{"id":119309,"date":"2022-05-04T14:53:55","date_gmt":"2022-05-04T09:23:55","guid":{"rendered":"https:\/\/www.mapsofindia.com\/my-india\/?p=119309"},"modified":"2022-05-04T14:55:28","modified_gmt":"2022-05-04T09:25:28","slug":"chapter-7-coordinate-geometry-questions-and-answers-ncert-solutions-for-class-10-mathematics","status":"publish","type":"post","link":"https:\/\/www.mapsofindia.com\/my-india\/education\/chapter-7-coordinate-geometry-questions-and-answers-ncert-solutions-for-class-10-mathematics","title":{"rendered":"Chapter 7 – Coordinate Geometry Questions and Answers: NCERT Solutions for Class 10 Mathematics"},"content":{"rendered":"

1. Find 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, \u2013 b)<\/h2>\n

Solution:
\nDistance formula to find the distance between two points (x1, y1) and (x2, y2) is, say d,<\/h3>\n

\"\"
\n2. Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2.<\/h2>\n

Solution:
\nLet us consider, town A at point (0, 0). Therefore, town B will be at point (36, 15).
\nDistance between points (0, 0) and (36, 15)<\/h3>\n

\"\"
\nIn section 7.2, A is (4, 0) and B is (6, 0)
\nAB2\u00a0= (6 \u2013 4)2\u00a0\u2013 (0 \u2013 0)2\u00a0= 4
\nThe distance between town A and B will be 39 km. The distance between the two towns A and B discussed in Section 7.2 is 4 km.<\/h3>\n

3. Determine if the points (1, 5), (2, 3) and (-2, -11) are collinear.<\/h2>\n

Solution:\u00a0The sum of the lengths of any two line segments is equal to the length of the third line segment then all three points are collinear.
\nConsider, A = (1, 5) B = (2, 3) and C = (-2, -11)
\nFind the distance between points; say AB, BC and CA<\/h3>\n

\"\"
\nSince AB + BC \u2260 CA
\nTherefore, the points (1, 5), (2, 3), and ( \u2013 2, \u2013 11) are not collinear.<\/h3>\n

4. Check whether (5, \u2013 2), (6, 4) and (7, \u2013 2) are the vertices of an isosceles triangle.<\/h2>\n

Solution:
\nSince two sides of any isosceles triangle are equal. To check whether given points are vertices of an isosceles triangle, we will find the distance between all the points.
\nLet the points (5, \u2013 2), (6, 4), and (7, \u2013 2) are representing the vertices A, B, and C respectively.<\/h3>\n

\"\"
\nThis implies, whether given points are vertices of an isosceles triangle.<\/h3>\n

5. In a classroom, 4 friends are seated at the points A, B, C and D as shown in Fig. 7.8. 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.<\/h2>\n

\"\"
\nSolution:
\nFrom figure, the coordinates of points A, B, C and D are (3, 4), (6, 7), (9, 4) and (6,1).
\nFind distance between points using distance formula, we get<\/h3>\n

\"\"
\nAll sides are of equal length. Therefore, ABCD is a square and hence, Champa was correct.<\/h3>\n

6. Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
\n(i) (- 1, \u2013 2), (1, 0), (- 1, 2), (- 3, 0)
\n(ii) (- 3, 5), (3, 1), (0, 3), (- 1, \u2013 4)
\n(iii) (4, 5), (7, 6), (4, 3), (1, 2)<\/h2>\n

Solution:
\n(i) Let the points (- 1, \u2013 2), (1, 0), ( \u2013 1, 2), and ( \u2013 3, 0) be representing the vertices A, B, C, and D of the given quadrilateral respectively.<\/h3>\n

\"\"
\nSide length = AB = BC = CD = DA = 2\u221a2
\nDiagonal Measure = AC = BD = 4
\nTherefore, the given points are the vertices of a square.<\/h3>\n

(ii) Let the points (- 3, 5), (3, 1), (0, 3), and ( \u2013 1, \u2013 4) be representing the vertices A, B, C, and D of the given quadrilateral respectively.<\/h3>\n

(iii) Let the points (4, 5), (7, 6), (4, 3), and (1, 2) be representing the vertices A, B, C, and D of the given quadrilateral respectively.<\/h3>\n

\"\"
\nOpposite sides of this quadrilateral are of the same length. However, the diagonals are of different lengths. Therefore, the given points are the vertices of a parallelogram.<\/h3>\n

7. Find the point on the x-axis which is equidistant from (2, \u2013 5) and (- 2, 9).<\/h2>\n

Solution:
\nTo find a point on x-axis. Therefore, its y-coordinate will be 0. Let the point on x-axis be (x,0).
\nConsider A = (x, 0); B = (2, \u2013 5) and C = (- 2, 9).<\/h3>\n

\"\"
\nSimplify the above equation,
\nRemove square root by taking square both the sides, we get
\n(2 \u2013 x)2\u00a0+ 25 = [-(2 + x)]2\u00a0+ 81
\n(2 \u2013 x)2\u00a0+ 25 =\u00a0(2 + x)2\u00a0+ 81
\nx2\u00a0+ 4 \u2013 4x + 25 = x2\u00a0+ 4 + 4x + 81
\n8x = 25 \u2013 81 = -56
\nx = -7
\nTherefore, the point is (- 7, 0).<\/h3>\n

\"\"
\nIts also seen that points A, B and C are collinear.
\nSo, the given points can only form 3 sides i.e, a triangle and not a quadrilateral which has 4 sides.
\nTherefore, the given points cannot form a general quadrilateral.<\/h3>\n

8. Find the values of y for which the distance between the points P (2, \u2013 3) and Q (10, y) is 10 units.<\/h2>\n

Solution:
\nGiven: Distance between (2, \u2013 3) and (10, y) is 10.
\nUsing distance formula,<\/h3>\n

\"\"
\nSimplify the above equation and find the value of y.
\nSquaring both sides,
\n64 + (y + 3)2\u00a0= 100
\n(y + 3)2\u00a0= 36
\ny + 3 = \u00b16
\ny + 3 = +6 or y + 3 = \u22126
\ny = 6 \u2013 3 = 3 or y = \u2013 6 \u2013 3 = -9
\nTherefore, y = 3 or -9.<\/h3>\n

9. If Q (0, 1) is equidistant from P (5, \u2013 3) and R (x, 6), find the values of x. Also find the distance QR and PR.<\/h2>\n

Solution:
\nGiven: Q (0, 1) is equidistant from P (5, \u2013 3) and R (x, 6), which means PQ = QR
\nStep 1: Find the distance between PQ and QR using distance formula,<\/h3>\n

\"\"
\nSquaring both the sides, to omit square root
\n41 = x2\u00a0+ 25
\nx2\u00a0= 16
\nx = \u00b1 4
\nx = 4 or x = -4
\nCoordinates of Point R will be R (4, 6) or R (-4, 6),
\nIf R (4, 6), then QR<\/h3>\n

\"\"
\n10. Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (- 3, 4).<\/h2>\n

Solution:
\nPoint (x, y) is equidistant from (3, 6) and ( \u2013 3, 4).<\/h3>\n

\"\"
\nSquaring both sides, (x \u2013 3)2+(y \u2013 6)2\u00a0= (x + 3)2\u00a0+(y \u2013 4)2
\nx2\u00a0+ 9 \u2013 6x + y2+ 36 \u2013 12y = x2\u00a0+ 9 + 6x + y2\u00a0+16 \u2013 8y
\n36 \u2013 16 = 6x + 6x + 12y \u2013 8y
\n20 = 12x + 4y
\n3x + y = 5
\n3x + y \u2013 5 = 0<\/h3>\n

Exercise 7.2<\/h2>\n

1. Find the coordinates of the point which divides the join of (- 1, 7) and (4, \u2013 3) in the ratio 2:3.<\/h2>\n

Solution:
\nLet P(x, y) be the required point. Using the section formula, we get
\nx = (2\u00d74 + 3\u00d7(-1))\/(2 + 3) = (8 \u2013 3)\/5 = 1
\ny = (2\u00d7-3 + 3\u00d77)\/(2 + 3) = (-6 + 21)\/5 = 3
\nTherefore, the point is (1, 3).<\/h3>\n

2. Find the coordinates of the points of trisection of the line segment joining (4, -1) and (-2, -3).<\/h2>\n

Solution:<\/h3>\n

\"\"
\nLet P (x1, y1) and Q (x2, y2) are the points of trisection of the line segment joining the given points i.e., AP = PQ = QB
\nTherefore, point P divides AB internally in the ratio 1:2.
\nx1\u00a0= (1\u00d7(-2) + 2\u00d74)\/3 = (-2 + 8)\/3 = 6\/3 = 2
\ny1\u00a0= (1\u00d7(-3) + 2\u00d7(-1))\/(1 + 2) = (-3 \u2013 2)\/3 = -5\/3
\nTherefore: P (x1, y1) = P(2, -5\/3)
\nPoint Q divides AB internally in the ratio 2:1.
\nx2\u00a0= (2\u00d7(-2) + 1\u00d74)\/(2 + 1) = (-4 + 4)\/3 = 0
\ny2\u00a0= (2\u00d7(-3) + 1\u00d7(-1))\/(2 + 1) = (-6 \u2013 1)\/3 = -7\/3
\nThe coordinates of the point Q is (0, -7\/3)<\/h3>\n

3. To 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 AD, as shown in the following figure. Niharika runs 1\/4 th the distance AD on the 2nd line and posts a green flag. Preet runs 1\/5th 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?<\/h2>\n

\"\"
\nSolution:
\nFrom the given instruction, we observed that Niharika posted the green flag at 1\/4th\u00a0of the distance AD i.e., (1\/4 \u00d7100) m = 25 m from the starting point of 2nd line. Therefore, the coordinates of this point are (2, 25).
\nSimilarly, Preet posted red flag at 1\/5 of the distance AD i.e., (1\/5 \u00d7100) m = 20m from the starting point of 8th line. Therefore, the coordinates of this point are (8, 20).
\nDistance between these flags can be calculated by using distance formula,<\/h3>\n

\"\"
\nThe point at which Rashmi should post her blue flag is the mid-point of the line joining these points. Let say this point be P(x, y).
\nx = (2 + 8)\/2 = 10\/2 = 5 and y = (20 + 25)\/2 = 45\/2
\nHence, P(\u00a0x,\u00a0y) = (5, 45\/2)
\nTherefore, Rashmi should post her blue flag at 45\/2 = 22.5m on 5th line.<\/h3>\n

4. Find the ratio in which the line segment joining the points (-3, 10) and (6, \u2013 8) is divided by (-1, 6).<\/h2>\n

Solution:
\nConsider the ratio in which the line segment joining ( -3, 10) and (6, -8) is divided by point ( -1, 6) be k :1.
\nTherefore, -1 = ( 6k-3)\/(k+1)
\n\u2013k\u00a0\u2013 1 = 6k\u00a0-3
\n7k\u00a0= 2
\nk\u00a0= 2\/7
\nTherefore, the required ratio is 2: 7.<\/h3>\n

5. Find the ratio in which the line segment joining A (1, \u2013 5) and B (- 4, 5) is divided by the x-axis. Also find the coordinates of the point of division.<\/h2>\n

Solution:
\nLet the ratio in which the line segment joining A (1, \u2013 5) and B ( \u2013 4, 5) is divided by x-axis be k : 1. Therefore, the coordinates of the point of division, say P(x, y) is ((-4k+1)\/(k+1), (5k-5)\/(k+1)).<\/h3>\n

\"\"
\nWe know that y-coordinate of any point on x-axis is 0.
\nTherefore, ( 5k \u2013 5)\/(k + 1) = 0
\n5k = 5
\nor k = 1
\nSo,\u00a0x-axis divides the line segment in the ratio 1:1.
\nNow, find the coordinates of the point of division:
\nP (x, y) = ((-4(1)+1)\/(1+1) , (5(1)-5)\/(1+1)) = (-3\/2 , 0)<\/h3>\n

6. If (1, 2), (4,\u00a0y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find\u00a0x\u00a0and\u00a0y.<\/h2>\n

Solution:
\nLet A,B,C and D be the points of a parallelogram : A(1, 2), B(4,\u00a0y), C(x, 6) and D(3, 5).<\/h3>\n

\"\"
\nSince the diagonals of a parallelogram bisect each other, the midpoint is same.
\nTo find the value of x and y, solve for midpoint first.
\nMidpoint of AC = ( (1+x)\/2 , (2+6)\/2 ) = ((1+x)\/2 , 4)
\nMidpoint of BD = ((4+3)\/2 , (5+y)\/2 ) = (7\/2 , (5+y)\/2)
\nMidpoint of AC and BD are same, this implies
\n(1+x)\/2 = 7\/2 and 4 = (5+y)\/2
\nx\u00a0+ 1 = 7 and 5 +\u00a0y\u00a0= 8
\nx\u00a0= 6 and\u00a0y\u00a0= 3<\/h3>\n

7. Find the coordinates of a point A, where AB is the diameter of circle whose centre is (2, \u2013 3) and B is (1, 4).<\/h2>\n

Solution:
\nLet the coordinates of point A be (x,\u00a0y).
\nMid-point of AB is (2, \u2013 3), which is the centre of the circle.
\nCoordinate of B = (1, 4)
\n(2, -3) =((x+1)\/2 , (y+4)\/2)
\n(x+1)\/2 = 2 and (y+4)\/2 = -3
\nx + 1 = 4 and y + 4 = -6
\nx = 3 and y = -10
\nThe coordinates of A(3,-10).<\/h3>\n

8. If A and B are (-2, -2) and (2, -4), respectively, find the coordinates of P such that AP = 3\/7 AB and P lies on the line segment AB.<\/h2>\n

Solution:<\/h3>\n

\"\"
\nThe coordinates of point A and B are (-2,-2) and (2,-4) respectively. Since AP = 3\/7 AB
\nTherefore, AP: PB = 3:4
\nPoint P divides the line segment AB in the ratio 3:4.<\/h3>\n

\"\"<\/p>\n

9. Find the coordinates of the points which divide the line segment joining A (- 2, 2) and B (2, 8) into four equal parts.<\/h2>\n

Solution:
\nDraw a figure, line dividing by 4 points.<\/h3>\n

\"\"
\nFrom the figure, it can be observed that points X, Y, Z are dividing the line segment in a ratio 1:3, 1:1, 3:1 respectively.<\/h3>\n

\"\"
\n10. Find the area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2,-1) taken in order.
\n[Hint: Area of a rhombus = 1\/2 (product of its diagonals)<\/h2>\n

Solution:
\nLet A(3, 0), B (4, 5), C( \u2013 1, 4) and D ( \u2013 2, \u2013 1) are the vertices of a rhombus ABCD.<\/h3>\n

\"\"<\/p>\n

\"\"
\nExercise 7.3<\/h2>\n

1. Find the area of the triangle whose vertices are:
\n(i) (2, 3), (-1, 0), (2, -4)
\n(ii) (-5, -1), (3, -5), (5, 2)<\/h2>\n

Solution:
\nArea of a triangle formula = 1\/2 \u00d7 [x1(y2\u00a0\u2013 y3) + x2(y3\u00a0\u2013 y1) + x3(y1\u00a0\u2013 y2)]
\n(i) Here,
\nx1\u00a0= 2, x2\u00a0= -1, x3\u00a0= 2, y1\u00a0= 3, y2\u00a0= 0 and y3\u00a0= -4
\nSubstitute all the values in the above formula, we get
\nArea of triangle = 1\/2 [2 {0- (-4)} + (-1) {(-4) \u2013 (3)} + 2 (3 \u2013 0)]
\n= 1\/2 {8 + 7 + 6}
\n= 21\/2
\nSo, area of triangle is 21\/2 square units.<\/h3>\n

(ii) Here,
\nx1\u00a0= -5, x2\u00a0= 3, x3\u00a0= 5, y1\u00a0= -1, y2\u00a0= -5 and y3\u00a0= 2
\nArea of the triangle = 1\/2 [-5 { (-5)- (2)} + 3(2-(-1)) + 5{-1 \u2013 (-5)}]
\n= 1\/2{35 + 9 + 20} = 32
\nTherefore, the area of the triangle is 32 square units.<\/h3>\n

2. In each of the following find the value of \u2018k\u2019, for which the points are collinear.
\n(i) (7, -2), (5, 1), (3, -k)
\n(ii) (8, 1), (k, -4), (2, -5)<\/h2>\n

Solution:
\n(i) For collinear points, area of triangle formed by them is always zero.
\nLet points (7, -2) (5, 1), and (3, k) are vertices of a triangle.
\nArea of triangle = 1\/2 [7 { 1- k} + 5(k-(-2)) + 3{(-2) \u2013 1}] = 0
\n7 \u2013 7k + 5k +10 -9 = 0
\n-2k + 8 = 0
\nk = 4<\/h3>\n

(ii) For collinear points, area of triangle formed by them is zero.
\nTherefore, for points (8, 1), (k, \u2013 4), and (2, \u2013 5), area = 0
\n1\/2 [8 { -4- (-5)} + k{(-5)-(1)} + 2{1 -(-4)}] = 0
\n8 \u2013 6k + 10 = 0
\n6k = 18
\nk = 3<\/h3>\n

3. Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, -1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.<\/h2>\n

Solution:
\nLet the vertices of the triangle be A (0, -1), B (2, 1), C (0, 3).
\nLet D, E, F be the mid-points of the sides of this triangle.
\nCoordinates of D, E, and F are given by
\nD = (0+2\/2, -1+1\/2 ) = (1, 0)
\nE = ( 0+0\/2, -1+3\/2 ) = (0, 1)
\nF = ( 0+2\/2, 3+1\/2 ) = (1, 2)<\/h3>\n

\"\"
\nArea of a triangle =\u00a01\/2 \u00d7 [x1(y2\u00a0\u2013 y3) + x2(y3\u00a0\u2013 y1) + x3(y1\u00a0\u2013 y2)]
\nArea of \u0394DEF = 1\/2 {1(2-1) + 1(1-0) + 0(0-2)} = 1\/2 (1+1) = 1
\nArea of \u0394DEF is 1 square units
\nArea of \u0394ABC = 1\/2 [0(1-3) + 2{3-(-1)} + 0(-1-1)] = 1\/2 {8} = 4
\nArea of \u0394ABC is 4 square units
\nTherefore, the required ratio is 1:4.<\/h3>\n

4. Find the area of the quadrilateral whose vertices, taken in order, are
\n(-4, -2), (-3, -5), (3, -2) and (2, 3).<\/h2>\n

Solution:
\nLet the vertices of the quadrilateral be A (- 4, \u2013 2), B ( \u2013 3, \u2013 5), C (3, \u2013 2), and D (2, 3).
\nJoin AC and divide the quadrilateral into two triangles.<\/h3>\n

\"\"
\nWe have two triangles \u0394ABC and \u0394ACD.
\nArea of a triangle =\u00a01\/2 \u00d7 [x1(y2\u00a0\u2013 y3) + x2(y3\u00a0\u2013 y1) + x3(y1\u00a0\u2013 y2)]
\nArea of \u0394ABC\u00a0= 1\/2 [(-4) {(-5) \u2013 (-2)} + (-3) {(-2) \u2013 (-2)} + 3 {(-2) \u2013 (-5)}]
\n= 1\/2 (12 + 0 + 9)
\n= 21\/2 square units
\nArea of \u0394ACD\u00a0= 1\/2 [(-4) {(-2) \u2013 (3)} + 3{(3) \u2013 (-2)} + 2 {(-2) \u2013 (-2)}]
\n= 1\/2 (20 + 15 + 0)
\n= 35\/2 square units
\nArea of quadrilateral ABCD = Area of \u0394ABC + Area of \u0394ACD
\n= (21\/2 + 35\/2) square units = 28 square units<\/h3>\n

5. You have studied in Class IX that a median of a triangle divides it into two triangles of equal areas. Verify this result for \u0394ABC whose vertices are A (4, \u2013 6), B (3, \u2013 2) and C (5, 2).<\/h2>\n

Solution:
\nLet the vertices of the triangle be A (4, -6), B (3, -2), and C (5, 2).<\/h3>\n

\"\"<\/p>\n

Let D be the mid-point of side BC of \u0394ABC. Therefore, AD is the median in \u0394ABC.
\nCoordinates of point D = Midpoint of BC = ((3+5)\/2, (-2+2)\/2) = (4, 0)
\nFormula, to find Area of a triangle =\u00a01\/2 \u00d7 [x1(y2\u00a0\u2013 y3) + x2(y3\u00a0\u2013 y1) + x3(y1\u00a0\u2013 y2)]
\nNow,\u00a0Area of \u0394ABD\u00a0= 1\/2 [(4) {(-2) \u2013 (0)} + 3{(0) \u2013 (-6)} + (4) {(-6) \u2013 (-2)}]
\n= 1\/2 (-8 + 18 \u2013 16)
\n= -3 square units
\nHowever, area cannot be negative. Therefore, area of \u0394ABD is 3 square units.
\nArea of \u0394ACD\u00a0= 1\/2 [(4) {0 \u2013 (2)} + 4{(2) \u2013 (-6)} + (5) {(-6) \u2013 (0)}]
\n= 1\/2 (-8 + 32 \u2013 30) = -3 square units
\nHowever, area cannot be negative. Therefore, the area of \u0394ACD is 3 square units.
\nThe area of both sides is the same. Thus, median AD has divided \u0394ABC in two triangles of equal areas.<\/h3>\n

Exercise 7.4<\/h2>\n

1. Determine the ratio in which the line 2x + y \u2013 4 = 0 divides the line segment joining the points A(2, \u20132) and B(3, 7).<\/h2>\n

Solution:
\nConsider line 2x + y \u2013 4 = 0 divides line AB joined by the two points A(2, -2) and B(3, 7) in k : 1 ratio.
\nCoordinates of point of division can be given as follows:
\nx = (2 + 3k)\/(k + 1) and y = (-2 + 7k)\/(k + 1)
\nSubstituting the values of x and y given equation, i.e. 2x + y \u2013 4 = 0, we have
\n2{(2 + 3k)\/(k + 1)} + {(-2 + 7k)\/(k + 1)} \u2013 4 = 0
\n(4 + 6k)\/(k + 1) + (-2 + 7k)\/(k + 1) = 4
\n4 + 6k \u2013 2 + 7k = 4(k+1)
\n-2 + 9k = 0
\nOr k = 2\/9
\nHence, the ratio is 2: 9.<\/h3>\n

2. Find the relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.<\/h2>\n

Solution:
\nIf given points are collinear then area of triangle formed by them must be zero.
\nLet (x, y), (1, 2) and (7, 0) are vertices of a triangle,
\nArea of a triangle =\u00a01\/2 \u00d7 [x1(y2\u00a0\u2013 y3) + x2(y3\u00a0\u2013 y1) + x3(y1\u00a0\u2013 y2)] = 0
\n[x(2 \u2013 0) + 1 (0 \u2013 y) + 7( y \u2013 2)] = 0
\n2x \u2013 y + 7y \u2013 14 = 0
\n2x + 6y \u2013 14 = 0
\nx + 3y \u2013 7 = 0.
\nWhich is the required result.<\/h3>\n

3. Find the centre of a circle passing through points (6, -6), (3, -7) and (3, 3).<\/h2>\n

Solution:
\nLet A = (6, -6), B = (3, -7), C = (3, 3) are the points on a circle.
\nIf O is the centre, then OA = OB = OC (radii are equal)
\nIf O = (x, y) then
\nOA = \u221a[(x \u2013 6)2\u00a0+ (y + 6)2]
\nOB =\u00a0\u221a[(x \u2013 3)2\u00a0+ (y + 7)2]
\nOC =\u00a0\u221a[(x \u2013 3)2\u00a0+ (y \u2013 3)2]
\nChoose: OA = OB, we have
\nAfter simplifying above, we get -6x = 2y \u2013 14 \u2026.(1)
\nSimilarly: OB = OC
\n(x \u2013 3)2\u00a0+ (y + 7)2\u00a0= (x \u2013 3)2\u00a0+ (y \u2013 3)2
\n(y + 7)2\u00a0= (y \u2013 3)2
\ny2\u00a0+ 14y + 49 = y2\u00a0\u2013 6y + 9
\n20y =-40
\nor y = -2
\nSubstituting the value of y in equation (1), we get;
\n-6x = 2y \u2013 14
\n-6x = -4 \u2013 14 = -18
\nx = 3
\nHence, the centre of the circle located at point (3,-2).<\/h3>\n

4. The two opposite vertices of a square are (-1, 2) and (3, 2). Find the coordinates of the other two vertices.<\/h2>\n

Solution:
\nLet ABCD is a square, where A(-1,2) and B(3,2). And Point O is the point of intersection of AC and BD
\nTo Find: Coordinate of points B and D.<\/h3>\n

\"\"
\nStep 1: Find distance between A and C and coordinates of point O.
\nWe know that, diagonals of a square are equal and bisect each other.
\nAC =\u00a0\u221a[(3 + 1)2\u00a0+ (2 \u2013 2)2] = 4
\nCoordinates of O can be calculated as follows:
\nx = (3 \u2013 1)\/2 = 1 and y = (2 + 2)\/2 = 2
\nSo, O(1,2)<\/h3>\n

Step 2: Find the side of the square using Pythagoras theorem
\nLet a be the side of square and AC = 4
\nFrom right triangle, ACD,
\na = 2\u221a2
\nHence, each side of square = 2\u221a2<\/h3>\n

Step 3: Find coordinates of point D
\nEquate length measure of AD and CD
\nSay, if coordinate of D are (x1, y1)
\nAD =\u00a0\u221a[(x1\u00a0+ 1)2\u00a0+ (y1\u00a0\u2013 2)2]
\nSquaring both sides,
\nAD2\u00a0= (x1\u00a0+ 1)2\u00a0+ (y1\u00a0\u2013 2)2
\nSimilarly, CD2\u00a0=\u00a0(x1\u00a0\u2013 3)2\u00a0+ (y1\u00a0\u2013 2)2
\nSince all sides of a square are equal, which means AD = CD
\n(x1\u00a0+ 1)2\u00a0+ (y1\u00a0\u2013 2)2\u00a0=\u00a0(x1\u00a0\u2013 3)2\u00a0+ (y1\u00a0\u2013 2)2
\nx12\u00a0+ 1 + 2×1\u00a0= x12\u00a0+ 9 \u2013 6×1
\n8×1\u00a0= 8
\nx1\u00a0= 1
\nValue of y1\u00a0can be calculated as follows by using the value of x.
\nFrom step 2: each side of square = 2\u221a2
\nCD2\u00a0=\u00a0(x1\u00a0\u2013 3)2\u00a0+ (y1\u00a0\u2013 2)2
\n8 =\u00a0(1\u00a0\u2013 3)2\u00a0+ (y1\u00a0\u2013 2)2
\n8 = 4 + (y1\u00a0\u2013 2)2
\ny1\u00a0\u2013 2 = 2
\ny1\u00a0= 4
\nHence, D = (1, 4)<\/h3>\n

Step 4: Find coordinates of point B
\nFrom line segment, BOD
\nCoordinates of B can be calculated using coordinates of O; as follows:
\nEarlier, we had calculated O = (1, 2)
\nSay B = (x2, y2)
\nFor BD;
\n1 = (x2\u00a0+ 1)\/2
\nx2\u00a0= 1
\nAnd 2 = (y2\u00a0+ 4)\/2
\n=> y2\u00a0= 0
\nTherefore, the coordinates of required points are B = (1,0) and D = (1,4)<\/h3>\n

5. The class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Saplings of Gulmohar are planted on the boundary at a distance of 1 m from each other. There is a triangular lawn in the plot as shown in the fig. 7.14. The students are to sow the seeds of flowering plants on the remaining area of the plot.
\n(i) Taking A as origin, find the coordinates of the vertices of the triangle.
\n(ii) What will be the coordinates of the vertices of triangle PQR if C is the origin?
\nAlso calculate the areas of the triangles in these cases. What do you observe?<\/h2>\n

\"\"
\nSolution:
\n(i) Taking A as origin, coordinates of the vertices P, Q and R are,
\nFrom figure: P = (4, 6), Q = (3, 2), R (6, 5)
\nHere AD is the x-axis and AB is the y-axis.<\/h3>\n

(ii) Taking C as origin,
\nCoordinates of vertices P, Q and R are ( 12, 2), (13, 6) and (10, 3) respectively.
\nHere CB is the x-axis and CD is the y-axis.
\nFind the area of triangles:
\nArea of triangle PQR in case of origin A:
\nUsing formula: Area of a triangle =\u00a01\/2 \u00d7 [x1(y2\u00a0\u2013 y3) + x2(y3\u00a0\u2013 y1) + x3(y1\u00a0\u2013 y2)]
\n= \u00bd [4(2 \u2013 5) + 3 (5 \u2013 6) + 6 (6 \u2013 2)]
\n= \u00bd (- 12 \u2013 3 + 24 )
\n= 9\/2 sq unit<\/h3>\n

(ii) Area of triangle PQR in case of origin C:
\nArea of a triangle =\u00a01\/2 \u00d7 [x1(y2\u00a0\u2013 y3) + x2(y3\u00a0\u2013 y1) + x3(y1\u00a0\u2013 y2)]
\n= \u00bd [ 12(6 \u2013 3) + 13 ( 3 \u2013 2) + 10( 2 \u2013 6)]
\n= \u00bd ( 36 + 13 \u2013 40)
\n= 9\/2 sq unit
\nThis implies, Area of triangle PQR at origin A = Area of triangle PQR at origin C
\nArea is same in both case because triangle remains the same no matter which point is considered as origin.<\/h3>\n

6. The vertices of a \u2206 ABC are A (4, 6), B (1, 5) and C (7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that AD\/AB = AE\/AC = 1\/4. Calculate the area of the \u2206 ADE and compare it with area of \u2206 ABC. (Recall Theorem 6.2 and Theorem 6.6)<\/h2>\n

Solution:
\nGiven: The vertices of a \u2206 ABC are A (4, 6), B (1, 5) and C (7, 2)<\/h3>\n

\"\"
\nAD\/AB = AE\/AC = 1\/4
\nAD\/(AD + BD) = AE\/(AE + EC) = 1\/4
\nPoint D and Point E divide AB and AC respectively in ratio 1 : 3.
\nCoordinates of D can be calculated as follows:
\nx = (m1x2\u00a0+ m2x1)\/(m1\u00a0+ m2) and y =\u00a0(m1y2\u00a0+ m2y1)\/(m1\u00a0+ m2)
\nHere m1\u00a0= 1 and m2\u00a0= 3
\nConsider line segment AB which is divided by the point D at the ratio 1:3.
\nx = [3(4) + 1(1)]\/4 = 13\/4
\ny = [3(6) + 1(5)]\/4 = 23\/4
\nSimilarly, Coordinates of E can be calculated as follows:
\nx = [1(7) + 3(4)]\/4 = 19\/4
\ny = [1(2) + 3(6)]\/4 = 20\/4 = 5
\nFind Area of triangle:
\nUsing formula: Area of a triangle =\u00a01\/2 \u00d7 [x1(y2\u00a0\u2013 y3) + x2(y3\u00a0\u2013 y1) + x3(y1\u00a0\u2013 y2)]
\nArea of triangle \u2206 ABC can be calculated as follows:
\n= \u00bd [4(5 \u2013 2) + 1( 2 \u2013 6) + 7( 6 \u2013 5)]
\n= \u00bd (12 \u2013 4 + 7) = 15\/2 sq unit
\nArea of \u2206 ADE can be calculated as follows:
\n= \u00bd [4(23\/4 \u2013 5) + 13\/4 (5 \u2013 6) + 19\/4 (6 \u2013 23\/4)]
\n= \u00bd (3 \u2013 13\/4 + 19\/16)
\n= \u00bd ( 15\/16 ) = 15\/32 sq unit
\nHence, ratio of area of triangle ADE to area of triangle ABC = 1 : 16.<\/h3>\n

7. Let A (4, 2), B (6, 5) and C (1, 4) be the vertices of \u2206 ABC.
\n(i) The median from A meets BC at D. Find the coordinates of point D.
\n(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1.
\n(iii) Find the coordinates of point Q and R on medians BE and CF respectively such that BQ : QE = 2:1 and CR : RF = 2 : 1.
\n(iv) What do you observe?
\n[Note: The point which is common to all the three medians is called the centroid
\nand this point divides each median in the ratio 2 : 1.]
\n(v) If A (x1, y1), B (x2, y2) and C (x3, y3) are the vertices of triangle ABC, find the coordinates of the centroid of the triangle.<\/h2>\n

Solution:<\/h3>\n

\"\"
\n(i) Coordinates of D can be calculated as follows:
\nCoordinates of D = ( (6+1)\/2, (5+4)\/2 ) = (7\/2, 9\/2)
\nSo, D is (7\/2, 9\/2)
\n(ii) Coordinates of P can be calculated as follows:
\nCoordinates of P = ( [2(7\/2) + 1(4)]\/(2 + 1), [2(9\/2) + 1(2)]\/(2 + 1) ) = (11\/3, 11\/3)
\nSo, P is (11\/3, 11\/3)
\n(iii) Coordinates of E can be calculated as follows:
\nCoordinates of E = ( (4+1)\/2, (2+4)\/2 ) = (5\/2, 6\/2) = (5\/2 , 3)
\nSo, E is (5\/2 , 3)
\nPoint Q and P would be coincident because medians of a triangle intersect each other at a common point called centroid. Coordinate of Q can be given as follows:
\nCoordinates of Q\u00a0=( [2(5\/2) + 1(6)]\/(2 + 1), [2(3) + 1(5)]\/(2 + 1) ) = (11\/3, 11\/3)
\nF is the mid- point of the side AB
\nCoordinates of F = ( (4+6)\/2, (2+5)\/2 ) = (5, 7\/2)
\nPoint R divides the side CF in ratio 2:1
\nCoordinates of R\u00a0= ( [2(5) + 1(1)]\/(2 + 1), [2(7\/2) + 1(4)]\/(2 + 1) ) = (11\/3, 11\/3)
\n(iv) Coordinates of P, Q and R are same which shows that medians intersect each other at a common point, i.e. centroid of the triangle.
\n(v) If A (x1, y1), B (x2, y2) and C (x3, y3) are the vertices of triangle ABC, the coordinates of centroid can be given as follows:
\nx = (x1\u00a0+ x2\u00a0+ x3)\/3 and y = (y1\u00a0+ y2\u00a0+ y3)\/3<\/h3>\n

8. ABCD is a rectangle formed by the points A (-1, \u2013 1), B (-1, 4), C (5, 4) and D (5, -1). P, Q, R and S are the midpoints of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.<\/h2>\n

Solution:<\/h3>\n

\"\"
\nP id the mid-point of side AB,
\nCoordinate of P = ( (-1 \u2013 1)\/2, (-1 + 4)\/2 ) = (-1, 3\/2)
\nSimilarly, Q, R and S are (As Q is mid-point of BC, R is midpoint of CD and S is midpoint of AD)
\nCoordinate of Q = (2, 4)
\nCoordinate of R = (5, 3\/2)
\nCoordinate of S = (2, -1)
\nNow,
\nLength of PQ = \u221a[(-1 \u2013 2)2\u00a0+ (3\/2 \u2013 4)2]\u00a0= \u221a(61\/4) =\u00a0\u221a61\/2
\nLength of SP =\u00a0\u221a[(2 + 1)2\u00a0+ (-1 \u2013 3\/2)2]\u00a0= \u221a(61\/4) =\u00a0\u221a61\/2
\nLength of QR =\u00a0\u221a[(2 \u2013 5)2\u00a0+ (4 \u2013 3\/2)2]\u00a0= \u221a(61\/4) =\u00a0\u221a61\/2
\nLength of RS =\u00a0\u221a[(5 \u2013 2)2\u00a0+ (3\/2 + 1)2]\u00a0= \u221a(61\/4) =\u00a0\u221a61\/2
\nLength of PR (diagonal) =\u00a0\u221a[(-1 \u2013 5)2\u00a0+ (3\/2 \u2013 3\/2)2]\u00a0= 6
\nLength of QS (diagonal) = \u221a[(2 \u2013 2)2\u00a0+ (4 + 1)2] = 5
\nThe above values show that, PQ = SP = QR = RS = \u221a61\/2, i.e. all sides are equal.
\nBut PR \u2260 QS i.e. diagonals are not of equal measure.
\nHence, the given figure is a rhombus.<\/h3>\n","protected":false},"excerpt":{"rendered":"

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