Chapter 7 Coordinate Geometry Additional Questions, Notes Pdf, KSEEB Solutions for Class 10 Maths\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka State Syllabus Class 10 Maths Chapter 7 Coordinate Geometry Additional Questions I. Multiple-choice Questions: Question 1. The co-ordinates …<\/p>\n","protected":false},"author":5,"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\/25460"}],"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=25460"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/25460\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=25460"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=25460"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=25460"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}