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4, , Acids,, Bases and, Coordinate, Salts, Geometry, 7., , Multiple Choice Questions (MCQs), DIRECTIONS : This section contains multiple choice, questions. Each question has 4 choices (a), (b), (c) and (d) out, of which only one is correct., 1., , If P = (2, 5), Q = (x, –7) and PQ = 13, what is the value of, ‘x’?, (a) 5 , (c) –3 , , 2., , (a) ab =, (c) a = b , 3., , (b), =, 1, (d) a = b1, , (b) –12, (d) –6, , (b) –15, (d) –10, , In what ratio is the line segment joining the points (3, 5), & (–4, 2) divided by y–axis?, (a) 3 : 2 , (c) 2 : 3 , , 6., , a1b, , If the mid point of the line joining (3, 4) and (k, 7) is, (x, y) and 2x + 2y + 1 = 0. Find the value of k., (a) 10 , (c) 15 , , 5., , ab1, , a , If , 4 is the midpoint of the line segment joining, 3 , A(–6, 5) and B(–2, 3), then what is the value of ‘a’?, (a) –4 , (c) 12 , , 4., , (b 3, (d) –5, , The points (a, b), (a1, b1) and (a – a1, b – b1) are collinear, if, a1b1 , , (b) 3 : 4, (d) 4 : 3, , In what ratio does the point (–2, 3) divide the line-segment, joining the points (–3, 5) and (4, –9) ?, (a) 2 : 3 , (c) 6 : 1 , , 8., , (b) 1 : 6, (d) 2 : 1, , 9., , If P (x , y) is any point on the line joining the points A (a, 0), and B (0, b), then, (a), , x y, + = 1 , b a, , (b), , x y, − =1, a b, , (c), , x y, + = 1 , a b, , (d), , x y, − =1, b a, , The perimeter of a triangle with vertices (0, 4),, (0, 0) and (3, 0)is, (a) 5 , , (b) 12, , (c) 11 , , (d), , 7+ 5, , The point P on x-axis equidistant from the points A(–1, 0), and B(5, 0) is, (a) (2, 0) , (c) (3, 0) , , (b) (0, 2), (d) (2, 2), , 10. The coordinates of the point which is reflection of point, (–3, 5) in x-axis are, (a) (3, 5) , , (b) (3, –5), , (c) (–3, –5) , , (d) (–3, 5), , 11. If the point P(6, 2) divides the line segment joining, A(6, 5) and B(4, y) in the ratio 3 : 1, then the value, of y is, (a) 4, , (b), , 3, , (c) 2, , (d) 1, , 12. P, Q, R are three collinear points. The coordinates of P, and R are (3, 4) and (11, 10) respectively and PQ is equal, to 2.5 units. Coordinates of Q are, (a) (5, 11/2) , , (b) (11, 5/2), , (c) (5, –11/2) , , (d) (–5, 11/2), , 13. C is the mid-point of PQ, if P is (4, x), C is (y, –1) and Q is, (–2, 4), then x and y respectively are, (a) – 6 and 1 , , (b) – 6 and 2, , (c) 6 and – 1 , , (d) 6 and – 2
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Mathematics, , M-30, 14. The ratio in which the point (2, y) divides the join of, (– 4, 3) and (6, 3) and hence the value of y is, (a) 2 : 3, y = 3 , , (b) 3 : 2, y = 4, , (c) 3 : 2, y = 3 , , (d) 3 : 2, y = 2, , 15. Ratio in which the line 3x + 4y = 7 divides the line segment, joining the points (1, 2) and (–2, 1) is, (a) 3 : 5 , , (b) 4 : 6, , (c) 4 : 9 , , (d) None of these, , 23., , (a) y = 5x + 1 , (c) y =, 24., , 16. The point on the X-axis which is equidistant from the points, A(–2, 3) and B(5, 4) is, (a) (0, 2) , , (b) (2, 0), , (c) (3, 0) , , (d) (–2, 0), , 17. The point which divides the line joining the points A (1, 2), and B(–1, 1) internally in the ratio 1 : 2 is, –1 5 , (a) , , 3 3, , 1 5, (b) , , 3 3, , (c) (–1, 5) , , (d) (1, 5), , 18. The centroid of the triangle whose vertices are (3, –7),, (–8, 6) and (5, 10) is, (a) (0, 9) , , (b) (0, 3), , (c) (1, 3) , , (d) (3, 5), , 19. The points A (– 4, – 1), B (–2, – 4), C (4, 0) and D (2, 3), are the vertices of a, (a) Parallelogram , , (b) Rectangle, , (c) Rhombus , , (d) Square, , 20. If the point P (p, q) is equidistant from the points, A (a + b, b – a) and B (a – b, a + b), then, (a) ap = by , , (b) bp = ay, , (c) ap + bq = 0 , , (d) bp + aq = 0, , 21. The distances of a point from the x-axis and the y-axis, are 5 and 4 respectively. The coordinates of the point, can be, (a) (5, 4) , , (b) (5, 0), , (c) (0, 4) , , (d) (4, 5), , 22. If the points (a, 0), (0, b) and (1, 1) are collinear then which, of the following is true :, , P is a point on the graph of y = 5x + 3. The coordinates of, a point Q are (3, –2). If M is the mid point of PQ, then M, must lie on the line represented by, , 25., , 5, 7, x – , 2, 2, , (b) y = 5x – 7, (d) y =, , 5, 1, x+, 2, 2, , A line l passing through the origin makes an angle q, 3, with positive direction of x-axis such that sin θ = . The, 5, coordinates of the point, which lies in the fourth quadrant, at a unit distance from the origin and on perpendicular to, l, are, 3 4, (a) ,− , 5 5, , 4 3, (b) ,− , 5 5, , (c) (3, –4) , , (d) (4, –3), , The centre of the circle passing through the ponts (6, – 6),, (3, – 7) and (3, 3) is, (a) (3, 2) , , (b) (–3, –2), , (c) (3, – 2) , , (d) (–3, 2), , 26. A circle passes through the vertices of a triangle ABC., If the vertices are A(–2, 5), B(–2, –3), C(2, –3), then the, centre of the circle is, (a) (0, 0) , , (b) (0, 1), , (c) (–2, 1) , , (d) (0, –3), , 27. Which of the following points is 10 units from the origin?, (a) (– 6, 8) , , (b) (– 4, 2), , (c) (– 6, 5) , , (d) (6, 4), , 28. The distance between which of the following two points, is 2 units?, (a) (–2, –3) and (–2, –4), , (b) (0, 4) and (0, 6), , (c) (7, 2) and (6, 2), , (d) (4, –3) and (2, 3), , 29. Which of the following is / are not correct ?, Three points will form :, (a) an equilateral triangle, if all the three sides are equal., (b) an isosceles triangle, if any two sides are equal., (c) a collinear or a line, if sum of two sides is equal to, third side., (d) a rhombus, if all the four sides are equal., 30. Which of the following is / are correct?, Four points will form :, , (a), , 1 1, + = 2 , a b, , (b), , 1 1, − =1, a b, , (a) a rectangle, if opposite sides and diagonals are not equal., , (c), , 1 1, − = 2 , a b, , (d), , 1 1, + =1, a b, , (c) a square, if all the four sides and diagonals are equal., , (b) a parallelogram, if opposite sides are not equal., (d) a right angle triangle, if sum of squares of any two, sides is equal to square of third largest side.
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Coordinate Geometry, , M-31, Assertion & Reason, , DIRECTIONS : Study the given Case/Passage and answer the, following questions., Case/Passage-I, In order to conduct Sports Day activities in your School,, lines have been drawn with chalk powder at a distance, of 1 m each, in a rectangular shaped ground ABCD, 100, flowerpots have been placed at a distance of 1 m from, each other along AD, as shown in given figure below., Niharika runs 1/4 th the distance AD on the 2nd line and, posts a green flag. Preet runs 1/5 th distance AD on the, eighth line and posts a red flag., [From CBSE Question Bank 2021], C, , D, , DIRECTIONS : Each of these questions contains an Assertion, followed by Reason. Read them carefully and answer the, question on the basis of following options. You have to select, the one that best describes the two statements., (a), , If both Assertion and Reason are correct and Reason is, the correct explanation of Assertion., , (b), , If both Assertion and Reason are correct, but Reason is, not the correct explanation of Assertion., , (c), , If Assertion is correct but Reason is incorrect., , (d), , If Assertion is incorrect but Reason is correct., , 36. Assertion : If A(2a, 4a) and B(2a, 6a) are two vertices of, an equilateral triangle ABC then, the vertex C is given by, (2a + a 3,5a) ., Reason : In equilateral triangle, all the coordinates of three, vertices can be rational., , G, R, , 37. Assertion : The points (k, 2 – 2k), (– k+ 1, 2k) and, (– 4 – k, 6 – 2k) are collinear if k =, , 2, 1, A, 1, , 31., , 32., , 33., , 34., , 35., , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 10, , Find the position of green flag, (a) (2, 25) , , (b) (2, 0.25), , (c) (25, 2) , , (d) (0, –25), , Find the position of red flag, (a) (8, 0) , , (b) (20, 8), , (c) (8, 20) , , (d) (8, 0.2), , 1, ., 2, , Reason : Three points A, B and C are collinear in same, straight line, if AB + BC = AC., 38. Assertion : Mid-point of a line segment divides line in the, ratio 1 : 1., Reason : If area of triangle is zero that means points are, collinear., , Match the Following, , What is the distance between both the flags?, (a) √41, , (b) √11, , (c) √61, , (d) √51, , If Rashmi has to post a blue flag exactly halfway between, the line segment joining the two flags, where should she, post her flag?, (a) (5, 22.5) , , (b) (10, 22), , (c) (2, 8.5) , , (d) (2.5, 20), , If Joy has to post a flag at one-fourth distance from green, flag ,in the line segment joining the green and red flags,, then where should he post his flag?, (a) (3.5, 24) , , (b) (0.5, 12.5), , (c) (2.25, 8.5) , , (d) (25, 20), , DIRECTIONS : Each question contains statements given in, two columns which have to be matched. Statements (A, B, C, D), in column-I have to be matched with statements (p, q, r, s) in column-II., 39. Column-II gives distance between pair of points given in, Column-I, Column-I Column-II, (A) (–5, 7), (–1, 3), , (p), , 17, , (B) (5, 6), (1, 3) , , (q), , 8, , (C) ( 3 + 1,1), (0, 3), , (r), , 6, , (D) (0,0) (− 3, 3), , (s), , 4 2
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Mathematics, , M-32, 40. Column-II gives the coordinates of the point P that divides, the line segment joining the points given in Column-I., Column-I Column-II, (A) A (–1, 3) and, , (p) (7, 3), , , B (5, –6) internally, , , (q) (0, 3), , , B (1, 4) internally, , , in the ratio 2 : 1, , (C) A (–1, 7) and, , (r), , (1, 3), , , B (4, –3) internally, , , (s) (1, 0), , , B (8, 5) internally, , , True / False, DIRECTIONS : Read the following statements and write your, answer as true or false., 51. The distance between P (x1, y1) and Q (x2, y2) is, , in the ratio 2 : 3, , (D) A (4, –3) and, , 49. Relation between x and y if the points (x, y), (1, 2) and, (7, 0) are collinear is ..........., 50. The distance of the point (x1, y1) from the origin is ............, , in the ratio 1 : 2, , (B) A (–2, 1) and, , 48. (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a, parallelogram taken in order, then the value of x and y are, ..........., , in the ratio 3 : 1, , Fill in the Blanks, DIRECTIONS : Complete the following statements with an, appropriate word / term to be filled in the blank space(s)., 41. Points (3, 2), (–2, –3) and (2, 3) form a ............. triangle., , ( x2 + x1 )2 + ( y2 + y1 )2, 52. The coordinates of the point P(x, y) which divides the, line segment joining the points A(x1, y1) and B(x2, y2), internally in the ratio m1 : m2 are, m1 x2 − m2 x1 m1 y2 − m2 y1 , m + m , m + m , 1, , 2, , 1, , 2, , 53. The mid-point of the line segment joining the points, x +x y + y , P (x1, y1) and Q (x2, y2) is 1 2 , 1 2 ., 2, 2 , , 42. If x – y = 2, then point (x, y) is equidistant from (7, 1) and, (.........), , 54. Points (1, 7), (4, 2), (–1, –1) and (– 4, 4) are the vertices, of a square., , 43. Distance between (2, 3) and (4, 1) is .............., , 55. Coordinates of the point which divides the join of (–1, 7), and (4, –3) in the ratio 2 : 3 is (1, 3)., , 44. Points (1, 5), (2, 3) and (– 2, – 11) are ..........., 45. (5, – 2) (6, 4) and (7, – 2) are the vertices of an .............., triangle., 46. Point on the X-axis which is equidistant from (2, –5) and, (–2, 9) is ............., 47. Point (– 4, 6) divide the line segment joining the points, A(– 6, 10) and B(3, – 8) in the ratio .............., , 56. Ratio in which the line segment joining the points (– 3, 10), and (6, – 8) is divided by (– 1, 6) is 3 : 7., 57. The ratio in which the point (3, 5) divides the join of, (1, 3) and (4, 6) is 2 : 1., 58. The distance of the point (5, 3) from the X-axis is 5 units, 59. The distance of a point (2, 3) from Y-axis is y-units.
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Coordinate Geometry, , M-33, , ANSWER KEY & SOLUTIONS, 1., , (c) PQ = 13 ⇒ PQ2 = 169, , ∴ Its x-cordinates = 0, , ⇒ (x – 2)2 + (–7 – 5)2 = 169, , ∴, , ⇒ x2 – 4x + 4 + 144 = 169, ⇒ x2 – 4x – 21 = 0, , ⇒K =, , ⇒ x2 – 7x + 3x – 21 = 0, , 6., , –2 =, , ⇒ ab1 – a1b = 0, , ⇒ – 2m1 – 2m 2 = 4m1 – 3m 2, , (b) Coordinates of mid-point are given by, , ⇒ m2 = 6 m1 ⇒ m1 : m2 =1:6, 7., , ⇒ a (b – y) + 0 (y – 0) + x (0 – b) = 0, ⇒ ab – ay – bx = 0 ⇒ bx + ay = ab, ⇒, , ∴ a = – 12, (b) Since (x, y) is midpoint of (3, 4) and (k, 7), ∴, , x=, , 8., , 3+ k, 4+7, and y =, 2, 2, , Also 2x + 2y + 1 = 0 putting values we get, 3+k+4+7+1=0, ⇒ k + 15 = 0 ⇒ k = – 15, , x=, , K(−4) + 3(1), K(2) + 5(1), ; y=, K +1, K +1, , Since, it lies on y-axis, , x y, + =1, a b, , (b) A(0, 4), B(0, 0), C(3, 0), AB =, , (0 − 0)2 + (0 − 4)2 = 4, , BC =, , (3 − 0)2 + (0 − 0)2 = 3, , CA =, , (0 − 3)2 + (4 − 0)2 = 5, , AB + BC + CA = 12, , (b) Let the required ratio be K : 1, ∴ The coordinates of the required point on the y-axis is, , (c) As the point P (x, y) lies on the line joining the points, A (a, 0) and B (0, b), the points A, B and P are collinear, , a , Here, coordinates of mid-point are , 4 , 3 , a −6 − 2, So, =, 3, 2, , 5., , m1 (4) + m 2 ( –3), m1 + m 2, , ⇒ ab1 = a1b., , x1 + x 2 y1 + y 2 , ,, , , 2 , 2, , 4., , (b) Suppose the required ratio is m1 : m2, Then, using the section formula, we get, , ⇒ 2ab1 – ab – a1b1 + ab – ab1 – a1b + a1b1 = 0, , 3., , 3, :1, 4, , ∴ ratio = 3 : 4, , ⇒ x = 7, –3, (b) We have,, a(b1 – b + b1) + a1(b – b1 – b) + (a – a1) (b – a1) (b – b1) = 0, , 3, 4, , ⇒ Required ratio =, , ⇒ (x – 7) (x + 3) = 0, , 2., , −4 K + 3, = 0 ⇒ −4 K + 3 = 0, K +1, , 9., , 5 −1 , (a) P(x, 0) = , , 0 = (2, 0), 2, , , [Q A and B both lies on x-axis], Three or more points lies in same line are called collinear.
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Mathematics, , M-34, 10., , 11., , (c) For reflection of a point with respect to x-axis change, sign of y-coordinate and with respect to y-axis change sign, of x-coordinate., , 4 × 3 + 1× 6 3 × y + 1× 5 , ,, (d) P(6, 2) = , , 3 +1 , 3 +1, 18, Q 6≠, , (Question is wrong), 4, , , , 2=, , 3y + 5, 4, , 19., , (b), , 20., , (b), , 21., , (d), , 4, , P (4, 5), 5, , 22., , ⇒ 3y + 5 = 8, , (d) As (a, 0), (0, b) and (1, 1) are collinear, \ a(b – 1) + 0(1 – 0) + 1(0 – b) = 0, , 3y = 3 ⇒ y = 1, , , ab – a – b = 0, , 12., , (a), , , ab = a + b, , 13., , (a) Since, C (y, – 1) is the mid-point of P (4, x) and Q (–2, 4)., , , 23., , 1=, , 1 1, +, a b, , (b) Let coordinate of point p be (h, 5h + 3), P(h, 5h + 3), , 4−2, 4+ x, We have,, = y and, = −1, 2, 2, ∴ y = 1 and x = – 6, 14., , M, , (c) Let the required ratio be k : 1, , Q (3, –2), , 6k − 4(1), 3, or k =, k +1, 2, 3, ∴ The required ratio is :1 or 3 : 2, 2, 3(3) + 2(3), Also, y =, =3, 3+ 2, Then, 2 =, , 3(1) + 4(2) − 7, 4 4, =, = −, −9 9, 3(−2) + 4(1) − 7, , 15., , (c) −, , 16., , (b) Let P (x, 0) be a point on X-axis such that AP = BP, , y = 5x + 3, , Since, M is the mid-point of PQ, therefore by mid-point, h + 3 5h + 3 − 2 , ,, formula, we have M = , ., 2, 2, Clearly by observing the options, we can say that M must, lie on the line, , y = 5x – 7, A, , ⇒ AP2 = BP2, ⇒ (x + 2)2 + (0 – 3)2 = (x – 5)2 + (0 + 4)2, ⇒ x2 + 4x + 4 + 9 = x2 – 10x + 25 + 16, , 24., , (a), , θ, , 90 – θ, O, , ⇒ 14x = 28 ⇒ x = 2, , 90 – θ, , Hence, required point is (2, 0)., 17., , (b), , 18., , x + x + x y + y + y3 , (b) Centroid is 1 2 3 , 1 2, , 3, 3, , , 3 + (–8) + 5 –7 + 6 + 10 0 9 , i.e. , ,, = 3 , 3 = (0, 3), 3, 3, , , , , B, y, , x, C, , ∠AOB = q, Q CO ^ OA
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Coordinate Geometry, , M-35, , \ ∠BOC = (90° – q), 3, 4 , sin θ =, cos θ =, Q cos θ = 1 −sin 2 θ , 5;, 5 , , , ⇒ x2 + 4 + 4x + (y + 3)2 = x2 + 4 – 4x + (y + 3)2, ⇒ 8x = 0 ⇒ x = 0, , −y, Now, sin(90° − θ) =, 1, ⇒ y = – cos q ⇒ y =, , \ centre of the circle is (0, 1)., −4, 5, , x, 3, cos(90° – q) =, ⇒ x = sin q, x =, 1, 5, 25., , (c) (x – 6)2 + (y + 6)2 = (x – 3)2 + (y + 7)2...(i), , , , ( x + 2) 2 + ( y + 3) 2 = ( x - 2) 2 + ( y + 3) 2, , ⇒, , Also, (x – 3)2 + (y – 3)2 = (x – 3)2 + (y + 7)2, , 27., , (a), , 28., , (b), , 29., , (d) , All the statements given in option ‘a’, ‘b’ and ‘c’ are, correct., , 30., , (c), , 31., , (a) (2, 25), , 1, , , ∵ x = 2, y = 4 × 100 = 25, , , , 32., , (c) (8, 20), , 1, , , ∵ x = 8, y = 5 × 100 = 20 , , , , 33., , (c), , 34., , 8 + 2 25 + 20 , ,, (a) , = (5, 22.5), 2, 2 , , 35., , 2 + 5 25 + 22.5 , ,, (a) , = (3.5, 24), 2, 2, , , 36., , (c) Let A(x1, y1), B(x2, y2) & C(x3, y3) are all rational, , , y2 – 6y + 9 = y2 + 14y + 49, , , – 20y = 40 ⇒ y = – 2, , Putting y = – 2 in equation (i), we have, (x – 6)2 + (4)2 = (x – 3)2 + (5)2, , x2, , – 12x + 36 + 16 =, , x2, , – 6x + 9 + 25, , –6x = – 18 ⇒ x = 3, 26., , (b), , A (–2, 5), , (8 − 2)2 + (25 − 20)2 = 36 + 25 = 61, , coordinates of a triangle ABC., , O, (x, y), B, (–2, 3), , , C, (2, –3), , Join OA, OB & OC., Q OA = OB = OC, \ OA2 = OB2, ⇒, , 2, , 2, , ( x + 2) + ( y - 5) = ( x + 2) + ( y + 3), , 2, , ⇒ x2 + 4 + 4x + y2 + 25 – 10y = x2 + 4 + 4x + y2 + 9 + 6x, ⇒ 16y = 16 ⇒ y = 1, Again: OB2 = OC2, , x1, 1, x2, 2, x3, , y1 1, y2 1, y3 1, , , , 3, [(x1 – x2)2 + (y1 – y2)2 ], 4, LHS = rational, RHS = irrational, , , , Hence, (x1, y1) (x2, y2) & (x3, y3) cannot be all rational., , , =, , Let O(x, y) is the centre of the given circle., , 2, , ar (∆ ABC) =, , 37., , (a) Both assertion and reason are correct. Reason is correct, explanation of assertion., , 38., , (b) Both statements are individually correct., , 39., , (A) → (s); (B) → (p); (C) → (q); (D) → (r), , 40., , (A) → (s); (B) → (q); (C) → (r); (D) → (p), , 41. right angle, 42. (3, 5)
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Mathematics, , M-36, 43., , 2 2 , , 44. Non-collinear, 45. isosceles , 46. (–7, 0), 47. 2 : 7 , 48. (6, 3), 49. x + 3y = 7, 50., , x12 + y12, , 51. False , 52. False, 53. True , 54. True, 55. True , 56. False , 57. True , 58. False, 59. False