{"id":3738,"date":"2021-02-24T18:45:51","date_gmt":"2021-02-24T13:15:51","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=3738"},"modified":"2021-09-15T11:40:41","modified_gmt":"2021-09-15T06:10:41","slug":"ap-ssc-10th-class-maths-solutions-chapter-7-ex-7-1","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/ap-ssc-10th-class-maths-solutions-chapter-7-ex-7-1\/","title":{"rendered":"AP SSC 10th Class Maths Solutions Chapter 7 Coordinate Geometry Ex 7.1"},"content":{"rendered":"

AP State Board Syllabus\u00a0AP SSC 10th Class Maths Textbook Solutions<\/a> Chapter 7 Coordinate Geometry Ex 7.1 Textbook Questions and Answers.<\/p>\n

AP State Syllabus SSC 10th Class Maths Solutions 7th Lesson Coordinate Geometry Exercise 7.1<\/h2>\n

10th Class Maths 7th Lesson Coordinate Geometry Ex 7.1 Textbook Questions and Answers<\/h3>\n

Question 1.
\nFind the distance between the following pair of points,
\n(i) (2, 3) and (4, 1)
\nAnswer:
\nDistance = \\(\\sqrt{\\left(x_{2}-x_{1}\\right)^{2}+\\left(y_{2}-y_{1}\\right)^{2}}\\)
\n= \\(\\sqrt{(4-2)^{2}+(1-3)^{2}}\\)
\n= \\(\\sqrt{4+4}\\)
\n= \u221a8 = 2\u221a2 units<\/p>\n

ii) (- 5, 7) and (-1, 3)
\nAnswer:
\nDistance = \\(\\sqrt{\\left(x_{2}-x_{1}\\right)^{2}+\\left(y_{2}-y_{1}\\right)^{2}}\\)
\n= \\(\\sqrt{(-1+5)^{2}+(3-7)^{2}}\\)
\n= \\(\\sqrt{4^{2}+(-4)^{2}}\\)
\n= \\(\\sqrt{16+16}\\)
\n= \u221a32 = 4\u221a2 units<\/p>\n

iii) (- 2, -3) and (3, 2)
\nAnswer:
\n\"AP<\/p>\n

iv) (a, b) and (- a, – b)
\nAnswer:
\n\"AP<\/p>\n

\"AP<\/p>\n

Question 2.
\nFind the distance between the points (0, 0) and (36, 15).
\nAnswer:
\nGiven: Origin O (0, 0) and a point P (36, 15).
\nDistance between any point and origin = \\(\\sqrt{x^{2}+y^{2}}\\)
\n\u2234 Distance = \\(\\sqrt{36^{2}+15^{2}}\\)
\n= \\(\\sqrt{1296+225}\\)
\n= \\(\\sqrt{1521}\\)
\n= 39 units
\n\"AP
\n\u2234 1521 = 32<\/sup> \u00d7 132<\/sup>
\n\\(\\sqrt{1521}\\) = 3 \u00d7 13 = 39<\/p>\n

Question 3.
\nVerify that the points (1, 5), (2, 3) and (-2, -1) are collinear or not.
\nAnswer:
\nGiven: A (1, 5), B (2, 3) and C (- 2, – 1)
\n\"AP
\nHere the sum of no two segments is equal to third segment.
\nHence the points are not collinear.
\n!! Slope of AB, m1<\/sub> = \\(\\frac{3-5}{2-1}\\) = -2
\nSlope of BC, m2<\/sub> = \\(\\frac{-1-3}{-2-2}\\) = 1
\nm1<\/sub>\u00a0\u2260 m2<\/sub>
\nHence A, B, C are not collinear.<\/p>\n

\"AP<\/p>\n

Question 4.
\nCheck whether (5, -2), (6, 4) and (7,-2) are the vertices of an isosceles triangle.
\nAnswer:
\nLet A = (5, – 2); B = (6, 4) and C = (7, – 2).
\n\"AP
\nNow we have, AB = BC.
\n\u2234 \u25b3ABC is an isosceles triangle,
\ni.e., given points are the vertices of an isosceles triangle.<\/p>\n

Question 5.
\nIn a classroom, 4 friends are seated at the points A, B, C and D as shown in figure. Jarina and Phani walk into the class and after observing for a few minutes Jarina asks Phani \u201cDon\u2019t you think ABCD is a square?\u201d Phani disagrees. Using distance formula, find which of them is correct. Why?
\n\"AP
\nAnswer:
\nGiven: Four friends are seated at A, B, C and D where A (3, 4), B (6, 7), C (9, 4) and D (6, 1).
\nNow distance
\n\"AP
\nBD = \\(\\sqrt{(6-6)^{2}+(1-7)^{2}}\\) = \u221a36 = 6
\nHence in \u25a1 ABCD four sides are equal
\ni.e., AB = BC = CD = DA
\n= 3\u221a2 units
\nand two diagonals are equal.
\ni.e., AC = BD = 6 units.
\n\u2234 \u25a1 ABCD forms a square.
\ni.e., Jarina is correct.<\/p>\n

\"AP<\/p>\n

Question 6.
\nShow that the following points form an equilateral triangle A(a, 0), B(- a, 0), C(0, a\u221a3).
\nAnswer:
\nGiven: A (a, 0), B (- a, 0), C (0, a\u221a3).
\n\"AP
\nNow, AB = BC = CA.
\n\u2234 \u25b3ABC is an equilateral triangle.<\/p>\n

Question 7.
\nProve that the points (-7, -3), (5, 10), (15, 8) and (3, -5) taken in order are the corners of a parallelogram.
\nAnswer:
\nTo show that the given points form a parallelogram.
\nWe have to show that the mid points of each diagonal are same. Since diagonals of a parallelogram bisect each other.
\nNow let A(-7, -3), B(5, 10), C(15, 8) and D(3, -5)
\nThen midpoint of diagonal
\n\"AP
\n\u2234 (1) = (2)
\nHence the given are vertices of a parallelogram.<\/p>\n

\"AP<\/p>\n

Question 8.
\nShow that the points (-4, -7), (-1, 2), (8, 5) and (5, -4) taken in order are the vertices of a rhombus. And find its area.
\n(Hint: Area of rhombus = \\(\\frac{1}{2}\\) \u00d7 product of its diagonals)
\nAnswer:
\nGiven in \u25b1 ABCD , A(-4, – 7), B (- 1, 2), C (8, 5) and D (5,-4)
\n\"AP
\n\"AP
\n\u2234\u00a0 In \u25b1 ABCD, AB = BC = CD = AD [from sides are equal]
\nHence \u25b1 ABCD is a rhombus.
\nArea of a rhombus = \\(\\frac{1}{2}\\) d1<\/sub>d2<\/sub>
\n= \\(\\frac{1}{2}\\) \u00d7 12\u221a2 \u00d7 6\u221a2
\n= 72 sq. units.<\/p>\n

Question 9.
\nName the type of quadrilateral formed, if any, by the following points, and give reasons for your answer.
\ni) (-1,-2), (1,0), (-1,2), (-3,0)
\nAnswer:
\nLet A (- 1, -2), B (1, 0), C (- 1, 2), D (- 3, 0) be the given points. Distance formula
\n\"AP
\nIn \u25b1 ABCD, AB = BC = CD = AD – four sides are equal.
\nAC = BD – diagonals are equal.
\nHence, the given points form a square,<\/p>\n

\"AP<\/p>\n

ii) (-3, 5), (1, 10), (3, 1), (-1,-4).
\nAnswer:
\nLet A(-3, 5), B(l,10), C(3, 1), D(-l, -4) then
\n\"AP
\nIn \u25b1 ABCD, \\(\\overline{\\mathrm{AB}}\\) = \\(\\overline{\\mathrm{CD}}\\) and \\(\\overline{\\mathrm{BC}}\\) = \\(\\overline{\\mathrm{AD}}\\) (i.e., both pairs of opposite sides are equal) and \\(\\overline{\\mathrm{AC}}\\) \u2260 \\(\\overline{\\mathrm{BD}}\\).
\nHence \u25b1 ABCD is a parallelogram,
\ni.e., The given points form a parallelogram.<\/p>\n

iii) (4, 5), (7, 6), (4, 3), (1, 2).
\nAnswer:
\nLet A (4, 5), B (7, 6), C (4, 3) and D (1, 2) be the given points.
\nDistance formula
\n\"AP
\nIn \u25b1 ABCD, AB = CD and BC = AD (i.e., both pairs of opposite sides are equal) and AC \u2260 BD.
\nHence \u25b1 ABCD is a parallelogram, i.e., The given points form a parallelogram.<\/p>\n

\"AP<\/p>\n

Question 10.
\nFind the point on the X-axis which is equidistant from (2, -5) and (-2,9).
\nAnswer:
\nGiven points, A (2, – 5), B (- 2, 9).
\nLet P (x, 0) be the point on X – axis which is equidistant from A and B. i.e., PA = PB.
\nDistance formula = \\(\\sqrt{\\left(x_{2}-x_{1}\\right)^{2}+\\left(y_{2}-y_{1}\\right)^{2}}\\)
\n\"AP
\nBut PA = PB.
\n\u21d2 \\(\\sqrt{x^{2}-4 x+29}=\\sqrt{x^{2}+4 x+85}\\)
\nSquaring on both sides, we get
\nx2<\/sup> – 4x + 29 = x2<\/sup> + 4x + 85
\n\u21d2 – 4x – 4x = 85 – 29
\n\u21d2 – 8x = 56
\n\u21d2 x = \\(\\frac{56}{-8}\\) = -7
\n\u2234 (x, 0) = (- 7, 0) is the point which is equidistant from the given points.<\/p>\n

Question 11.
\nIf the distance between two points (x, 7) and (1, 15) is 10, find the value of x.
\nAnswer:
\nFormula for distance between two points = \\(\\sqrt{\\left(x_{1}-x_{2}\\right)^{2}+\\left(y_{1}-y_{2}\\right)^{2}}\\)
\nNow distance between (x, 7) and (1,15) is 10.
\n\u2234 \\(\\sqrt{(x-1)^{2}+(7-15)^{2}}\\) = 10
\n\u2234 (x – l)2<\/sup> + (-8)2<\/sup> = 102<\/sup>
\n\u21d2 (x – l)2<\/sup> = 100 – 64 = 36
\n\u2234 x – 1 = \u221a36 = \u00b1 6
\n\u2234 x – 1 = 6 or x – 1 = -6
\n\u21d2 x = 6 + 1 = 7 or x = -6 + 1 = -5
\n\u2234 x = 7 or x = – 5<\/p>\n

\"AP<\/p>\n

Question 12.
\nFind the values of y for which the distance between the points P(2, -3) and Q(10, y) is 10 units.
\nAnswer:
\nGiven: P (2, – 3), Q (10, y) and
\n\\(\\overline{\\mathrm{PQ}}\\) = 10.
\nDistance formula = \\(\\sqrt{\\left(x_{2}-x_{1}\\right)^{2}+\\left(y_{2}-y_{1}\\right)^{2}}\\)
\n\"AP
\n\u21d2 y2<\/sup> + 6y + 73 = 100
\n\u21d2 y2<\/sup> + 6y – 27 = 0
\n\u21d2 y2<\/sup> + 9y – 3y – 27 = 0
\n\u21d2 y (y + 9) – 3 (y + 9) = 0
\n\u21d2 (y + 9) (y – 3) = 0
\n\u21d2 y + 9 = 0 or y – 3 = 0
\n\u21d2 y = -9 or y = 3
\n\u21d2 y = – 9 or 3.<\/p>\n

Question 13.
\nFind the radius of the circle whose centre is (3, 2) and passes through (-5,6).
\nAnswer:
\n\"AP
\nGiven: A circle with centre A (3, 2) passing through B (- 5, 6).
\nRadius = AB
\n[\u2235 Distance of a point from the centre of the circle]
\nDistance formula = \\(\\sqrt{\\left(x_{2}-x_{1}\\right)^{2}+\\left(y_{2}-y_{1}\\right)^{2}}\\)
\n\"AP<\/p>\n

\"AP<\/p>\n

Question 14.
\nCan you draw a triangle with vertices (1, 5), (5, 8) and (13, 14)? Give reason.
\nAnswer:
\nLet A (1, 5), B (5, 8) and C (13, 14) be the given points.
\nDistance formula
\n\"AP
\nHere, AC = AB + BC.
\n\u2234 \u25b3ABC can\u2019t be formed with the given vertices.
\n[\u2235 Sum of the any two sides of a triangle must be greater than the third side].<\/p>\n

Question 15.
\nFind a relation between x and y such that the point (x, y) is equidistant from the points (-2, 8) and (-3, -5).
\nAnswer:
\nLet A (- 2, 8), B (- 3, – 5) and P (x, y). If P is equidistant from A, B, then PA = PB.
\nDistance formula =
\n\"AP
\n\"AP
\nSquaring on both sides we get, x2<\/sup> + y2<\/sup> + 4x – 16y + 68
\n= x2<\/sup> + y2<\/sup> + 6x +10y + 34
\n\u21d2 4x – 16y – 6x – 10y = 34-68
\n\u21d2 – 2x – 26y = -34
\n\u21d2 x + 13y = 17 is the required condition.<\/p>\n","protected":false},"excerpt":{"rendered":"

AP State Board Syllabus\u00a0AP SSC 10th Class Maths Textbook Solutions Chapter 7 Coordinate Geometry Ex 7.1 Textbook Questions and Answers. AP State Syllabus SSC 10th Class Maths Solutions 7th Lesson Coordinate Geometry Exercise 7.1 10th Class Maths 7th Lesson Coordinate Geometry Ex 7.1 Textbook Questions and Answers Question 1. Find the distance between the following …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[2],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/3738"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=3738"}],"version-history":[{"count":0,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/3738\/revisions"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=3738"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=3738"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=3738"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}