NCERT Solutions for Class 10 Maths<\/a> Chapter 7 Coordinate Geometry Ex 7.1 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\nNCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.1<\/h2>\n <\/p>\n
Question 1. \nFind the distance between the following pairs of points: \n(i) (2, 3), (4, 1) \n(ii) (-5, 7), (-1, 3) \n(iii) (a, b), (-a, -b) \nSolution: \n(i) Let P (2, 3) and Q (4, 1) be the two points. Therefore, by distance formula we know that \n \nTherefore distance between points (2, 3) and (4, 11) is 2\\(\\sqrt{2}\\) unit.<\/p>\n
(ii) Let P (- 5, 7) and Q (- 1, 3) be two points. \nTherefor by distance formula we know that \n \nTherefore, distance between points (- 5, 7) and (- 1, 3) is 4\\(\\sqrt{2}\\) unit.<\/p>\n
(iii) Let P (a, b) and Q (-a, – b) be the points. \nTherefore, by distance formula we know that \n \nTherefore, distance between points (a. b) and (- a, – b) is 2\\(\\sqrt{a^{2}+b^{2}}\\) unit.<\/p>\n
Question 2. \nFind the distance between the points (0, 0) and (36, 15). \nSolution: \nThe distance between the points (0,0) and (36,15) by using distance formula, the distance of point P (x1<\/sub>, y1<\/sub>) from (0, 0) are \n \nThe distance between two towns is 39 km.<\/p>\n <\/p>\n
Question 3. \nDetermine if the points (1, 5), (2, 3) and (-2, -11) are collinear. \nSolution: \nLet points be A (1, 5), B (2, 3) and C (-2, -11) \n <\/p>\n
Question 4. \nCheck whether (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle. \nSolution: \nLet points be A(5, -2), B (6, 4) and C (7, -2) \n \nTherefore, points (1,5), (2, 3) and (-2, -11) are not collinear.<\/p>\n
Question 5. \nIn a classroom, 4 friends are seated at the points A, B, C and D as shown in given figure. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, \u201cDon\u2019t you think ABCD is a square?\u201d Chameli disagrees. Using distance formula, find which of them is correct. \nSolution: \nPoints A (3, 4), B (6, 7), C (9, 4) and D (6, 1) \n \nBy distance formula we come to know that Champa is correct. The quadrilateral ABCD is a square \nBecause AB = BC = CD = DA = 3\\(\\sqrt{2}\\)<\/p>\n
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Question 6. \nName the type of quadrilateral formed, if any, by the following points, and give reasons for your answer. \n(i) (-1, -2), (1, 0), (-1, 2), (-3, 0) \n(ii) (-3, 5), (3, 1), (0, 3), (-1, -4) \n(iii) (4, 5), (7, 6), (4, 3), (1, 2) \nSolution: \n(i) Let points be A (-1, -2), B (1, 0), C (-1, 2) and D (-3, 0) be the vertices of quadrilateral. \nTherefore, by distance formula, we know that \n \nAB = BC = CD = AD and diagonals AC = BD \nSo, the quadrilateral formed by points (-1, -2), (1, 0), (-1,2) and (-3, 0) is a square.<\/p>\n
(ii) Let A (-3, 5), B (3, 1), C = (0, 3) and D = (-1, -4) be the vertices of quadrilateral. \nTherefore, by distance formula, we know that \n \nThere is no quadrilateral from the given points.<\/p>\n
(iii) Let A (4, 5), B (7, 6), C (4,3) and D (1,2) be the vertices of quadrilateral ABCD. \nTherefore, by distance formula, we know that \n \nThe points (4,5), (7,6), (4,3) and (1,2) form a parallelogram.<\/p>\n
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Question 7. \nFind the point on the x-axis which is equidistant from (2, -5) and (-2, 9). \nSolution: \nLet P (x, 0) be a point on x-axis which is equidistant from the points. A (2, -5) and B (-2, 9) \nUsing the distance formula, we have \n \nHence, the required point on the x-axis is P (x, 0) = P (-7, 0)<\/p>\n
Question 8. \nFind the values of y for which the distance between the points P (2, -3) and Q (10, y) is 10 units. \nSolution: \nBy distance formula we know that \n \nTherefore, the values of y are – 9 and 3<\/p>\n
Question 9. \nIF Q (0, 1) is equidistant from P (5, – 3) and R (x, 6), find the value of x. Also find the distances QR and PR. \nSolution: \nBy distance formula we know that. \n \nAccording the question, \n \nNow the points Q (0,1), P (5, -3) and R (4, 6) by using of distance formula. \n <\/p>\n
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Question 10. \nFind the relation between between x and y such that the point (x, y) is equidistant from the point (3, 6) and (3, 4). \nSolution: \nLet the points be P (x, y), Q (3, 6) and R (- 3, 4). By using distance formula. \n <\/p>\n","protected":false},"excerpt":{"rendered":"
These NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.1 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.1 Question 1. Find the distance between the following pairs of points: (i) (2, 3), (4, 1) (ii) (-5, 7), …<\/p>\n
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.1<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[2],"tags":[],"yoast_head":"\nNCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.1 - MCQ Questions<\/title>\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n