{"id":47086,"date":"2022-05-31T22:00:35","date_gmt":"2022-05-31T16:30:35","guid":{"rendered":"https:\/\/cbselibrary.com\/?p=47086"},"modified":"2023-01-25T11:18:38","modified_gmt":"2023-01-25T05:48:38","slug":"ml-aggarwal-class-9-solutions-for-icse-maths-chapter-19-chapter-test","status":"publish","type":"post","link":"https:\/\/cbselibrary.com\/ml-aggarwal-class-9-solutions-for-icse-maths-chapter-19-chapter-test\/","title":{"rendered":"ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 19 Coordinate Geometry Chapter Test"},"content":{"rendered":"

ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 19 Coordinate Geometry Chapter Test<\/h2>\n

Question 1.
\nThree vertices of a rectangle are A (2, -1), B (2, 7) and C(4, 7). Plot these points on a graph and hence use it to find the co-ordinates of the fourth vertex D Also find the co-ordinates of
\n(i) the mid-point of BC
\n(ii) the mid point of CD
\n(iii) the point of intersection of the diagonals. What is the area of the rectangle ?
\nAnswer:
\nGiven three vertices of a rectangle are A (2, -1), B (2, 7) and C (4, 7)
\nFrom graph the co-ordinates of the fourth vertex D (4, -1)
\n(i) mid-point of BC is (3, 7)
\n(ii) mid-point of CD is (4, 3)
\n(iii) The point of intersection of the diagonals (3, 3). Area of rectangle ABCD = AB \u00d7 BC = 8 \u00d7 2 sq. units = 16 sq. units.
\n\"ML<\/p>\n

Question 2.
\nThree vertices of a parallelogram are A (3, 5), B (3, -1) and C (-1, -3). Plot these points on a graph paper and hence use it to find the coordinates of the fourth vertex D. Also find the coordinates of the mid-point of the side CD. Waht is the area of the parallelogram?
\nAnswer:
\nThe vertices A, B and C of parallelogram are A (3, 5), B (3, -1) and C (-1,-3)
\nD is the fourth vertex of the parallelogram which is (-1, 3)
\nE is the mid-piont of CD whose coordinates are (-1, 0)
\nNow area of the parallelogram ABCD
\n= Base \u00d7 Height = AB \u00d7 EF = 6 \u00d7 4 = 24 sq. units
\n\"ML<\/p>\n

Question 3.
\nDraw the graphs of the following linear equations.
\n(i) y = 2x – 1
\n(ii) 2x + 3y = 6
\n(iii) 2x – 3y – 4.
\nAlso find slope and j-intercept of these lines.
\nAnswer:
\n(i) y = 2x – 1
\n\"ML
\nm = 2 and c = -1
\n(ii) 2x + 3y = 6 or 3y = 6 – 2x
\n\"ML
\n\"ML
\n\"ML
\n\"ML<\/p>\n

Question 4.
\nDraw the graph of the equation 3x – 4y = 12. From the graph, find :
\n(i) the value of y when x = -4
\n(ii) the value of x when y = 3.
\nAnswer:
\nGiven equation is 3x – 4y = 12
\n\"ML
\n\"ML
\n(i) when x = -4 then value of y = -6
\n(ii) when y = 3 then value of x = 8.<\/p>\n

Question 5.
\nSolve graphically, the simultaneous equations: 2x – 3y = 1; x + 6y = 11.
\nSolution:
\n2x – 3y = 7, x + 6y = 11
\n2x – 3y = 7 \u21d2 2x = 3y + 7
\n\u21d2 x = \\(\\frac{3 y+7}{2}\\)
\nGiving some different value to y, we get corresponding value of x.
\n\"ML
\nPlot the points (5, 1), (2, -1) and (-1, -3) on the graph and join them to get a line. Similarly in
\nx + 6y = 11 \u21d2 x = 11 – 6y
\n\"ML
\nNow plot the points (5, 1), (-1, 2) and (11, 0) and join them to get another line.
\nWe see that there two lines intersect at (5, 1)
\nHence x = 5, y = 1
\n\"ML<\/p>\n

Question 6.
\nSolve the following system of equations graphically: x – 2y – 4 = 0, 2x + y – 3 = 0.
\nSolution:
\nx – 2y – 4 = 0 and 2x + y – 3 = 0
\nx – 2y – 4 = 0 \u21d2 x = 2y + 4
\nGiving some different value toy, we get corresponding values of x
\n\"ML
\nPlot the points (4, 0), (2, -1) and (0, -2) on the graph and join them to get a line.
\nSimilarly in 2x + y – 3 = 0 \u21d2 y = 3 – 2x
\n\"ML
\nPlot the points (0, 3), (1, 1) and (2, -1) and join them to get another line.
\nWe see that these lines intersect each other at x = 2, y = -1
\n\"ML<\/p>\n

Question 7.
\nUsing a scale of 1 cm to 1 unit for both the axes, draw the graphs of the following equations : 6p = 5x + 10, y = 5x – 15. From the graph, find
\n(i) the coordinates of the point where the two lines intersect.
\n(ii) the area of the triangle between the lines and the x-axis.
\nSolution:
\n6y = 5x + 10, y = 5x – 15
\n6y = 5x + 10 \u21d2 y = \\(\\frac{5 x+10}{6}\\)
\nGiving some different values to x, we get corresponding values of y
\n\"ML
\nPlot the points (1, 2, 5), (-2, 0) and (4, 5) on the graph and join them to get a line.
\nSimilarly in y = 5x – 15
\n\"ML
\nPlot the points (2, -5), (3, 0) and (4, 5) on the graph and join them to get a line.
\nWe see that three two lines intersect each other at
\n\"ML<\/p>\n

Question 8.
\nFind, graphically, the coordinates of the vertices of the triangle formed by the lines : 8y – 3x + 7 = 0, 2x – y + 4 = 0 and 5x + 4 y = 29.
\nSolution:
\n8y – 3x + 7 = 0 \u21d2 8y = 3x – 7
\n\u21d2 y = \\(\\frac{3 x-7}{8}\\)
\nGiving some different values to x, we get corresponding values of y
\n\"ML
\nthe graph and join them to get a line
\n2x – xy + 4 = 0 \u21d2 2x = y – 4
\n\u21d2 x = \\(\\frac{y-4}{2}\\)
\nGiving some different values to y, we get corresponding value of x.
\n\"ML
\nPlot the point on the graph and join them to get a line
\nand 5x + 4y = 29 \u21d2 5x = 29 – 4y
\n\u21d2 x = \\(\\frac{29-4y}{5}\\)
\n\"ML
\nPlot the points on the graph and join them to get another line.
\nWe see that there three lines intersect each other at (-3, -2), (1, 5) and (1, 6) respectively Therefore vertices of (-3, -2, (1, 5), (1, 6).
\n\"ML<\/p>\n

Question 9.
\nFind graphically the coordinates of the vertices of the triangle formed by the lines y – 2 = 0, 2y + x = 0 and y + 1 = 3(x – 2). Hence, find the area of the triangle formed by these lines.
\nSolution:
\ny – 2 = 0
\ny = 2, which is parallel to x-axis
\n\"ML
\nPlot the points (0, 0), (-2, 1) and (-4, 2) on the graph and join them to get a line.
\ny + 1 = 3(x – 2) = 3x – 6
\nGiving some different values to x, we get corresponding the value of y
\n\"ML
\nPlot the points (1, -4), (2, -1) and (3, 2) on the graph and join them to get another line. Now we see that three lines intersect each other.
\nCoordinates of the vertices of the triangle are (2, -1), (3, 2), (-4, 2) and
\n\u2234 Area of triangle = \\(\\frac{\\mathrm{BC} \\times \\mathrm{AB}}{2}\\)
\n= \\(\\frac{7 \\times 3}{2}=\\frac{21}{2}\\) = 10.5 cm2<\/sup><\/p>\n

Question 10.
\nA line segment is of length 10 units and one of its end is (-2, 3). If the ordinate of the other end is 9, find the abscissa of the other end.
\nSolution:
\nOrdinates of the point on the other end (y) = 9
\nLet abscissa (x) = x
\nThen distance between the two ends (-2, 3)
\n\"ML
\n\u21d2 x2<\/sup> + 4x + 4 + 36 = 100
\n\u21d2 x2<\/sup> + 4x = 100 – 36 – 4 = 60
\n\u21d2 x2<\/sup> + 4x – 60 = 0
\n\u21d2 x2<\/sup> + 10x – 6x – 60 = 0
\n\u21d2 x(x + 10) – 6(x + 10) = 0
\n\u21d2 (x+ 10) (x – 6) = 0
\nEither x + 10 = 0, then x = -10
\nor x – 6 = 0, then x = 6
\n\u2234 Abscissa will be -10 or 6<\/p>\n

Question 11.
\nA (-4, -1), B (-1, 2) and C (\u03b1, 5) are the vertices of an isosceles triangle. Find the value of \u03b1, given that AB is the unequal side.
\nSolution:
\nA (-4, -1), B (-1, 2) and C (a, 5) are vertices of an isosceles triangle. AB is the unequal side.
\n\u2234 AC = BC
\n\"ML
\n\"ML
\nSquaring both sides,
\n(\u03b1 + 4)2<\/sup> + 36 = (\u03b1 + 1)2<\/sup> + 9
\na2<\/sup> + 8\u03b1 + 16 + 36 = \u03b12<\/sup> + 2\u03b1 + 1 + 9
\n8\u03b1 – 2\u03b1 = 1 + 9 – 16 – 36
\n6\u03b1 = -42 \u21d2 \u03b1 = \\(\\frac{-42}{6}\\) = -7
\n\u2234 \u03b1 = -7<\/p>\n

Question 12.
\nIf A (-3, 2), B (\u03b1, \u03b2) and C (-1, 4) are the vertices of an isosceles triangle, prove that \u03b1 + \u03b2 = 1, given AB = BC.
\nSolution:
\nA (-3, 2), B (\u03b1, \u03b2) and C (-1, 4) are the value of an isosceles triangle AB = BC
\n\"ML
\nSquaring both sides,
\n(\u03b1 + 3)2<\/sup> + (\u03b2 – 2)2<\/sup> = (\u03b1 + 1) + (\u03b2 – 4)2<\/sup>
\n\u21d2 \u03b12<\/sup> + 6\u03b1 + 9 + \u03b22<\/sup> – 4\u03b2 + 4 = \u03b12<\/sup> + 2\u03b1 + 1 + \u03b22<\/sup> – 8\u03b2 + 16
\n6\u03b1 – 2\u03b1 – 4\u03b2 + 8\u03b2 = 16 – 9 – 4 + 1
\n4\u03b1 + 4\u03b2 = 4 \u21d2 \u03b1 + \u03b2 = 1 (dividing by 4)
\nHence \u03b1 + \u03b2 = 1<\/p>\n

Question 13.
\nProve that the points (3, 0), (6, 4) and (-1, 3) are the vertices of a right angled isosceles triangle.
\nSolution:
\nLet points A (3, 0), B (6, 4) and (-1, 3) are the vertices of a right angled.
\n\"ML
\n\"ML<\/p>\n

Question 14.
\n(i) Show that the points (2, 1), (0, 3), (-2, 1) and (0, -1), taken in order, are the vertices of a square. Also find the area of the square.
\n(ii) Show that the points (-3, 2), (-5, -5), (2, -3) and (4, 4), taken in order, are the vertices of rhombus. Also find its area. Do the given points form a square?
\nSolution:
\n(i) Let points A (2, 1), B (0, 3), C (-2, 1) and D (0, -1) taking in order, are the vertices of the square
\n\"ML
\n\"ML
\n(ii) Let the given points are A (-3, 2), B (-5, -5), C(2, -3) and D (4, 4)
\n\"ML
\n\"ML
\n\"ML<\/p>\n

Question 15.
\nThe ends of a diagonal of a square have co-ordinates (-2, p) and (p, 2). Find p if the area of the square is 40 sq. units.
\nSolution:
\nEnds of a diagonal of a square are (-2, p) and (p, 2)
\nArea of square = 40 sq. units
\n\u2234 Side = \\( \\sqrt{{40}} \\) units = 2\\( \\sqrt{{10}} \\) units
\nand diagonal = \\( \\sqrt{{2}} \\) \u00d7 side
\n= \\( \\sqrt{{2}} \\) \u00d7 \\( \\sqrt{{40}} \\) = \\( \\sqrt{{80}} \\) = 4\\( \\sqrt{{5}} \\) unit
\nDiagonal = AC = \\(\\sqrt{\\left(x_{2}-x_{1}\\right)^{2} \\times\\left(y_{2}-y_{1}\\right)^{2}}\\)
\n= \\(\\sqrt{(p+2)^{2}+(2-p)^{2}}=4 \\sqrt{5}\\)
\nSquaring both side,
\n(p + 2)2<\/sup> + (2 – p)2<\/sup> = 16 \u00d7 5 = 80
\n\u21d2 p2<\/sup>2 + 4p + 4 + 4 – 4p + p2<\/sup> = 80
\n\u21d2 2p2<\/sup> + 8 = 80 \u21d2 2p2<\/sup> = 80 – 8 = 72
\n\u21d2 p2<\/sup> = \\(\\frac{72}{2}\\) = 36 = (\u00b16)2<\/sup>
\n\u2234 p = \u00b16
\n\u2234 P = 6, -6<\/p>\n

Question 16.
\nWhat type of quadrilateral do the points A (2, -2), B (7, 3), C (11, -1) and D (6, -6), taken in the order, form?
\nSolution:
\nVertices of a quadrilateral ABCD are A (2, -2), B (7, 3), C (11, -1), D (6, -6) taken on order.
\n\"ML
\n\"ML<\/p>\n

Question 17.
\nFind the coordinates of the centre of the circle passing through the three given points A (5, 1), B (-3, -7) and C (7, -1).
\nSolution:
\nLet coordinates of the centre of the circle be (x, y)
\nPoints A (5, 1), B (-3, -7) and C (7, -1) are on the circle
\n\u2234 OA = OB = OC
\n\"ML
\nOA2<\/sup> = OB2<\/sup> and OA2<\/sup> = OC2<\/sup>
\n\u2234 (x – 5)2<\/sup> + (y – 1)2<\/sup> = (x + 3)2<\/sup> + (y + 7)2<\/sup>
\n\u21d2 x2<\/sup> – 10x + 25 + y2<\/sup> – 2y + 1 = x2<\/sup> + 6x + 9 + y2<\/sup> + 14y + 49
\n\u21d2 6x + 14y+ 10x + 2y = -9 – 49 + 25 + 1
\n\u21d2 16x + 16y = -32
\n\u21d2 x + y = -2
\n\u21d2 x = -2 – y …(i)
\nNow OA2<\/sup> = OC2<\/sup>
\n(x – 5)2<\/sup> + (y – 1)2<\/sup> = (x – 7)2<\/sup> + (y + 1)2<\/sup>
\n\u21d2 x2<\/sup> – 10x + 25 + y2<\/sup> – 2y + 1 = x2<\/sup> – 14x + 49 + y2<\/sup> + 1 + 2y
\n\u21d2 -10x + 14x – 2y – 2y = 49 + 1 – 25 – 1
\n\u21d2 4x – 4y = 24
\n\u21d2 x – y = 6 …(ii)
\n(Taking 4 common)
\nNow substitute the value of (i) in (ii), we get
\n\u21d2 (-2 – y) – y = 6
\n\u21d2 -2 – y – y = 6
\n\u21d2 -2y = 6 + 2 \u21d2 y = \\(\\frac{-8}{2}\\) \u21d2 y = -4
\nNow put the value of y = -4 in equation (i)
\nx = -2 – 7 = -2 – (-4)
\n= -2 + 4 = 2
\n\u2234 The coordinates of the centre of the circle are (2, -4)<\/p>\n

ML Aggarwal Class 9 Solutions for ICSE Maths<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"

ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 19 Coordinate Geometry Chapter Test Question 1. Three vertices of a rectangle are A (2, -1), B (2, 7) and C(4, 7). Plot these points on a graph and hence use it to find the co-ordinates of the fourth vertex D Also find the co-ordinates of … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[3034],"tags":[],"yoast_head":"\nML Aggarwal Class 9 Solutions for ICSE Maths Chapter 19 Coordinate Geometry Chapter Test - CBSE Library<\/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:\/\/cbselibrary.com\/ml-aggarwal-class-9-solutions-for-icse-maths-chapter-19-chapter-test\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 19 Coordinate Geometry Chapter Test\" \/>\n<meta property=\"og:description\" content=\"ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 19 Coordinate Geometry Chapter Test Question 1. 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Three vertices of a rectangle are A (2, -1), B (2, 7) and C(4, 7). Plot these points on a graph and hence use it to find the co-ordinates of the fourth vertex D Also find the co-ordinates of ... 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