{"id":12861,"date":"2022-06-05T23:00:08","date_gmt":"2022-06-05T17:30:08","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=12861"},"modified":"2022-05-23T16:10:59","modified_gmt":"2022-05-23T10:40:59","slug":"coordinate-geometry-class-10-extra-questions","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/coordinate-geometry-class-10-extra-questions\/","title":{"rendered":"Coordinate Geometry Class 10 Extra Questions Maths Chapter 7 with Solutions Answers"},"content":{"rendered":"

Here you will find Coordinate Geometry Class 10 Extra Questions Maths Chapter 7 with Answers Solutions, Extra Questions for Class 10 Maths<\/a> are solved by experts and will guide students in the right direction.<\/p>\n

Extra Questions for Class 10 Maths Coordinate Geometry with Answers Solutions<\/h2>\n

Extra Questions for Class 10 Maths Chapter 7 Coordinate Geometry with Solutions Answers<\/strong><\/p>\n

Coordinate Geometry Class 10 Extra Questions Objective Type<\/h3>\n

Question 1.
\nThe point (-13, -14) lies in :
\n(a) 1st quadrant
\n(b) 2nd quadrant
\n(c) 3rd quadrant
\n(d) None of these.
\nAnswer:
\n(c) 3rd quadrant<\/p>\n

Question 2.
\nPoint (8, -8) is in which quadrant?
\n(a) First
\n(b) Second
\n(c) Third
\n(d) Fourth.
\nAnswer:
\n(d) Fourth.<\/p>\n

Question 3.
\nThe position of the point with abscissa = -5 and ordinate = +4 will be :
\n(a) In the first quadrant
\n(b) In the second quadrant
\n(c) In the third quadrant
\n(d) None of these.
\nAnswer:
\n(b) In the second quadrant<\/p>\n

\"Coordinate<\/p>\n

Question 4.
\nDistance of point (-6, 8) from origin is :
\n(a) 8
\n(b) 2\u221a7
\n(c) 10
\n(d) 6
\nAnswer:
\n(c) 10
\nSolution.
\nDistance of point (-6, 8) from origin (0, 0) is
\n\"Coordinate
\n\u2234 Choice (c) is correct.<\/p>\n

Question 5.
\nCoordinates of a point on X-axis which is equal distance from the points A(2, -5) and B(-2, 9) will be :
\n(a) (-7, 0)
\n(b)(0, 7)
\n(c) 0, -7)
\n(d) (7, 0).
\nAnswer:
\n(a) (-7, 0)
\nSolution.
\nAny point on X-axis is P(x, 0)
\n\u2234 AP = BP \u21d2 AP2 = BP2
\n= (x – 2)2<\/sup> + (-5 – 0)2<\/sup> = (x + 2)2<\/sup> + (9 – 0)2<\/sup>
\n= x2<\/sup> – 4x + 4 + 25 = x2<\/sup> + 4x + 4 + 81
\nx2<\/sup> – 4x + 29 = x2<\/sup> + 4x + 85
\n– 8x = 85 – 29 = 56
\nx = \\(\\frac{56}{-8}\\) = -7
\n\u2234 point P on x – axis is (-7, 0)
\nHence choice (a) is correct.<\/p>\n

Question 6.
\nThe coordinate of a point on Y-axis which is equidistant from the point A(6, 5) and B(-4, 3) will be :
\n(a) (0, 9)
\n(b) (0, -9)
\n(c) (0, 5)
\n(d) (0, 3)
\nAnswer:
\n(a) (0, 9
\nSolution.
\nLet point P is on Y-axis whose coordinate is (0, y)
\nPA = PB
\n\u21d2 PA2<\/sup> = PB2<\/sup>
\n(6 – 0)2<\/sup> + (5 – y)2<\/sup> = (-4 -0)2<\/sup> + (3 – y)2<\/sup>
\n\u21d2 36 + 25 + y2<\/sup> – 10y = 16 + 9 + y2<\/sup> – 6y
\n\u21d2 61 – 10y = 25 – 6y
\n\u21d2 61 – 25 = – 6y + 10y
\n\u21d2 4y = 36
\n\u21d2 y = 9
\nRequired point P(0, 9)
\nHence choice (a) is correct. (a)<\/p>\n

Question 7.
\nA point lies in third quadrant. Co-ordinates of this point may be :
\n(a) (5, 5)
\n(b) (5, -5)
\n(c) (-5, -5)
\n(d) (-5, 5).
\nAnswer:
\n(c) (-5, -5)<\/p>\n

Question 8.
\nThe distance between the points (5, – 3) and (8, 1) is:
\n(a) 5 units
\n(b) 6 units
\n(c) 25 units
\n(d) 10 units.
\nAnswer:
\n(a) 5 units<\/p>\n

\"Coordinate<\/p>\n

Question 9.
\nThe distance between the points (2, 0) and (-1, 4) is :
\n(a) 5 units
\n(b) 25 units
\n(c)- 25 units
\n(d) 10 units.
\nAnswer:
\n(a) 5 units<\/p>\n

Question 10.
\nThe distance between the points (-6, 5) and (-1, 7) is :
\n(a) 169 units
\n(b) \u221a119 units
\n(c) \u221a13 units
\n(d) None of these.
\nAnswer:
\n(d) None of these.<\/p>\n

Question 11.
\nThe co-ordinates of a point Mare (3, 4). Its distance from the origin is :
\n(a) 7 units
\n(b) 11 units
\n(c) 5 units
\n(d) 12 units.
\nAnswer:
\n(c) 5 units<\/p>\n

Question 12.
\nThe distance between points (7, 3) and (-5, -2) is :
\n(a) 10 units
\n(b) 13 units
\n(c) 16 units
\n(d) 18 units.
\nAnswer:
\n(b) 13 units<\/p>\n

Question 13.
\nThe distance between origin and (15, 8) is :
\n(a) 18 units
\n(b) 15 units
\n(c) 17 units
\n(d) 28 units.
\nAnswer:
\n(c) 17 units<\/p>\n

Question 14.
\nThe distance between (-1, -3) and (3, 0) is :
\n(a) 4 units
\n(b) 5 units
\n(c) 13 units
\n(d) 16 units.
\nAnswer:
\n(b) 5 units<\/p>\n

\"Coordinate<\/p>\n

Question 15.
\nThe distance between the points (-3, 4) and (0, 0) will be :
\n(a) 5
\n(b) 4
\n(c) 13
\n(d) 1
\nAnswer:
\n(a) 5<\/p>\n

Question 16.
\nThe third vertex of an equilateral triangle whose other two vertices are (1, 1) and (1, -1) respectively is :
\n(a) (\u221a3, – \u221a3)
\n(b) (- \u221a3, \u221a3)
\n(c) both (a) and (b)
\n(d) none of these.
\nAnswer:
\n(c) both (a) and (b)<\/p>\n

Question 17.
\nThe coordinates of two points are (6, 0) and (0, 8). The coordinates of the mid-point are :
\n(a) (3, 4)
\n(b) (6, 8)
\n(c) (0, 0)
\n(d) None of these.
\nAnswer:
\n(a) (3, 4)<\/p>\n

Question 18.
\nThe co-ordinates of two points are (9, 4) and (3, 8). The co-ordinates of the mid-point are :
\n(a) (6, 0)
\n(b) (0, 6)
\n(c) (6, 6)
\n(d) (-6, -6).
\nAnswer:
\n(c) (6, 6)<\/p>\n

\"Coordinate<\/p>\n

Question 19.
\nThe co-ordinates of two points are (-8, 0) and (0, -8). The co-ordinates of the midpoint of the line segment joining them will be :
\n(a) (-8, 4)
\n(b) (4, -8)
\n(c) (-4, -4)
\n(d) (-4, 4)
\nAnswer:
\n(c) (-4, -4)<\/p>\n

Question 20.
\nThe abscissa of the point which divides the line joining points (0, 4) and (0, 8) internally in the ratio of 3 : 1 will be :
\n(a) 0
\n(b) 4
\n(c) 6
\n(d) 5
\nAnswer:
\n(a) 0<\/p>\n

Coordinate Geometry Class 10 Extra Questions Very Short Answer Type<\/h3>\n

Question 1.
\nFind the coordinate of a point A. Where AB is the diameter of a circle whose centre is (2, -3) and coordinate of B is (1, 4).
\nSolution.
\nAs centre is the mid point of AB.
\n\"Coordinate
\nLet coordinate of A is (x, y)
\n\u2234 \\(\\frac{x+1}{2}\\) = 2
\n\u21d2 2 x + 1 = 4
\n\u21d2 x = 4 – 1 = 3
\nand \\(\\frac{y+4}{2}\\) = -3 = y + 4 = -6
\ny = – 10
\nHence coordinate of A is (3, -10)<\/p>\n

\"Coordinate<\/p>\n

Question 2.
\nWhat are the ordinate of the point whose abscissa is 10 and whose distance from (2, -3) is 10 units?
\nSolution.
\nLet y be the ordinate, then the point is (10, y). The distance between (10, y) and (2, -3)
\n\"Coordinate
\n\"Coordinate
\n\u21d2 y2<\/sup> + 6y + 73 = 100
\n\u21d2 y2<\/sup> + 6y – 27 = 0
\n\u21d2 (y +9)(y – 3) = 0
\n\u21d2 y = -9, y = 3.<\/p>\n

Question 3.
\nFind the distance between the points A(a sin \u03b8, a cos \u03b8) and B(a cos \u03b8, – a sin \u03b8).
\nSolution.
\nHere x1 <\/sub>= a sin \u03b8, x2<\/sub> = a cos \u03b8 y1<\/sub> = a cos \u03b8, y2<\/sup> = – a sin AB
\nDistance AB
\n\"Coordinate<\/p>\n

Question 4.
\nProve that the point (4, 1) is the centre of the circle on whose circumference lie the points (-2, 9), (10, -7) and (12, -5).
\nSolution.
\nLet A(-2, 9), B(10, -7), C(12, – 5) and O(4, 1). Then,
\n\"Coordinate
\n\u2234 OA = OB = 0C.
\nHence O is the centre and A, B, C are the points on the circumference of the circle.<\/p>\n

\"Coordinate<\/p>\n

Question 5.
\nFind that ratio in which the point \\(P\\left(\\frac{3}{4}, \\frac{5}{12}\\right)\\) divides the line segment joining the points \\(A\\left(\\frac{1}{2}, \\frac{3}{2}\\right)\\) and B(2, -5).
\nSolution.
\nLet P divides AB in the ratio \u03bb : 1
\n\"Coordinate
\n\u21d2 3\u03bb + 3 = 8\u03bb + 2
\n\u21d2 5\u03bb = 1
\n\u03bb = \\(\\frac {1}{5}\\)
\nHence P divides AB in ratio 1 : 5<\/p>\n

Question 6.
\nFind the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B.
\nSolution.
\nLet M (0, 0) and N (36, 15) be the given points.
\nHere, x1<\/sub> = 0, y1<\/sub> = 0 and x2<\/sub> = 36, y2<\/sub> = 15
\n\"Coordinate
\nSince, the position of towns A and B are given (0, 0) and (36, 15), respectively and so, the distance between them is 39 km.<\/p>\n

Question 7.
\nFind the coordinates of that point which divides the line segment joining two given points (3, 5) and (8, 10) internally in the ratio of 2 : 3.
\nSolution.
\nHere, x1<\/sub> = 3, x2<\/sup> = 8, y1<\/sub> = 5, y2<\/sup> = 10, m = 2, n = 3.
\n\"Coordinate
\n\"Coordinate
\nHence, the co-ordinates of the required point are (5, 7).<\/p>\n

\"Coordinate<\/p>\n

Question 8.
\nIf the area of \u2206ABC formed by A(x, y), B(1, 2) and C(2, 1) is 6 square units, then prove x + y = 15.
\nSolution.
\nArea of \u2206 = \\(\\frac{1}{2}\\) [x1<\/sub> (y2<\/sub> – y3<\/sub>) + x2<\/sub>(y3<\/sub> – y1<\/sub>)+ x3<\/sub> y1<\/sub> – y2<\/sub>)]
\n\u21d2 6 = \\(\\frac{1}{2}\\)[x(2 – 1) + 1(1 – y) + 2 (y – 2)]
\n\u21d2 12 = x – y + 1 + 2y – 4
\n\u21d2 12 = x + y = 3
\n\u2234 x + y = 15<\/p>\n

Question 9.
\nFind the coordinates of point which divide line segment AB joining points A(-2, 2) and B( 2, 8) into four equal parts.
\nSolution.
\nClearly point a is the midpoint of AB
\n\"Coordinate<\/p>\n

Question 10.
\nThe coordinates of one endpoint of a diameter of a circle are (3, 5). If the coordinates of the centre be (6, 6), find the coordinates of the other endpoint of the diameter.
\nSolution.
\nLet AB be the given diameter and O the centre of the circle. Then, the coordinates of A and O and (3, 5) and (6, 6) respectively. Let (a, b) be the coordinates of B.
\nThen \\(\\frac{3+a}{2}\\) = 6 and \\(\\frac{5+b}{2}\\) = 6
\n\u2234 a = 9 and 6 = 7.
\nHence, the required point is (9, 7).<\/p>\n

\"Coordinate<\/p>\n

Question 11.
\nIf the points (6, 9),(0, x)and (-6, -7) are collinear then find the value of x.
\nSolution.
\nx1<\/sub> = 6, y1<\/sub> = 9
\nx2<\/sub> = 0, y2<\/sub> = x
\nx3<\/sub> = -6, y3<\/sub> = -7
\nThe three points are collinear if the area of the triangle formed by them is zero.
\nNow,
\narea = \\(\\frac{1}{2}\\)[x1<\/sub>(y2<\/sub> – y3<\/sub>) + x2<\/sub>(y3<\/sub> – y1<\/sub>)
\n+ x3<\/sub>(y1<\/sub> – y2<\/sub>)] = 0
\n\u21d2 \\(\\frac{1}{2}\\)[6(x + 7) + 0(-7 – 9) + (-6)(9 – x)) = 0
\n\u21d2 6x + 42 – 54 + 6x = 0
\n\u21d2 12x = 12
\n\u21d2 x = 1.<\/p>\n

Question 12.
\nFind the value of the points (a, 1), (1, -1) and (11, 4) are collinear.
\nSolution.
\nHere, A(a, 1) = (x1<\/sub>, y1<\/sub>) B(1, -1) = (x2<\/sub>, y2<\/sub>,) andC(11, 4) = (x3<\/sub>, y3<\/sub>)
\nCondition for collinear points
\n\u21d2 x1<\/sub>(y2<\/sub> – y3<\/sub>) + x2<\/sub>(y3<\/sub> – y1<\/sub>) + x3<\/sub>(y1<\/sub> – y2<\/sub>)
\n\u21d2 a(-1 -4) + 1(4 – 1) + 11(1 + 1) = 0
\n\u21d2 -5a + 3 + 22 = 0
\n\u21d2 -5a = -25
\n\u21d2 a = 5
\n\u2234 The given three points are collinear.<\/p>\n

Coordinate Geometry Class 10 Extra Questions Short Answer Type<\/h3>\n

Question 1.
\nShow that the points (2, 1),(5, 2), (6, 4) and (3, 3) form a parallelogram.
\nSolution.
\nLet A(2, 1), B(5, 2), C(6, 4) and D(3, 3).
\n\"Coordinate
\n\"Coordinate
\nThus, AB = CD, BC = DA.
\nSince, opposite sides are equal, hence ABCD is a parallelogram.<\/p>\n

\"Coordinate<\/p>\n

Question 2.
\nProve that the points A(1, -2), B(3, 0) and C(1, 2) are the vertices of an isosceles right-angled triangle.
\nSolution.
\nLet A = (1, -2), B = (3, 0) and C = (1, 2).
\n\"Coordinate
\nCA = 4
\nNow, AB2<\/sup> + BC2<\/sup> = (2\u221a2)2<\/sup> + (2\u221a2)2<\/sup>
\n= 8 + 8 = 16
\nAB2<\/sup> + BC2<\/sup> = (AC)2<\/sup> …(4)
\nHence, from result (1), (2) and (4) \u2206ABC is an isosceles right-angled triangle.<\/p>\n

Question 3.
\nFind the centre of the circle which passes through the points (-3, -2), (-2, 3) and (3, 2).
\nSolution.
\nLet O(x, y) be the centre of the circle and the points
\nA(-3, – 2), B(-2, 3) and C(3, 2)
\nlie on the circle, Then, we have
\nOA2<\/sup> = (x + 3)2<\/sup> + (y + 2)2<\/sup>
\nOB2<\/sup> = (x + 2)2<\/sup> + (y – 3)2<\/sup>
\nand OC2<\/sup> = (x – 3)2<\/sup> + (y – 2)2<\/sup>
\n\u2235 OA = OB
\n\u21d2 OA2<\/sup> = OB2<\/sup>
\n\u2234(x + 3)2<\/sup> + (y + 2)2<\/sup> = (x + 2)2<\/sup> + (y – 3)2<\/sup>
\n\u21d2 x2<\/sup> + y2<\/sup> + 6x + 4y + 13
\n= x2<\/sup> + y2<\/sup> + 4x – 6y + 13
\n\u21d2 6x + 4y = 4x – 6y
\n\u21d2 2x + 10y = 0
\n\u21d2 x + 5y = 0 ……(i)
\nAgain, OB = OC
\n\u21d2 OB2<\/sup> = OC2<\/sup>
\n\u2234 (x + 2)2<\/sup> + (y – 3)2<\/sup> = (x – 3)2<\/sup> + (y – 2)2<\/sup>
\n\u21d2 x2<\/sup> + y2<\/sup> + 4x – 6y + 13
\n= x2<\/sup> + y2<\/sup> – 6x – 4y + 13
\n\u21d2 4x – 6y = – 6x – 4y
\n\u21d2 10x – 2y = 0
\n\u21d2 5x – y = 0 …..(ii)
\nOn solving (i) and (ii), we get
\nx = 0 and y = 0.
\nHence, the centre of the circle is (0, 0).<\/p>\n

\"Coordinate<\/p>\n

Question 4.
\nProve that the triangle whose vertices are respectively (a, a), (-a, -a) and (-a\u221a3, a\u221a3) is an equilateral triangle.
\nSolution.
\nLet points P(a, a), Q(-a, -a) and R(-a\u221a3, a\u221a3) are given, then
\nand
\n\"Coordinate
\n= a\u221a8 = 2a\u221a2
\nThus, PQ = QR = RP
\n\u2234 \u2206PQR is an equilateral triangle.<\/p>\n

Question 5.
\nFind the point on the X-axis which is equidistant from (2, -5) and (-2, 9).
\nSolution.
\nSince, the point on the X-axis.
\n\u2234 it’s ordinate = 0
\nSo, A(x, 0) is any point on the X-axis.
\nSince, A(x, 0) is equidistant from B(2, -5) and C(-2, 9).
\nAB = AC
\n\u21d2 AB2<\/sup> = AC2<\/sup>
\n\u21d2 (x – 2)2<\/sup> + (0 + 5)2<\/sup> = (x + 2)2<\/sup> + (0 – 9)2<\/sup>
\n\"Coordinate
\n\u21d2 x2<\/sup> – 4x + 4 + 25 = x2<\/sup> + 4x + 4 + 81
\n\u21d2 -4x – 4x = 81 – 25
\n\u21d2 -8x = 56
\n\u21d2 x = –\\(\\frac{56}{8}\\) = -7
\nSo, the point equidistant from given points on the x-asis is (-7, 0).<\/p>\n

\"Coordinate<\/p>\n

Question 6.
\nFind the values of y for which the distance between the points P(2, -3) and Q(10, y) is 10 units.
\nSolution.
\nAccording to question,
\nPQ = 10
\n\"Coordinate
\nSquaring both sides, we get
\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 or y = 3
\n\u21d2 y = -9 or 3<\/p>\n

Question 7.
\nFind a relation between x and y such that the point (x, y) is equidistant from the points (3, 6) and (-3, 4).
\nSolution.
\nLet the point A(x, y) be equidistant from the points B(3, 6) and (-3, 4).
\nAB = AC
\nAB2<\/sup> = AC2<\/sup>
\n\u21d2 (x – 3)2<\/sup> + (y – 6)2<\/sup> = (x + 3)2<\/sup> + (y – 4)2<\/sup>
\n\"Coordinate
\n\u21d2 x2<\/sup> – 6x + 9 + y2<\/sup> – 12y + 36
\n\u21d2 x2<\/sup> + 6x + 9 + y2<\/sup> – 8y + 16
\n\u21d2 -6x – 6x – 12y + 8y + 36 – 16 = 0
\n\u21d2 -12x – 4y + 20 = 0
\n\u21d2 -4(3x + y – 5) = 0
\n\u21d2 3x + y – 5 = 0
\n(\u2235 -4 \u2260 0)<\/p>\n

\"Coordinate<\/p>\n

Question 8.
\nIf the vertices of a \u2206ABC, are (sin2<\/sup> \u03b8, cos2<\/sup> \u03b8), (cos2<\/sup> \u03b8, sin2<\/sup> \u03b8) and (0, 0) then find the coordinates of the centroid of the triangle.
\nSolution.
\nHere A = (sin2<\/sup> \u03b8, cos2<\/sup> \u03b8)
\n= (x1<\/sub>, y1<\/sub>)
\nB = (cos2<\/sup> \u03b8, sin2<\/sup> \u03b8)
\n= (x2<\/sub>, y2<\/sub>)
\nC = (0, 0)
\n= (x3<\/sub>, y3<\/sub>)
\nThe co-ordinates of the centroid be
\n\"Coordinate<\/p>\n

Question 9.
\nFind the ratio in which the point (11, 15) divides the line segment joining the points (15, 5) and (9, 20).
\nSolution.
\nA(x1<\/sub>, y1<\/sub>) = (15, 5)
\nB(x2<\/sub>, y2<\/sub>) = (9, 20)
\nP(x, y) = (11, 15)
\nm1<\/sub> : m2<\/sub> = ?
\n\"Coordinate
\n\u21d2 11(m1<\/sub> + m2<\/sub>) = 9m1<\/sub> + 15m2<\/sub>
\n\u21d2 15(m1<\/sub> + m2) = 20m1<\/sub> + 5m2<\/sub>
\n\u21d2 2m1<\/sub> = 4m2<\/sub>
\n\u21d2 5m1<\/sub> = 10m2<\/sub>
\n\u21d2 m1<\/sub> = 2m2<\/sub>
\n\u21d2 m1<\/sub> = 2m2<\/sub>
\n\u21d2 \\(\\frac{m_{1}}{m_{2}}\\) = \\(\\frac{2}{1}\\)
\n= i.e., m1<\/sub> = m2<\/sub> = 2 : 1.<\/p>\n

Question 10.
\nFind the coordinates of the point which divides the line segment joining the points (a, b) and (b, a) internally in the ratio of (a – b) : (a + b).
\nSolution.
\nFormula,
\n\"Coordinate<\/p>\n

Question 11.
\nThe mid-points of the sides of a triangle are (1, 2), (0, -1) and (2, -1). Find the coordinates of the vertices of the triangle.
\nSolution.
\nLet A(x1<\/sub>, y1<\/sub>), B(x2<\/sub>, y2<\/sub>) and C(x3<\/sub>, y3<\/sub>) be the vertices of \u2206ABC. Let (1, 2), (0, -1) and (2, -1) be the mid-points of AB, BC and CA respectively.
\nThen
\n\"Coordinate
\nx1<\/sub> + x2<\/sub> = 2 …..(i)
\nx2<\/sup> + x3<\/sub> = 0 …..(ii)
\nx3<\/sub> + x1<\/sub> = 4 …..(iii)
\ny1<\/sub> + y2<\/sub> = 4 ……(iv)
\ny2<\/sub> + y3<\/sub> = -2 …..(v)
\ny3<\/sub> + y1<\/sub> = -2 ……(vi)
\nAdding (i), (ii) and (iii) we get
\nx1<\/sub> + x2<\/sub> + x3<\/sub> = 3.
\n\u2234 x3<\/sub> = 1, x1<\/sub> = 3 and x2<\/sub> = -1.
\nAdding (iv), (v) and (vi) we get
\ny1<\/sub> + y2<\/sub> + y3<\/sub> = 0.
\n\u2234 y3<\/sub>= – 4, y1<\/sub> = 2 and y2<\/sub> = 2.
\nHence the vertices of the triangle are A(3, 2), B(-1, 2) and C(1, -4) respectively.<\/p>\n

\"Coordinate<\/p>\n

Question 12.
\nThe three vertices of a parallelogram taken in order are (0, -1), (1, 3) and (2, 2) respectively. Find the coordinates of the forth vertex.
\nSolution.
\nLet A(0, -1), B(1, 3), C(2, 2) and D(x, y) be the vertices of a parallelogram. Then the mid-points of the diagonals AC and BD have the same co-ordinates.
\n\"Coordinate
\nHence, the forth vertex of the parallelogram is (1, -2).<\/p>\n

Question 13.
\nFind the coordinates of the point which divides the join of (-1, 7) and (4, -3) in the ratio 2 : 3.
\nSolution.
\nLet the coordinates of a point are (x, y).
\nwe have, x1<\/sub> = -1, y1<\/sub> = 7;
\nx2<\/sub> = 4, y2<\/sub> = -3
\nand m1<\/sub> = 2, m2<\/sub> = 3
\n\u2234 By using section formula,
\n\"Coordinate
\nHence, coordinates of the point are (1, 3)<\/p>\n

Question 14.
\nIf (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
\nSolution.
\nLet A(1, 2), B(4, y), C(x, 6) and D(3, 5) are the vertices of a parallelogram.
\n\"Coordinate
\nSince, ABCD is a parallelogram.
\n\u2234 AC and BD will bisect each other.
\nHence, mid-point of AC and mid-point of BD are same point.
\n\"Coordinate
\n\u21d2 1 + x = 7 and 8 = 5 + y
\n\u2234 x = 6 and y = 3.<\/p>\n

Question 15.
\nProve that the three points (1, 2), (3, 3) and (-1, 1) are collinear.
\nSolution.
\nThree points are collinear if the area of the triangle formed by them is zero.
\nHere
\nx1<\/sub> = 1, y1<\/sub> = 2
\nx2<\/sub> = 3, Y2<\/sub> = 3
\nx3<\/sub> = -1, y3<\/sub> = 1
\nNow, area
\n\"Coordinate
\n\u2234 The given three points are collinear.<\/p>\n

Question 16.
\nIf the points A(x, y), B(2, 3) and C(3, 4) are collinear, then prove that x + 5y = 17.
\nSolution.
\nHere x1<\/sub> = x, x2<\/sub> = 2, x3<\/sub> = -3, y1<\/sub> = y, y2<\/sub> = 3, y3<\/sub> = 4
\n\"Coordinate
\nThe points A, B, C are collinear, therefore \u2206 = 0.
\n\u2234 \\(\\frac{1}{2}\\) (- x – 5y + 17) = 0
\n\u21d2 – x – 5y + 17 = 0
\n\u21d2 x + 5y = 17.<\/p>\n

\"Coordinate<\/p>\n

Question 17.
\nFor what value of a, the points (1, 4) (a, -2) and (-3, 16) are collinear?
\nSolution.
\nThe given points are collinear if the area of triangle formed by these three points is zero.
\nHere, x1<\/sub> = 1, x2<\/sub> = a, x3<\/sub> = -3
\ny1<\/sub> = 4, y2<\/sub>\u00a0= -2, y3<\/sub> = 16
\n\"Coordinate<\/p>\n

Question 18.
\nFind the area of the triangle whose vertices are
\n(i) (2, 3), (-1, 0), (2, -4)
\n(ii) (-5, -1), (3, -5), (5, 2)
\n\"Coordinate
\n\"Coordinate<\/p>\n

Question 19.
\nFind the equation of the locus of points equidistant from (-2, 2) and (-4, -2).
\nSolution.
\nLet A(-2, 2) and B(-4, -2) be the given points and let P(x, y) be the variable point, Then.
\nPA = PB PA2<\/sup> = PB2<\/sup>
\n\u21d2 (x + 2)2<\/sup> + (y – 2)2<\/sup> = (x + 4)2<\/sup> + (y + 2)2<\/sup>
\n\u21d2 x2<\/sup> + 2x + 4 + y2<\/sup> – 2y + 4 = x2<\/sup> + 8x + 16 + y2<\/sup> + 4y + 4
\n\u21d2 2x – 2y + 8 = 8x + 4y + 20
\n\u21d2 – 6x – 6y – 12 = 0
\n\u21d2 x + y + 2 = 0
\nwhich is the required equation of the locus.<\/p>\n

Question 20.
\nFind the equation of the locus of all points equidistant from (3, 5) and the x-axis.
\nSolution.
\nLet P(x, y) be the variable point, then its distance from x-axis is y and its distance from (3, 5) is
\n\"Coordinate
\nBy the given condition we have
\n\"Coordinate
\nSquaring on both sides we get
\ny2<\/sup> = x2<\/sup> – 6x + 9 + y2<\/sup> – 10y + 25
\n0 = x2<\/sup> – 6x – 10y + 25
\n\u2234 x2<\/sup> – 6x – 10y + 25 = 0
\nwhich is the required equation of the locus.<\/p>\n

\"Coordinate<\/p>\n

Question 21.
\nProve that the equation of the locus of a point which moves in such a way that its distance from a point (-g, -f) is always equal to a, is
\nx2<\/sup> + y2<\/sup> + 2gx + 2fy + c = 0,
\nwhere x = (g2<\/sup> + f2<\/sup> – a2<\/sup>).
\nSolution.
\nLet P(x, y) be the variable point on the locus. Then,
\n\"Coordinate
\n\u21d2 (x + g)2<\/sup> + (y + f)2<\/sup> = a2<\/sup>
\n\u21d2 x2<\/sup> + y2<\/sup> + 2gx + 2fy + (g2<\/sup> + f2<\/sup> – a2<\/sup>) = 0
\n\u21d2 x2<\/sup> + y2<\/sup> + 2gx + 2fy + c = 0,
\nwhere c = (g2<\/sup> + f2<\/sup> – a2<\/sup>).<\/p>\n

Coordinate Geometry Class 10 Extra Questions Long Answer Type<\/h3>\n

Question 1.
\nIn a classroom, 4 friends are seated at the points, A, B, C and Das shown in figure. Champ and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, \u201cDon’t you think ABCD is a square ?\u201d Chameli disagrees. Using distance formula, find which of them is correct.
\n\"Coordinate
\nSolution.
\nFrom the given figure, the coordinates of points A, B, C and D are (3, 4), (6, 7), (9, 4) and (6, 1).
\n\"Coordinate
\nSince, the four sides and diagonals are equal. Hence, ABCD is a square.
\nSo, Champa is correct.<\/p>\n

\"Coordinate<\/p>\n

Question 2.
\nIf Q(0, 1) is equidistant from P(5, -3) and R(x, 6), find the values of x. Also, find the distance QR and PR.
\nSolution.
\nSince, the point Q (0, 1) is equidistant from P(5, 3) and R(x, 6)
\n\u2234 QP = QR
\n\u2234 QP2<\/sup> = QR2<\/sup>
\n= (5 – 0)2<\/sup> + (-3 – 1)2<\/sup> = (x – 0)2 <\/sup>+ (6 – 1)2<\/sup>
\n\"Coordinate
\n\u21d2 52<\/sup> + 42<\/sup> = x2<\/sup> + 52<\/sup>
\n\u21d2 25 + 16 = x2<\/sup> + 25
\n\u21d2 x2<\/sup> = 16
\n\u21d2 x = \u00b14
\nThus, R is (4, 6) or (-4, 6).
\nNow, QR = Distance between Q(0, 1) and R(4, 6)
\n\"Coordinate<\/p>\n

Question 3.
\nFind the third vertex of the triangle ABC, if its two vertices are (3,5), (-4, -6) and the coordinates of the centroid are (4, 3).
\n\"Coordinate
\nSolution.
\nLet the co-ordinates of the third vertex be C(x, y).
\nHere x1<\/sub> = 3, x2<\/sub> = – 4, x3<\/sub> = x
\ny1<\/sub> = 5, y2<\/sub>\u00a0= -6, y3<\/sub> = y
\n\"Coordinate
\n\u21d2 -1 + x = 12; – 1 + y = 9
\n\u21d2 x = 13; y = 10
\n\u2234 Co-ordinates of Care (13, 10).<\/p>\n

Question 4.
\nThe endpoints of a line segment AB are A(2, 5) and B(1, 9). In what ratio the line segment is drawn through A and B is divided by x-axis and y-axis?
\nSolution.
\nFor x-axis :
\nThe coordinates of any point on the x-axis are (x, 0).
\nLet the required ratio be m1<\/sub> : m2<\/sub>
\n\"Coordinate
\nHence, the required ratio is 5 : 9. The negative sign represents external division.
\nFor y-axis :
\nThe coordinates of any point on the y-axis are (0, y).
\nLet the required ratio be m1<\/sub> : m2<\/sub>
\n\"Coordinate
\nHence, the required ratio is 2 : 1. The negative sign represents external division.<\/p>\n

\"Coordinate<\/p>\n

Question 5.
\nA point (2, 3) divides the distance between the points (x, y) and (-5, 4) internally in the ratio 1 : 2. Find the values of x and y.
\nSolution.
\n\"Coordinate<\/p>\n

Question 6.
\nUse analytical geometry to prove that the mid-point of the hypotenuse of a right-angled triangle is equidistant from its vertices.
\nSolution.
\nLet AOB be the given right-angled triangle with base OA taken along the x-axis and the perpendicular OB taken along the y-axis.
\nLet OA = a and OB = b.
\nLet D be the mid-point of hypotenuse AB.
\nThen, the coordinates of these points are
\n\"Coordinate
\n\"Coordinate
\n\u21d2 DA = DB = DO.<\/p>\n

Question 7.
\nIf the centroid of any triangle ABC be G, then using analytical geometry, prove that AB2<\/sup> + BC2<\/sup> + CA2<\/sup> = 3(GA2<\/sup> + GB2<\/sup> + GC2<\/sup>).
\nSolution.
\nLet B be the origin, BC be X-axis and through B perpendicular on BC be Y-axis. Then coordinates of B are (0, 0).
\n\"Coordinate
\nLet BC = a, then co-ordinates of Care (a, 0). Let the coordinates of A be (x1<\/sub>, y1<\/sub>), then co-ordinates of the centroid G of \u2206ABC are :
\n\"Coordinate
\n\"Coordinate<\/p>\n

\"Coordinate<\/p>\n

Question 8.
\nFind the coordinates of the points of trisection of the line segment joining (4, -1) and (-2, -3).
\nSolution.
\nLet A(4, -1) and B(-2, -3) be the line segments and points of trisection of the line segment be P and Question Then,
\nAP = PQ = BQ = k (Say)
\n\u2234 PB = PQ + QB = 2k
\nand AQ = AP + PQ = 2k
\n\"Coordinate
\n\u21d2 AP : PB = k : 2k = 1 : 2
\nand AQ : QB = 2k : k = 2 : 1
\nSince, P divides AB internally in the ratio 1 : 2. So, the coordinate of P are
\n\"Coordinate
\nand Q divides AB internally in the ratio 2 : 1. So, the coordinates of Q are
\n\"Coordinate
\nSo, the two points of trisection are
\n\"Coordinate<\/p>\n

Question 9.
\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 1 m from each other along with AD, as shown in figure. Niharika runs \\(\\frac{1}{4}\\)th the distance AD on the 2nd line and posts a green flag. Preet runs \\(\\frac{1}{5}\\)th the distance AD on the eighth line 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?
\n\"Coordinate
\nSolution.
\nFrom the above figure, the position of green flag posted by Niharika is M
\n\"Coordinate
\nLet P be the position of the blue flag posted by Rashmi in the halfway of line segment MN.
\n\"Coordinate
\nHence, the blue flag is on the fifth line at a distance of 22.5 m above it.<\/p>\n

Question 10.
\nFind the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by ( 1, 6).
\nSolution.
\nLet the point A(-1, 6) divide the line joining B(-3, 10) and C(6, -8) in the ratio k : 1. Then, the coordinates of A are
\n\"Coordinate
\n\u21d2 6k – 3 = – k – 1 and – 8k + 10 = 6k + 6
\n\u21d2 6k + k = – 1 + 3 and – 8k – 6k = 6 – 10
\n\u21d2 7k = 2 and – 14k = -4 \u21d2 k = \\(\\frac{2}{7}\\)
\nSo, the point A divides BC in the ratio 2 : 7.<\/p>\n

\"Coordinate<\/p>\n

Question 11.
\nFind the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, -3) and Bis (1, 4).
\nSolution.
\nSuppose, AB be a diameter of the circle having its centre at C(2, -3) and coordinates of endpoint B are (1, 4).
\nLet the coordinates of A be (x, y).
\n\"Coordinate
\nSince, AB is diameter.
\n\u2234 C is the mid-point of AB.
\n\"Coordinate
\n\u21d2 y + 4 = -6
\n\u21d2 y = -10<\/p>\n

Question 12.
\nIf A and B are (-2, -2) and (2, -4), respectively, find the coordinates of P such that
\nAP = \\(\\frac{3}{7}\\)AB and P lies on the line segment AB.
\nSolution.
\nAccording to the question,
\n\"Coordinate
\nSuppose, P(x, y) be the point which divides the line segment joining the points A (-2, -2) and B (2, -4) in the ratio 3 : 4.
\n\"Coordinate
\n\"Coordinate<\/p>\n

Question 13.
\nFind the coordinates of the points which divide the line segment joining A(-2, 2) and B(2, 8) into four equal parts.
\nSolution.
\nLet P, Q and R be the points on line segment AB such that
\nAP = PQ = QR = RB
\nLet AP= PQ = QR = RB = k .
\n\"Coordinate
\nQ is the mid-point of AB.
\n\"Coordinate<\/p>\n

\"Coordinate<\/p>\n

Question 14.
\nThe coordinates of three vertices of a \u2206ABC are A(3, 4), B(2, -1) and C(4, -6) and the middle points of the three sides BC, CA and AB are respectively, P, Q and R. Prove that:
\nArea of \u2206ABC = 4 (area of \u2206PQR).
\nSolution.
\nThe middle point of BC is
\n\"Coordinate
\n\"Coordinate
\nFrom (i) and (ii), we get
\nArea of \u2206ABC = 4 x area of \u2206PQR.<\/p>\n

Question 15.
\nFind the area of a triangle, the coordinates of the mid-points of whose sides are (-2, -1), (1, 6) and (5, 3) respectively.
\nSolution.
\nLet ABC be the given triangle and D(-2, -1), E(1, 6) and F(5, 3) be the midpoints of AB, BC and AC respectively.
\n\"Coordinate
\nThe co-ordinates of mid-points D, E and F are
\nHere
\nx1<\/sub> = -2, x2<\/sub> = 1, x3<\/sub> = 5
\ny1<\/sub> = -1, y2<\/sub> = 6, y3<\/sub> = 3
\nArea of \u2206DEF
\n\"Coordinate
\n= \\(\\frac{37}{2}\\) square unit.
\nArea of \u2206ABC = 4 \u00d7 ar (\u2206DEF)
\n= 4 \u00d7 \\(\\frac{37}{2}\\)
\n= 74 square units.<\/p>\n

Question 16.
\nMedian of a triangle divides it into two triangles of equal areas. Verify this result for \u2206ABC whose vertices are A (4, -6), B (3, – 2) and C (5, 2).
\nSolution.
\nAccording to the question, AD is the median of \u2206ABC, therefore D is the midpoint of BC.
\n\"Coordinate
\n\"Coordinate
\n(\u2235 Area of triangle is positive) .
\n\u2235 Area of \u2206ADC = Area of \u2206ABD
\nHence, the median of the triangle divides it into two triangles of equal areas.<\/p>\n

Question 17.
\nFind the locus of the point P(x, y) when :
\n(i) x = 3
\n(ii) y = 5
\n(iii) x = y
\nSolution.
\n(i) The given condition is x = 3.
\nAs y is not restricted, it can take any real value between – \u221e to \u221e i.e., – \u221e < y < \u221e.
\n\"Coordinate
\nHence, in the figure the line AB is the required locus.<\/p>\n

(ii) The condition given in the question is y = 5. As x is not restricted, it can take any real value between -\u221e and \u221e i.e., -\u221e < x < \u221e.
\n\"Coordinate
\nHence, in the figure the line PQ is the required locus.<\/p>\n

(iii) The given condition is x = y. If x takes the values – 2, -1, 0, 1, 2, then y will also take the values -2, -1, 0, 1, 2 respectively. Therefore, the points (-2, -2), (-1, -1), (0, 0), (1, 1) and (2, 2) will satisfy the above condition. Hence, AB will be the required locus.
\n\"Coordinate<\/p>\n

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

Here you will find Coordinate Geometry Class 10 Extra Questions Maths Chapter 7 with Answers Solutions, Extra Questions for Class 10 Maths are solved by experts and will guide students in the right direction. Extra Questions for Class 10 Maths Coordinate Geometry with Answers Solutions Extra Questions for Class 10 Maths Chapter 7 Coordinate Geometry …<\/p>\n

Coordinate Geometry Class 10 Extra Questions Maths Chapter 7 with Solutions Answers<\/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":"\nCoordinate Geometry Class 10 Extra Questions Maths Chapter 7 with Solutions Answers - MCQ Questions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mcq-questions.com\/coordinate-geometry-class-10-extra-questions\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Coordinate Geometry Class 10 Extra Questions Maths Chapter 7 with Solutions Answers - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"Here you will find Coordinate Geometry Class 10 Extra Questions Maths Chapter 7 with Answers Solutions, Extra Questions for Class 10 Maths are solved by experts and will guide students in the right direction. Extra Questions for Class 10 Maths Coordinate Geometry with Answers Solutions Extra Questions for Class 10 Maths Chapter 7 Coordinate Geometry … Coordinate Geometry Class 10 Extra Questions Maths Chapter 7 with Solutions Answers Read More »\" \/>\n<meta property=\"og:url\" content=\"https:\/\/mcq-questions.com\/coordinate-geometry-class-10-extra-questions\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ Questions\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/NCERTSolutionsGuru\/\" \/>\n<meta property=\"article:published_time\" content=\"2022-06-05T17:30:08+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2022-05-23T10:40:59+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/mcq-questions.com\/wp-content\/uploads\/2020\/05\/NCERT-Solutions.png\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:creator\" content=\"@ncertsolguru\" \/>\n<meta name=\"twitter:site\" content=\"@ncertsolguru\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"Prasanna\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"20 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"WebSite\",\"@id\":\"https:\/\/mcq-questions.com\/#website\",\"url\":\"https:\/\/mcq-questions.com\/\",\"name\":\"MCQ Questions\",\"description\":\"MCQ Questions for Class 1 to 12\",\"potentialAction\":[{\"@type\":\"SearchAction\",\"target\":{\"@type\":\"EntryPoint\",\"urlTemplate\":\"https:\/\/mcq-questions.com\/?s={search_term_string}\"},\"query-input\":\"required name=search_term_string\"}],\"inLanguage\":\"en-US\"},{\"@type\":\"ImageObject\",\"@id\":\"https:\/\/mcq-questions.com\/coordinate-geometry-class-10-extra-questions\/#primaryimage\",\"inLanguage\":\"en-US\",\"url\":\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2020\/05\/NCERT-Solutions.png?fit=460%2C45&ssl=1\",\"contentUrl\":\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2020\/05\/NCERT-Solutions.png?fit=460%2C45&ssl=1\",\"width\":460,\"height\":45,\"caption\":\"NCERT Solutions\"},{\"@type\":\"WebPage\",\"@id\":\"https:\/\/mcq-questions.com\/coordinate-geometry-class-10-extra-questions\/#webpage\",\"url\":\"https:\/\/mcq-questions.com\/coordinate-geometry-class-10-extra-questions\/\",\"name\":\"Coordinate Geometry Class 10 Extra Questions Maths Chapter 7 with Solutions Answers - MCQ Questions\",\"isPartOf\":{\"@id\":\"https:\/\/mcq-questions.com\/#website\"},\"primaryImageOfPage\":{\"@id\":\"https:\/\/mcq-questions.com\/coordinate-geometry-class-10-extra-questions\/#primaryimage\"},\"datePublished\":\"2022-06-05T17:30:08+00:00\",\"dateModified\":\"2022-05-23T10:40:59+00:00\",\"author\":{\"@id\":\"https:\/\/mcq-questions.com\/#\/schema\/person\/4ba9570f32f2057e70e670c7885e47f3\"},\"breadcrumb\":{\"@id\":\"https:\/\/mcq-questions.com\/coordinate-geometry-class-10-extra-questions\/#breadcrumb\"},\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"ReadAction\",\"target\":[\"https:\/\/mcq-questions.com\/coordinate-geometry-class-10-extra-questions\/\"]}]},{\"@type\":\"BreadcrumbList\",\"@id\":\"https:\/\/mcq-questions.com\/coordinate-geometry-class-10-extra-questions\/#breadcrumb\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Home\",\"item\":\"https:\/\/mcq-questions.com\/\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"Coordinate Geometry Class 10 Extra Questions Maths Chapter 7 with Solutions Answers\"}]},{\"@type\":\"Person\",\"@id\":\"https:\/\/mcq-questions.com\/#\/schema\/person\/4ba9570f32f2057e70e670c7885e47f3\",\"name\":\"Prasanna\",\"image\":{\"@type\":\"ImageObject\",\"@id\":\"https:\/\/mcq-questions.com\/#personlogo\",\"inLanguage\":\"en-US\",\"url\":\"https:\/\/secure.gravatar.com\/avatar\/174540ad43736c7d1a4c4f83c775e74d?s=96&d=mm&r=g\",\"contentUrl\":\"https:\/\/secure.gravatar.com\/avatar\/174540ad43736c7d1a4c4f83c775e74d?s=96&d=mm&r=g\",\"caption\":\"Prasanna\"},\"url\":\"https:\/\/mcq-questions.com\/author\/prasanna\/\"}]}<\/script>\n<!-- \/ Yoast SEO plugin. -->","yoast_head_json":{"title":"Coordinate Geometry Class 10 Extra Questions Maths Chapter 7 with Solutions Answers - MCQ Questions","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/mcq-questions.com\/coordinate-geometry-class-10-extra-questions\/","og_locale":"en_US","og_type":"article","og_title":"Coordinate Geometry Class 10 Extra Questions Maths Chapter 7 with Solutions Answers - MCQ Questions","og_description":"Here you will find Coordinate Geometry Class 10 Extra Questions Maths Chapter 7 with Answers Solutions, Extra Questions for Class 10 Maths are solved by experts and will guide students in the right direction. Extra Questions for Class 10 Maths Coordinate Geometry with Answers Solutions Extra Questions for Class 10 Maths Chapter 7 Coordinate Geometry … Coordinate Geometry Class 10 Extra Questions Maths Chapter 7 with Solutions Answers Read More »","og_url":"https:\/\/mcq-questions.com\/coordinate-geometry-class-10-extra-questions\/","og_site_name":"MCQ Questions","article_publisher":"https:\/\/www.facebook.com\/NCERTSolutionsGuru\/","article_published_time":"2022-06-05T17:30:08+00:00","article_modified_time":"2022-05-23T10:40:59+00:00","og_image":[{"url":"https:\/\/mcq-questions.com\/wp-content\/uploads\/2020\/05\/NCERT-Solutions.png"}],"twitter_card":"summary_large_image","twitter_creator":"@ncertsolguru","twitter_site":"@ncertsolguru","twitter_misc":{"Written by":"Prasanna","Est. reading time":"20 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebSite","@id":"https:\/\/mcq-questions.com\/#website","url":"https:\/\/mcq-questions.com\/","name":"MCQ Questions","description":"MCQ Questions for Class 1 to 12","potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/mcq-questions.com\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"ImageObject","@id":"https:\/\/mcq-questions.com\/coordinate-geometry-class-10-extra-questions\/#primaryimage","inLanguage":"en-US","url":"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2020\/05\/NCERT-Solutions.png?fit=460%2C45&ssl=1","contentUrl":"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2020\/05\/NCERT-Solutions.png?fit=460%2C45&ssl=1","width":460,"height":45,"caption":"NCERT Solutions"},{"@type":"WebPage","@id":"https:\/\/mcq-questions.com\/coordinate-geometry-class-10-extra-questions\/#webpage","url":"https:\/\/mcq-questions.com\/coordinate-geometry-class-10-extra-questions\/","name":"Coordinate Geometry Class 10 Extra Questions Maths Chapter 7 with Solutions Answers - MCQ Questions","isPartOf":{"@id":"https:\/\/mcq-questions.com\/#website"},"primaryImageOfPage":{"@id":"https:\/\/mcq-questions.com\/coordinate-geometry-class-10-extra-questions\/#primaryimage"},"datePublished":"2022-06-05T17:30:08+00:00","dateModified":"2022-05-23T10:40:59+00:00","author":{"@id":"https:\/\/mcq-questions.com\/#\/schema\/person\/4ba9570f32f2057e70e670c7885e47f3"},"breadcrumb":{"@id":"https:\/\/mcq-questions.com\/coordinate-geometry-class-10-extra-questions\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/mcq-questions.com\/coordinate-geometry-class-10-extra-questions\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/mcq-questions.com\/coordinate-geometry-class-10-extra-questions\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/mcq-questions.com\/"},{"@type":"ListItem","position":2,"name":"Coordinate Geometry Class 10 Extra Questions Maths Chapter 7 with Solutions Answers"}]},{"@type":"Person","@id":"https:\/\/mcq-questions.com\/#\/schema\/person\/4ba9570f32f2057e70e670c7885e47f3","name":"Prasanna","image":{"@type":"ImageObject","@id":"https:\/\/mcq-questions.com\/#personlogo","inLanguage":"en-US","url":"https:\/\/secure.gravatar.com\/avatar\/174540ad43736c7d1a4c4f83c775e74d?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/174540ad43736c7d1a4c4f83c775e74d?s=96&d=mm&r=g","caption":"Prasanna"},"url":"https:\/\/mcq-questions.com\/author\/prasanna\/"}]}},"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/mcq-questions.com\/wp-json\/wp\/v2\/posts\/12861"}],"collection":[{"href":"https:\/\/mcq-questions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/mcq-questions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/mcq-questions.com\/wp-json\/wp\/v2\/users\/9"}],"replies":[{"embeddable":true,"href":"https:\/\/mcq-questions.com\/wp-json\/wp\/v2\/comments?post=12861"}],"version-history":[{"count":1,"href":"https:\/\/mcq-questions.com\/wp-json\/wp\/v2\/posts\/12861\/revisions"}],"predecessor-version":[{"id":35148,"href":"https:\/\/mcq-questions.com\/wp-json\/wp\/v2\/posts\/12861\/revisions\/35148"}],"wp:attachment":[{"href":"https:\/\/mcq-questions.com\/wp-json\/wp\/v2\/media?parent=12861"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mcq-questions.com\/wp-json\/wp\/v2\/categories?post=12861"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mcq-questions.com\/wp-json\/wp\/v2\/tags?post=12861"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}