2025 DPP : 02 Straight Line Practice Paper PDF

2025 DPP : 02 STRAIGHT LINE DPP – Daily Practice Paper PRACTICE KING NIMCET | CUET | CET MAH| JAMIA | VIT 21500+ QUESTIONS ASPIRE STUDY PH. NO. 8400072444, 7007286637 ADD: 21 LAKHANPUR, KANPUR UP ...

2025 DPP : 02 STRAIGHT LINE DPP – Daily Practice Paper PRACTICE KING NIMCET | CUET | CET MAH| JAMIA | VIT 21500+ QUESTIONS ASPIRE STUDY PH. NO. 8400072444, 7007286637 ADD: 21 LAKHANPUR, KANPUR UP  Aspire Study MCA Entrance Classes Visit www.aspirestudy.in Call Us: 8400072444, 7007286637 51. If the distance of any point (𝑥, 𝑦) from the origin is defined as 𝑑(𝑥, 𝑦) =max {|𝑥|, |𝑦|}, 𝑑(𝑥, 𝑦) = 𝑎, non- zero constant, then the locus is a) A circle b) A straight line c) A square d) A triangle 52. The pair of lines joining origin to the points of intersection of the two curves 𝑎𝑥 2 + 2ℎ𝑥𝑦 + 𝑏𝑦 2 + 2g𝑥 = 0 and 𝑎′ 𝑥 2 + 2ℎ′ 𝑥𝑦 + 𝑏 ′ 𝑦 2 + 2g ′ 𝑥 = 0 will be at right angles, if a) (𝑎′ + 𝑏 ′ )g ′ = (𝑎 + 𝑏)g b) (𝑎 + 𝑏)g ′ = (𝑎′ + 𝑏 ′ )g 2 2 c) ℎ2 − 𝑎𝑏 = ℎ′ − 𝑎′𝑏′ d) 𝑎 + 𝑏 + ℎ2 = 𝑎′ + 𝑏 ′ + ℎ′ 53. Two vertices of a triangle are (5, −1) and (−2,3). If the orthocentre of the triangle is the origin, then coordinates of the third vertex are a) (4,7) b) (−4, −7) c) (−4, 7) d) None of these 54. The distance between the parallel lines 𝑦 = 𝑥 + 𝑎, 𝑦 = 𝑥 + 𝑏 is |𝑏 − 𝑎| |𝑏 + 𝑎| a) b) |𝑎 − 𝑏| c) |𝑎 + 𝑏| d) √2 √2 55. Consider the fourteen lines in the plane given by 𝑦 = 𝑥 + 𝑟, 𝑦 = −𝑥 + 𝑟, where 𝑟 ∈ {0, 1, 2, 3, 4, 5, 6}. The number of squares formed by these lines, whose sides are of length √2, is a) 9 b) 16 c) 25 d) 36 56. The line (𝑝 + 2 𝑞)𝑥 + (𝑝 − 3 𝑞)𝑦 = 𝑝 − 𝑞 for different values of p and q passes through the fixed point a) (3/2, 5/2) b) (2/5,2/5) c) (3/5,3/5) d) (2/5,3/5) 57. A line passing through origin and is perpendicular to two given lines 2𝑥 + 𝑦 + 6 = 0 and 4𝑥 + 2𝑦 − 9 = 0. The ratio in which the origin divides this line, is a) 1: 2 b) 2: 1 c) 4: 2 d) 4: 3 58. A straight line through the point (1, 1) meets the 𝑥-axis at ‘𝐴’ and 𝑦-axis at ‘𝐵’. The locus of the mid point of 𝐴𝐵 is a) 2𝑥𝑦 + 𝑥 + 𝑦 = 0 b) 𝑥 + 𝑦 − 2𝑥𝑦 = 0 c) 𝑥 + 𝑦 + 2 = 0 d) 𝑥 + 𝑦 − 2 = 0 59. The distance between the lines 3𝑥 + 4𝑦 = 9 and 6𝑥 + 8𝑦 = 15 is 3 3 c) 6 d) None of these a) b) 2 10 60. The equation of the pair of straight lines parallel to 𝑥-axis and touching the circle 𝑥 2 + 𝑦 2 − 6𝑥 − 4𝑦 − 12 = 0 is a) 𝑦 2 − 4𝑦 − 21 = 0 b) 𝑦 2 + 4𝑦 − 21 = 0 c) 𝑦 2 − 4𝑦 + 21 = 0 d) 𝑦 2 + 4𝑦 + 21 = 0 61. If the area of the triangle formed by the pair of lines given by 8𝑥 − 6𝑥𝑦 + 𝑦 = 0 and the line 2𝑥 + 2 2 3𝑦 = 𝑎 is 7, then 𝑎 = a) 14 b) 14√2 c) 28 d) None of these 62. The equation of the line which is such that the portion of line segment intercepted between the coordinate axes is bisected at (4, −3), is a) 3𝑥 + 4𝑦 = 24 b) 3𝑥 − 4𝑦 = 12 c) 3𝑥 − 4𝑦 = 24 d) 4𝑥 − 3𝑦 = 24 63. Let α be the distance between lines – 𝑥 + 𝑦 = 2 and 𝑥 − 𝑦 = 2 and β be the distance between the lines 4𝑥 − 3𝑦 = 5 and 6𝑦 − 8𝑥 = 1, then a) 20√2β = 11α b) 20√2α = 11β c) 11√2β = 20α d) None of these 64. If the lines 𝑥 2 + 2𝑥𝑦 − 35𝑦 2 − 4𝑥 + 44𝑦 − 12 = 0 and 5𝑥 + 𝜆𝑦 − 8 = 0 are concurrent, then the value of λ is 21, Lakhanpur , Kanpur, Uttar Pradesh 208024 Page No. 1 8400072444, 7007286637 Download Aspire Study APP www.myaspirestudy.com  Aspire Study MCA Entrance Classes Visit www.aspirestudy.in Call Us: 8400072444, 7007286637 a) 0 b) 1 c) −1 d) 2 65. The straight line 3𝑥 + 4𝑦 − 5 = 0 and 4𝑥 = 3𝑦 + 15 intersect at the point 𝑃. On these lines the points 𝒬 and 𝑅 are chosen so that 𝑃𝒬 = 𝑃𝑅. The slopes of the lines 𝒬𝑅 passing through (1, 2) are a) −7, 1/7 b) 7,1/7 c) 7, −1/7 d) 3, −1/3 66. The vertices of a triangle are 𝐴(3,7), 𝐵(3,4) and 𝐶(5,4). The equation of the bisector of the angle 𝐴𝐵𝐶 is a) 𝑦 = 𝑥 + 1 b) 𝑦 = 𝑥 − 1 c) 𝑦 = 3𝑥 − 5 d) 𝑦 = 𝑥 67. The position of reflection of the point (4, 1) about the line 𝑦 = 𝑥 − 1 is a) (1, 2) b) (3, 4) c) (−1, 0) d) (2, 3) 68. A straight line through the point 𝐴(3,4) is such that its intercept between the axes is bisected at 𝐴. Its equation is a) 3𝑥 − 4𝑦 + 7 = 0 b) 4𝑥 + 3𝑦 = 24 c) 3𝑥 + 4𝑦 = 25 d) 𝑥 + 𝑦 = 7 69. If (−4, 5) is one vertex and 7𝑥 − 𝑦 + 8 = 0 is one diagonal of a square, then the equation of second diagonal is a) 𝑥 + 3𝑦 = 21 b) 2𝑥 − 3𝑦 = 7 c) 𝑥 + 7𝑦 = 31 d) 2𝑥 + 3𝑦 = 21 70. The pair of lines joining origin to the points of intersection of the two curves 𝑎𝑥 2 + 2ℎ𝑥𝑦 + 𝑏𝑦 2 + 2g𝑥 = 0 and 𝑎′ 𝑥 2 + 2ℎ′ 𝑥𝑦 + 𝑏 ′ 𝑦 2 + 2g ′ 𝑥 = 0 will be at right angles, if a) (𝑎′ + 𝑏 ′ )g ′ = (𝑎 + 𝑏)g b) (𝑎 + 𝑏)g ′ = (𝑎′ + 𝑏 ′ )g 2 2 c) ℎ2 − 𝑎𝑏 = ℎ′ − 𝑎′𝑏′ d) 𝑎 + 𝑏 + ℎ2 = 𝑎′ + 𝑏 ′ + ℎ′ 71. If a variable line passes through the point of intersection of the lines 𝑥 + 2𝑦 − 1 = 0 and 2𝑥 − 𝑦 − 1 = 0 and meets the coordinates axes in 𝐴 and 𝐵, then the locus of the mid point of 𝐴𝐵 is a) 𝑥 + 3𝑦 = 0 b) 𝑥 + 3𝑦 = 10 c) 𝑥 + 3𝑦 = 10𝑥𝑦 d) None of these 72. Distance between the two parallel lines 𝑦 = 2𝑥 + 7 and 𝑦 = 2𝑥 + 5 is a) √5/2 b) 2/5 c) 2/√5 d) 1/√5 73. If 𝑃 is the length of the perpendicular from the origin on the line whose intercepts on the axes are 𝑎 and 𝑏, then 1 1 1 1 1 1 a) 𝑝2 = 𝑎2 + 𝑏 2 b) 𝑝2 = 𝑎2 − 𝑏 2 c) 2 = 2 + 2 d) 2 = 2 − 2 𝑝 𝑎 𝑏 𝑝 𝑎 𝑏 74. If the diagonals of a parallelogram 𝐴𝐵𝐶𝐷 are along the lines 𝑥 + 5𝑦 = 7 and 10𝑥 − 2𝑦 = 9, then 𝐴𝐵𝐶𝐷 must be a a) Rectangle b) Square c) Cyclic quadrilateral d) Rhombus 75. The value of 𝑘 such that the lines 2𝑥 − 3𝑦 + 𝑘 = 0, 3𝑥 − 4𝑦 − 13 = 0 and 8𝑥 − 11𝑦 − 33 = 0 are concurrent, is a) 20 b) −7 c) 7 d) −20 76. The area (in square units) of the quadrilateral formed by two pairs of lines 𝑙 𝑥 − 𝑚2 𝑦 2 − 2 2 𝑛(𝑙𝑥 + 𝑚𝑦) = 0 and 𝑙 2 𝑥 2 − 𝑚2 𝑦 2 + 𝑛(𝑙𝑥 − 𝑚𝑦) = 0, is 𝑛2 𝑛2 𝑛 𝑛2 a) b) c) d) 2 |𝑙𝑚| |𝑙𝑚| 2 |𝑙𝑚| 4|𝑙𝑚| 77. A square of side 𝑎 lies above the 𝑥-axis and has one vertex at origin. The side passing through the π origin makes an angle α (0 < 𝛼 < 4 ) with the positive direction of 𝑥-axis. The equation of its diagonal not passing through the origin is 21, Lakhanpur , Kanpur, Uttar Pradesh 208024 Page No. 2 8400072444, 7007286637 Download Aspire Study APP www.myaspirestudy.com  Aspire Study MCA Entrance Classes Visit www.aspirestudy.in Call Us: 8400072444, 7007286637 a) 𝑦(cos α − sin α) − 𝑥(sin α − cos α) = 𝑎 b) 𝑦(cos α + sin α) + 𝑥(sin α − cos α) = 𝑎 c) 𝑦(cos α + sin α) + 𝑥(sin α + cos α) = 𝑎 d) 𝑦(cos α + sin α) + 𝑥(cos α − sin α) = 𝑎 78. The equation of one of the lines parallel to 4𝑥 − 3𝑦 = 5 and at a unit distance from the point (−1, −4) is a) 3𝑥 + 4𝑦 − 3 = 0 b) 3𝑥 + 4𝑦 + 3 = 0 c) 4𝑥 − 3𝑦 + 3 = 0 d) 4𝑥 − 3𝑦 − 3 = 0 79. The point 𝑃(𝑎, 𝑏) lies on the straight line 3𝑥 + 2𝑦 = 13 and the point 𝒬(𝑏, 𝑎) lies on the straight line 4𝑥 − 𝑦 = 5, then equation of the line 𝑃𝒬 is a) 𝑥 − 5 = 5 b) 𝑥 + 𝑦 = 5 c) 𝑥 + 𝑦 = −5 d) 𝑥 − 𝑦 = −5 80. The equation 𝑥 2 + 𝑘𝑥𝑦 + 𝑦 2 − 5𝑥 − 7𝑦 + 6 = 0 represents a pair of straight lines, then 𝑘 is a) 5/3 b) 10/3 c) 3/2 d) 3/10 81. 2 If 𝑡1 and 𝑡2 are roots of the equation 𝑡 + 𝜆𝑡 + 1 = 0, where λ is an arbitrary constant. Then, the line joining the points (𝑎𝑡12 , 2𝑎𝑡1 ) and (𝑎𝑡22 , 2𝑎𝑡2 ) always passes through a fixed point whose coordinates are a) (𝑎, 0) b) (−𝑎, 0) c) (0, 𝑎) d) (0, −𝑎) 82. A straight line through the point (2, 2) intersects the lines √3𝑥 + 𝑦 = 0 and √3𝑥 − 𝑦 = 0 at the point 𝐴 and 𝐵. The equation to the line 𝐴𝐵 so that the ∆ 𝑂𝐴𝐵 is equilateral is a) 𝑥 − 2 = 0 b) 𝑦 − 2 = 0 c) 𝑥 + 𝑦 − 4 = 0 d) None of these 83. In order to eliminate the first degree terms from the equation 2𝑥 2 + 4𝑥𝑦 + 5𝑦 2 − 4𝑥 − 22𝑦 + 7 = 0, the point to which origin is to be shifted, is a) (1, −3) b) (2, 3) c) (−2, 3) d) (1, 3) 84. If the lines given by 𝑎𝑥 + 2ℎ𝑥𝑦 + 𝑏𝑦 = 0 are equally inclined to the lines given by 𝑎𝑥 2 + 2ℎ𝑥𝑦 + 2 2 𝑏𝑦 2 + 𝜆(𝑥 2 + 𝑦 2 ) = 0, then a) 𝜆 is any real number b) 𝜆 = 2 c) 𝜆 = 1 d) None of these 85. The equation of the line bisecting perpendicularly the segment joining the points (−4, 6) and (8, 8), is a) 6𝑥 + 𝑦 − 19 = 0 b) 𝑦 = 7 c) 6𝑥 + 2𝑦 − 19 = 0 d) 𝑥 + 2𝑦 − 7 = 0 86. The value of λ for which the lines 3𝑥 + 4𝑦 = 5, 5𝑥 + 4 = 4 and 𝜆𝑥 + 4𝑦 = 6 meet at a point is a) 2 b) 1 c) 4 d) 3 87. The parallelism condition for two straight lines one of which is specified by the equation 𝑎𝑥 + 𝑏𝑦 + 𝑐 = 0 and the other being represented parametrically by 𝑥 = α𝑡 + β, 𝑦 = γ𝑡 + δ, is given by a) 𝑎γ + bα = 0, β = δ = c = 0 b) 𝑎α − 𝑏γ = 0, β = δ = 0 c) 𝑎α + 𝑏γ = 0 d) 𝑎γ = 𝑏α = 0 88. Origin containing angle bisector of two lines 𝐿1 ≡ 𝑎1 𝑥 + 𝑏1 𝑦 + 𝑐1 = 0 and 𝐿2 ≡ 𝑎2 𝑥 + 𝑏2 𝑦 + 𝑐2 = 0 (where 𝑐1 𝑐2 < 0) is 𝑎1 𝑥 + 𝑏1 𝑦 + 𝑐1 𝑎2 𝑥 + 𝑏2 𝑦 + 𝑐2 𝑎1 𝑥 + 𝑏1 𝑦 + 𝑐1 𝑎2 𝑥 + 𝑏2 𝑦 + 𝑐2 a) 2 2 = 2 2 b) 2 2 =− √𝑎1 + 𝑏1 √𝑎2 + 𝑏2 √𝑎1 + 𝑏1 √𝑎22 + 𝑏22 𝑎1 𝑥 + 𝑏1 𝑦 + 𝑐1 𝑎2 𝑥 + 𝑏2 𝑦 + 𝑐2 c) = d) Depends on the value of 𝑐1 and 𝑐2 𝑎12 + 𝑏12 𝑎22 + 𝑏22 89. The point of intersection of the two lines given by 2 𝑥 2 − 5 𝑥𝑦 + 2𝑦 2 − 3 𝑥 + 3 𝑦 + 1 = 0 is a) (1/2,1/3) b) (−1/7, −1/7) c) (−1/3,1/3) d) None of these 90. 2 2 If the slope of one of the lines given by 𝑎𝑥 − 6 𝑥𝑦 + 𝑦 = 0 is square of the other, then 𝑎 = a) 8, −27 b) −8,27 c) 1,8 d) −8, −27 21, Lakhanpur , Kanpur, Uttar Pradesh 208024 Page No. 3 8400072444, 7007286637 Download Aspire Study APP www.myaspirestudy.com  Aspire Study MCA Entrance Classes Visit www.aspirestudy.in Call Us: 8400072444, 7007286637 91. Equation of straight line belonging to families of straight lines (𝑥 + 2𝑦) + 𝜆(3𝑥 + 2𝑦 + 1) = 0 and (𝑥 − 2𝑦) + μ(𝑥 − 𝑦 + 1) = 0 is a) 6𝑥 + 5𝑦 = 2 b) 5𝑥 − 6𝑦 + 4 = 0 c) 5𝑥 + 6𝑦 = 4 d) None of these 92. A straight line through 𝑃(1, 2) is such that intercept between the axes is bisected at 𝑃. Its equation is a) 𝑥 + 𝑦 = −1 b) 𝑥 + 𝑦 = 3 c) 𝑥 + 2𝑦 = 5 d) 2𝑥 + 𝑦 = 4 93. The equation 𝑥 2 + 2√2𝑥𝑦 + 2𝑦 2 + 4𝑥 + 4√2 𝑦 + 1 = 0 represents a pair of lines which are parallel to each other. The distance between them is a) 4 units b) 2√3 units c) 4√3 units d) 2 units 94. A straight rod of length 9 units slides with its ends 𝐴, 𝐵 always on the 𝑋 and 𝑌 axis respectively. then, the locus of the centroid of ∆ 𝑂𝐴𝐵 is a) 𝑥 2 + 𝑦 2 = 3 b) 𝑥 2 + 𝑦 2 = 9 c) 𝑥 2 + 𝑦 2 = 1 d) 𝑥 2 + 𝑦 2 = 81 95. If the point (1, 𝛼) always remains in the interior of the triangle formed by the lines 𝑦 = 𝑥, 𝑦 = 0 and 𝑥 + 𝑦 = 4, then 𝛼 lies in the interval a) (0,1) b) [0,1] c) [0,4] d) None of these 96. The angle between the pair of straight lines 𝑦 2 sin2 θ − 𝑥𝑦sin2 θ + 𝑥 2 (cos2 θ − 1) = 0 is a) π/3 b) π/4 c) π/6 d) π/2 97. The area of a pentagon whose vertices are (4,1), (3,6), (−5,1), (−3, −3) and (−3,0) is a) 30 sq. units b) 60 sq. units c) 9 sq. units d) None of these 98. Let 𝑃𝑆 be the median of the triangle with vertices 𝑃(2, 2), 𝑄(6, −1) and 𝑅(7, 3). The equation of the line passing through (1, −1) and parallel to 𝑃𝑆 is a) 2𝑥 − 9𝑦 − 7 = 0 b) 2𝑥 − 9𝑦 − 11 = 0 c) 2𝑥 + 9𝑦 − 11 = 0 d) 2𝑥 + 9𝑦 + 7 = 0 99. If a line passes through the point (2,2) and encloses a triangle of area 𝐴 square units with the coordinate axes, then the intercepts made by the line on the coordinate axes are the roots of the equations a) 𝑥 2 ± 𝐴𝑥 ∓ 2𝐴 = 0 b) 𝑥 2 ± 𝐴𝑥 ± 2𝐴 = 0 c) 𝑥 2 ± 2𝐴𝑥 ± 𝐴 = 0 d) 𝑥 2 ± 2𝐴𝑥 ∓ 𝐴 = 0 100. Joint equation of the diagonals of the square formed the pairs of lines 𝑥𝑦 + 4𝑥 − 3𝑦 − 12 = 0 and 𝑥𝑦 − 3𝑥 + 4𝑦 − 12 = 0, is a) 𝑥 2 − 𝑦 2 + 𝑥 − 𝑦 = 0 b) 𝑥 2 − 𝑦 2 + 𝑥 + 𝑦 = 0 c) 𝑥 2 + 2𝑥𝑦 + 𝑦 2 + 𝑥 + 𝑦 = 0 d) 𝑥 2 − 2 𝑥𝑦 + 𝑦 2 + 𝑥 − 𝑦 = 0 21, Lakhanpur , Kanpur, Uttar Pradesh 208024 Page No. 4 8400072444, 7007286637 Download Aspire Study APP www.myaspirestudy.com  Aspire Study MCA Entrance Classes Visit www.aspirestudy.in Call Us: 8400072444, 7007286637 Answer Key Ques. 51 52 53 54 55 56 57 58 59 60 Ans. B B B A C D D B B A Ques. 61 62 63 64 65 66 67 68 69 70 Ans. C C A D A A D B C B Ques. 71 72 73 74 75 76 77 78 79 80 Ans. C C C D B A D D B B Ques. 81 82 83 84 85 86 87 88 89 90 Ans. B B C A A B C B D A Ques. 91 92 93 94 95 96 97 98 99 100 Ans. B D D B B D A D A A 21, Lakhanpur , Kanpur, Uttar Pradesh 208024 Page No. 5 8400072444, 7007286637 Download Aspire Study APP www.myaspirestudy.com  Aspire Study MCA Entrance Classes Visit www.aspirestudy.in Call Us: 8400072444, 7007286637 Solution 51 (b) 𝑑(𝑥, 𝑦) = max {|𝑥|, |𝑦|} …(i) but 𝑑(𝑥, 𝑦) = 𝑎 …(ii) From, Eqs. (i) and (ii), 𝑎 = max {|𝑥|, |𝑦|} If |𝑥| > |𝑦|, then 𝑎 = |𝑥| ∴ 𝑥 = ±𝑎 and if |𝑦| > |𝑥|, then 𝑎 = |𝑦| ∴ 𝑦 = ±𝑎 Therefore, locus represents a straight line 52 (b) The intersection of two curves 𝑎𝑥 2 + 2ℎ𝑥𝑦 + 𝑏𝑦 2 + 2g𝑥 + λ(𝑎′ 𝑥 2 + 2ℎ′ 𝑥𝑦 + 𝑏′𝑦 2 + 2𝑔′𝑥) = 0 ⇒ 𝑥 2 (𝑎 + 𝑎′ λ) + 2𝑥𝑦(ℎ + ℎ′ λ) + 𝑦 2 (𝑏 + λ𝑏 ′ ) + 2𝑥(𝑔 + λ𝑔′ ) = 0 For making homogeneous equating, g + λg ′ = 0 g ⇒ λ=− ′ g Since, lines are perpendicular. ∴ Coefficient of 𝑥 2 + Coefficient of 𝑦 2 = 0 ⇒ 𝑎 + 𝑎′ λ + 𝑏 + 𝑏 ′ λ = 0 g ⇒ 𝑎 + 𝑏 = −(𝑎′ + 𝑏 ′ ) (− ′ ) g ⇒ (𝑎 + 𝑏)g ′ = (𝑎′ + 𝑏 ′ )g 53 (b) Let the coordinates of the third vertex 𝐴 be (ℎ, 𝑘). Then, 𝐴𝐷 ⊥ 𝐵𝐶 𝑘−0 4 ⇒ 𝑂𝐴 ⊥ 𝐵𝐶 ⇒ × = −1 ⇒ 7 ℎ = 4 𝑘 … (i) ℎ − 0 −7 𝑘 − 3 −1 and, 𝑂𝐵 ⊥ 𝐴𝐶 ⇒ × = −1 ⇒ 5 ℎ − 𝑘 + 13 = 0 … (ii) ℎ + 2 −5 Solving (i) and (ii), we get ℎ = −4, 𝑘 = −7 Hence, the coordinates of the third vertex are (−4, −7) 21, Lakhanpur , Kanpur, Uttar Pradesh 208024 Page No. 6 8400072444, 7007286637 Download Aspire Study APP www.myaspirestudy.com  Aspire Study MCA Entrance Classes Visit www.aspirestudy.in Call Us: 8400072444, 7007286637 54 (a) |𝑏 − 𝑎| |𝑏 − 𝑎| Required distance = = √12 +12 √2 55 (c) The given lines are perpendicular to each other. |𝑟1 − 𝑟2 | ∴ Perpendicular distance = = √2 √2 ⇒ 𝑟1 − 𝑟2 = 2 The difference between the 𝑦-intercepts = 2 This can happen for five combinations {(0, 2), (1, 3), (2, 4), (3, 5), (4, 6)}. The difference between the 𝑥-intercepts = 2 This can happen for five combinations. Hence, total number of squares = 5 × 5 = 25 56 (d) We have, (𝑝 + 2 𝑞)𝑥 + (𝑝 − 3 𝑞)𝑦 − 𝑝 + 𝑞 = 0 ⇒ 𝑝(𝑥 + 𝑦 − 1) + 𝑞(2 𝑥 − 3 𝑦 + 1) = 0, Clearly, it represents a family of lines passing through the intersection of the lines 𝑥 + 𝑦 − 1 = 0 and 2 𝑥 − 3 𝑦 + 1 = 0. The coordinates of the point of the intersection these two lines are (2/5, 3/5) 57 (d) Equation of line perpendicular to 2𝑥 + 𝑦 + 6 = 0 and passes through origin is 𝑥 − 2𝑦 = 0 12 6 Now, point of intersection of 2𝑥 + 𝑦 + 6 = 0 and 𝑥 − 2𝑦 = 0 is (− , − ) 5 5 9 9 Similarly, point of intersection of 𝑥 − 2𝑦 = 0 and 4𝑥 + 2𝑦 − 9 = 0 is ( , ) 5 10 Let the origin divide the line 𝑥 − 2𝑦 = 0 in the ratio 𝜆: 1 9 12 𝜆− ∴ 𝑥= 5 5 = 0 ⇒ 9 𝜆 = 12 𝜆+1 5 5 12 4 ⇒ 𝜆= = 9 3 58 (b) Let the straight line meets the 𝑥-axis at 𝐴(𝑎, 0) and the 𝑦-axis at 𝐵(0, 𝑏). The equation of this straight line will be 21, Lakhanpur , Kanpur, Uttar Pradesh 208024 Page No. 7 8400072444, 7007286637 Download Aspire Study APP www.myaspirestudy.com  Aspire Study MCA Entrance Classes Visit www.aspirestudy.in Call Us: 8400072444, 7007286637 𝑥 𝑦 + = 1 …(i) 𝑎 𝑏 Since, it passes through 𝑃(1, 1) 1 1 ∴ 𝑎 + 𝑏 = 1 ⇒ 𝑎 + 𝑏 = 𝑎𝑏 …(ii) Let the coordinates of the mid point 𝑀 of 𝐴𝐵 are (ℎ, 𝑘) 𝑎+0 ∴ ℎ= ⇒ 𝑎 = 2ℎ 2 0+𝑏 and 𝑘 = 2 ⇒ 𝑏 = 2𝑘 substitute the values of 𝑎 and 𝑏 in Eq. (ii), we get 2ℎ + 2𝑘 = 2ℎ × 2𝑘 ⇒ ℎ + 𝑘 = 2ℎ𝑘 Hence, the equation of the locus of mid point 𝑀(ℎ, 𝑘) will be 𝑥 + 𝑦 − 2𝑥𝑦 = 0 59 (b) Given lines are 3𝑥 + 4𝑦 = 9 …(i) and 6𝑥 + 8𝑦 = 15 15 ⇒ 3𝑥 + 4𝑦 = 2 …(ii) ∵ Both lines are parallel, therefore the distance between two 15 | 2 − 9| 3 3 lines = = = √32 + 42 2 ∙ 5 10 60 (a) Let the lines are 𝑦 = 𝑚1 𝑥 + 𝑐1 and 𝑦 = 𝑚2 𝑥 + 𝑐2. Since, pair of straight lines are parallel to 𝑥-axis ∴ 𝑚1 = 𝑚2 = 0 Hence, the lines will be 𝑦 = 𝑐1 and 𝑦 = 𝑐2. Given circle is 𝑥 2 + 𝑦 2 − 6𝑥 − 4𝑦 − 12 = 0 ∴ Centre (3, 2) and radius = 5 Here, the perpendicular drawn from centre to the lines are 𝐶𝑃 and 𝐶𝑃′′ 2 − 𝑐1 ∴ 𝐶𝑃 = = ±5 √1 ⇒ 𝑐1 = 7 and 𝑐1 = −3 Hence, the lines are 𝑦 − 7 = 0, 𝑦+3=0 𝑖𝑒, (𝑦 − 7)(𝑦 + 3) = 0 or 𝑦 2 − 4𝑦 − 21 = 0 21, Lakhanpur , Kanpur, Uttar Pradesh 208024 Page No. 8 8400072444, 7007286637 Download Aspire Study APP www.myaspirestudy.com  Aspire Study MCA Entrance Classes Visit www.aspirestudy.in Call Us: 8400072444, 7007286637 62 (c) Let the coordinates of point 𝐴 and 𝐵 are (𝑎, 0) and (0, −𝑏) 𝑎 ∴ =4 ⇒ 𝑎=8 2 𝑏 and − 2 = −3 ⇒ 𝑏 = 6 𝑥 𝑦 ∴ Equation of line is + =1 8 −6 ⇒ 3𝑥 − 4𝑦 = 24 63 (a) Given, α be the distance between lines 𝑥 − 𝑦 + 2 = 0 and 𝑥 − 𝑦 − 2 = 0 |2 + 2| |4| ∴ α= = = 2√2 √1 + 1 √2 and β be the distance between the lines 1 4𝑥 − 3𝑦 − 5 = 0 and 4𝑥 − 3𝑦 + = 0 2 1 |5 + 2| |11| 11 ∴ β= = = √(4)2 + (3)2 2√25 10 α 2√2 20√2 Now, = = β 11/10 11 ⇒ 20√2β = 11α 64 (d) Given line is 𝑥 2 + 2𝑥𝑦 − 35𝑦 2 − 4𝑥 + 44𝑦 − 12 = 0 Here, 𝑎 = 1, 𝑏 = −35, 𝑐 = −12, ℎ = 1, 𝑓 = 22 22 − 70 −2 − 22 4 2 ∴ Point of intersection = ( , )=( , ) −35 − 1 −35 − 1 3 3 4 2 If the lines are concurrent. The point (3 , 3) will be on the line 5𝑥 + 𝜆𝑦 − 8 = 0 4 2 ∴ 5( ) + 𝜆( ) −8 = 0 3 3 21, Lakhanpur , Kanpur, Uttar Pradesh 208024 Page No. 9 8400072444, 7007286637 Download Aspire Study APP www.myaspirestudy.com  Aspire Study MCA Entrance Classes Visit www.aspirestudy.in Call Us: 8400072444, 7007286637 2 20 4 ⇒ 𝜆 =8− = ⇒𝜆=2 3 3 3 65 (a) The given equation are 3𝑥 + 4𝑦 − 5 = 0 …(i) and 4𝑥 − 3𝑦 − 15 = 0 …(ii) Since, these lines are perpendicular to each other so ∠𝒬𝑃𝑅 is right angle and 𝑃𝒬 = 𝑃𝑅. Hence, ∆𝑃𝒬𝑅 is a right angle isosceles triangle. ∠ 𝑃𝒬𝑅 = ∠𝑃𝑅𝒬 = 45° 3 4 Slope of 𝑃𝒬 = − 4 and slope of 𝑃𝑅 = 3 Let slope of 𝒬𝑅 = 𝑚 4 −𝑚 ∴ tan 45° = ± | 3 | 4 1 + 3𝑚 1 ⇒ 𝑚 = , −7 7 66 (a) Required line is passing through (3, 4) and having slope 1. ∴ Equation of required line is 𝑦 − 4 = 1(𝑥 − 3) ⇒ 𝑥−𝑦+1=0 ⇒ 𝑦 =𝑥+1 67 (d) Let 𝒬(𝑥, 𝑦) be the image of the point 𝑃(4, 1) to the line 𝑦 − 𝑥 + 1 = 0 Then, 𝑃𝒬 is perpendicular to 𝑦 − 𝑥 + 1 = 0 𝑦+1 ∴ × 1 = −1 𝑥−4 ⇒ 𝑦 + 𝑥 = 4 + 1 = 5 …(i) 4+𝑥 𝑦+1 Also, mid point of 𝑃𝒬, 𝑖𝑒, ( , ) lies on 𝑦 − 𝑥 + 1 = 0 2 2 𝑦 + 1 (4 + 𝑥) ∴ − +1=0 2 2 ⇒ 𝑦 − 𝑥 − 1 = 0 …(ii) On solving Eqs. (i) and (ii), we get the required point (2, 3) 68 (b) Since, 𝐴 is mid point of line 𝑃𝒬 21, Lakhanpur , Kanpur, Uttar Pradesh 208024 Page No. 10 8400072444, 7007286637 Download Aspire Study APP www.myaspirestudy.com  Aspire Study MCA Entrance Classes Visit www.aspirestudy.in Call Us: 8400072444, 7007286637 𝑎+0 ∴ 3= =𝑎=6 2 0+𝑏 and 4 = 2 ⇒ 𝑏 = 8 Thus, equation of line is 𝑥 𝑦 + = 1 ⇒ 4𝑥 + 3𝑦 = 24 6 8 70 (b) The intersection of two curves 𝑎𝑥 2 + 2ℎ𝑥𝑦 + 𝑏𝑦 2 + 2g𝑥 + λ(𝑎′ 𝑥 2 + 2ℎ′ 𝑥𝑦 + 𝑏′𝑦 2 + 2𝑔′𝑥) = 0 ⇒ 𝑥 2 (𝑎 + 𝑎′ λ) + 2𝑥𝑦(ℎ + ℎ′ λ) + 𝑦 2 (𝑏 + λ𝑏 ′ ) + 2𝑥(𝑔 + λ𝑔′ ) = 0 For making homogeneous equating, g + λg ′ = 0 g ⇒ λ=− ′ g Since, lines are perpendicular. ∴ Coefficient of 𝑥 2 + Coefficient of 𝑦 2 = 0 ⇒ 𝑎 + 𝑎′ λ + 𝑏 + 𝑏 ′ λ = 0 g ⇒ 𝑎 + 𝑏 = −(𝑎′ + 𝑏 ′ ) (− ′ ) g (𝑎 ′ (𝑎 ′ ′ )g ⇒ + 𝑏)g = +𝑏 71 (c) The equation of line passing through the point of intersection of 𝑥 + 2𝑦 − 1 = 0 and 2𝑥 − 𝑦 − 1 = 0 is (𝑥 + 2𝑦 − 1) + 𝜆(2𝑥 − 𝑦 − 1) = 0 ⇒ 𝑥(1 + 2𝜆) + 𝑦(2 − 𝜆) − 1 − 𝜆 = 0 1+𝜆) 𝜆+1 This meets the coordinate axes at 𝐴 (2𝜆+1 , 0) and 𝐵 (0, 2−𝜆) Let (ℎ, 𝑘) be the mid point of 𝐴𝐵, then 1 1+𝜆 1 𝜆+1 ℎ= ( ),𝑘 = ( ) 2 2𝜆 + 1 2 2−𝜆 On eliminating 𝜆 from the these equations, we get ℎ + 3𝑘 = 10ℎ𝑘 Thus, the locus of (ℎ, 𝑘) is 𝑥 + 3𝑦 = 10𝑥𝑦 72 (c) On comparing the given lines with 𝑦 = 𝑚1 𝑥 + 𝑐1 and 𝑦 = 𝑚2 𝑥 + 𝑐2 , we get 21, Lakhanpur , Kanpur, Uttar Pradesh 208024 Page No. 11 8400072444, 7007286637 Download Aspire Study APP www.myaspirestudy.com  Aspire Study MCA Entrance Classes Visit www.aspirestudy.in Call Us: 8400072444, 7007286637 𝑚1 = 2 and 𝑐1 = 7 and 𝑚2 = 2 and 𝑐2 = 5 |𝑐1 − 𝑐2 | ∴ Required distance = √(𝑚)2 + 1 |7 − 5| 2 = = √(2)2 + 1 √5 73 (c) 𝑥 𝑦 Here the equation of 𝐴𝐵 is + = 1 𝑎 𝑏 From the figure, 𝑂𝑃 ⊥ 𝐴𝐵, 1 1 0 (𝑎) + 0 ( ) − 1 ∴ 𝑂𝑃 = || 𝑏 || 1 √ 2+ 2 1 𝑎 𝑏 1 ⇒𝑝= √ 12 + 12 𝑎 𝑏 1 ⇒ 𝑝2 = [squaring both sides] 1 1 + 𝑎2 𝑏 2 1 1 1 ⇒ 2= 2+ 2 𝑝 𝑎 𝑏 74 (d) Clearly, diagonals are perpendicular So, 𝐴𝐵𝐶𝐷 must be a rhombus 75 (b) Given lines are concurrent 2 −3 𝑘 ∴ |3 −4 −13| = 0 8 −11 −33 ⇒ 2(132 − 143) + 3(−99 + 104) + 𝑘(−33 + 32) = 0 ⇒ −22 + 15 − 𝑘 = 0 ⇒ 𝑘 = −7 76 (a) The equations of the sides of the quadrilateral are given by 𝑙 2 𝑥 2 − 𝑚2 𝑦 2 − 𝑛(𝑙𝑥 + 𝑚𝑦) = 0 and, 𝑙 2 𝑥 2 = 𝑚2 𝑦 2 + 𝑛(𝑙𝑥 + 𝑚𝑦) = 0 21, Lakhanpur , Kanpur, Uttar Pradesh 208024 Page No. 12 8400072444, 7007286637 Download Aspire Study APP www.myaspirestudy.com  Aspire Study MCA Entrance Classes Visit www.aspirestudy.in Call Us: 8400072444, 7007286637 ⇒ (𝑙𝑥 + 𝑚𝑦)(𝑙𝑥 − 𝑚𝑦 − 𝑛) = 0 and (𝑙𝑥 − 𝑚𝑦)(𝑙𝑥 + 𝑚𝑦 + 𝑛) = 0 ⇒ 𝑙𝑥 + 𝑚𝑦 = 0, 𝑙𝑥 − 𝑚𝑦 − 𝑛 = 0, 𝑙𝑥 − 𝑚𝑦 = 0, 𝑙𝑥 + 𝑚𝑦 + 𝑛 = 0 Clearly, the lines form a parallelogram whose are is {0 − (−𝑛)}{0 − 𝑛} 𝑛2 | |= 𝑙 𝑚 2|𝑙𝑚| | | 𝑙 −𝑚 77 (d) Since, line 𝑂𝐴 makes an angle α with 𝑥-axis and given 𝑂𝐴 = 𝑎, then coordinates of 𝐴 are (𝑎 cos α , 𝑎 sin α). Also, 𝑂𝐵 ⊥ 𝑂𝐴, then 𝑂𝐵 makes an angle (90° + α) with 𝑥-axis, then coordinates of 𝐵 are [𝑎 cos(90° + α) , 𝑎 sin(90° + α)] 𝑖𝑒, (−𝑎 sin α , 𝑎 cos α) Equation of the diagonal 𝐴𝐵 not passing through the origin is 𝑎 cos α − 𝑎 sin α (𝑦 − 𝑎 sin α) = (𝑥 − 𝑎 cos α) −𝑎 sin α − 𝑎 cos α ⇒ (sin α + cos α)(𝑦 − 𝑎 sin α) = (sin α − cos α)(𝑥 − 𝑎 cos α) ⇒ 𝑦(sin α + cos α) + 𝑥(cos α − sin α) = 𝑎 sin α (sin α + cos α) − 𝑎 cos α (sin α − cos α) = 𝑎 (sin2 α + sin α cos α − cos α sin α + cos 2 α) ⇒ 𝑦(sin α + cos α) + 𝑥(cos α − sin α) = 𝑎 78 (d) Required equation can be 4𝑥 − 3𝑦 − 𝐾 = 0 4 × −1 − 3 × −4 − 𝐾 ∴| |=1 √42 + (−3)2 −4 + 12 − 𝐾 ⇒ = ±1 5 ⇒ 8 − 𝐾 = ±5 ⇒ 𝐾 = 3 or 𝐾 = 13 ∴ Equation of lines are 4𝑥 − 3𝑦 − 3 = 0 and 4𝑥 − 3𝑦 − 13 = 0 79 (b) ∵ Point 𝑃(𝑎, 𝑏) lies on 3𝑥 + 2𝑦 = 13 ∴ 3𝑎 + 2𝑏 = 13...(i) and point 𝒬(𝑏, 𝑎) is lies on 4𝑥 − 𝑦 = 5 ∴ 4𝑏 − 𝑎 = 5 …(ii) On solving Eqs. (i) and (ii), we get 𝑎 = 3, 𝑏 = 2 Therefore, the coordinates of 𝑃 and 𝒬 are (3, 2) and (2, 3) respectively. Now, equation of 𝑃𝒬 is 3−2 𝑦−2= (𝑥 − 3) ⇒ 𝑥 + 𝑦 = 5 2−3 80 (b) The given equation 21, Lakhanpur , Kanpur, Uttar Pradesh 208024 Page No. 13 8400072444, 7007286637 Download Aspire Study APP www.myaspirestudy.com  Aspire Study MCA Entrance Classes Visit www.aspirestudy.in Call Us: 8400072444, 7007286637 𝑥 2 + 𝑘𝑥𝑦 + 𝑦 2 − 5𝑥 − 7𝑦 + 6 = 0 is compared with 𝑎𝑥 2 + 2ℎ𝑥𝑦 + 𝑏𝑦 2 + 2g𝑥 + 2𝑓𝑦 + 𝑐 = 0, we get 𝑘 −5 −7 𝑎 = 1, 𝑏 = 1, ℎ = , g = ,𝑓 = ,𝑐 = 6 2 2 2 This equation represents a pair of straight lines, 𝑎 ℎ g if |ℎ 𝑏 𝑓 | = 0 g 𝑓 𝑐 1 𝑘/2 −5/2 ⇒ | 𝑘/2 1 −7/2| = 0 −5/2 −7/2 6 49 𝑘 6𝑘 35 5 7𝑘 5 ⇒ 1 (6 − ) − ( − ) − (− + )=0 4 2 2 4 2 4 2 24 − 49 𝑘 12𝑘 − 35 5 −7𝑘 + 10 ⇒( )− ( )− ( )=0 4 2 4 2 4 ⇒ −50 − 12𝑘 2 + 35𝑘 + 35𝑘 − 50 = 0 ⇒ −12𝑘 2 + 70𝑘 − 100 = 0 ⇒ 6𝑘 2 − 35𝑘 + 50 = 0 10 ⇒𝑘= 3 81 (b) Since, 𝑡1 , 𝑡2 are the roots of the equation 𝑡 2 + 𝜆𝑡 + 1 = 0 ∴ 𝑡1 + 𝑡2 = −𝜆, 𝑡1 𝑡2 = 1 The equation of a line passing through (𝑎𝑡12 , 2𝑎𝑡1 ) and (𝑎𝑡22 , 2𝑎𝑡2 ) is 2 𝑦 − 2𝑎𝑡2 = (𝑥 − 𝑎𝑡22 ) 𝑡1 + 𝑡2 2 ⇒ 𝑦 − 2𝑎𝑡2 = − (𝑥 − 𝑎𝑡22 ) 𝜆 ⇒ 𝜆𝑦 − 2𝑎𝜆𝑡2 = −2𝑥 + 2𝑎𝑡22 ⇒ 𝜆𝑦 + 2𝑥 = 2𝑎(𝜆𝑡2 + 𝑡22 ) ⇒ 𝜆𝑦 + 2𝑥 = 2𝑎(−1) ⇒ 2(𝑥 + 𝑎) + 𝜆𝑦 = 0 ∴ Fixed point is (−𝑎, 0) 82 (b) √3𝑥 + 𝑦 = 0 makes an angle of 120° with 𝑂𝑋 and √3𝑥 − 𝑦 = 0 makes an angle 60° with 𝑂𝑋. So, the required line is 𝑦 − 2 = 0 83 (c) Here, 𝑎 = 2, 𝑏 = 5, 𝑐 = 7, ℎ = 2, g = −2, 𝑓 = −11 To eliminate 1st degree terms origin is to be shifted to the point ℎ𝑓 − 𝑏𝑔 𝑔ℎ − 𝑎𝑓 −22 + 10 −4 + 22 ( , ) = ( , ) = (−2, 3) 𝑎𝑏 − ℎ2 𝑎𝑏 − ℎ2 10 − 4 10 − 4 84 (a) 21, Lakhanpur , Kanpur, Uttar Pradesh 208024 Page No. 14 8400072444, 7007286637 Download Aspire Study APP www.myaspirestudy.com  Aspire Study MCA Entrance Classes Visit www.aspirestudy.in Call Us: 8400072444, 7007286637 If the lines given by 𝑎𝑥 2 + 2ℎ𝑥𝑦 + 𝑏𝑦 2 = 0 are equally inclined to the lines given by 𝑎𝑥 2 + 2ℎ𝑥𝑦 + 𝑏𝑦 2 + 𝜆(𝑥 2 + 𝑦 2 ) = 0, then the two pairs have same bisectors. Therefore, equations 𝑥 2 −𝑦 2 𝑥𝑦 𝑥 2 −𝑦2 𝑥𝑦 = and (𝑎+𝜆)−(𝑏+𝜆) = 𝑎−𝑏 ℎ ℎ represent same pair of lines. Clearly, these two equations are identical for all values of 𝜆 85 (a) Equation of the line passing through (−4, 6) and (8, 8) is 8−6 𝑦−6=( ) (𝑥 + 4) 8+4 2 ⇒𝑦−6= (𝑥 + 4) 12 ⇒ 6𝑦 − 36 = 𝑥 + 4 ⇒ 6𝑦 − 𝑥 − 40 = 0 …(i) Now, equation of any line perpendicular to the Eq. (i), is 6𝑥 + 𝑦 + 𝜆 = 0 …(ii) This line passes through the mid point of (−4, 6) and (8, 8) is −4 + 8 6 + 8 ( , ) , 𝑖𝑒, (2, 7) 2 2 ∴6×2+7+𝜆 =0 ⇒ 19 + 𝜆 = 0 ⇒ 𝜆 = −19 On putting 𝜆 = −19 in Eq. (ii), we get the equation of required line which is 6𝑥 + 𝑦 − 19 = 0 86 (b) Given lines are 3𝑥 + 4𝑦 = 5, 5𝑥 + 4𝑦 = 4 and 𝜆𝑥 + 4𝑦 = 6. These three lines meet at point, if the point of intersection of first two lines lies on the third line 1 13 Now, point of intersection of line 3𝑥 + 4𝑦 = 5 and 5𝑥 + 4𝑦 = 4 is (− 2 , 8 ) 1 13 The line 𝜆𝑥 + 4𝑦 = 6 passes through the point (− 2 , 8 ) 1 13 ∴ 𝜆 (− ) + 4 ( ) = 6 2 8 ⇒ −𝜆 + 13 = 12 ⇒ 𝜆=1 87 (c) Give lines are 𝑎𝑥 + 𝑏𝑦 + 𝑐 = 0 …(i) 𝑥 = α𝑡 + β …(ii) and 𝑦 = γ𝑡 + δ …(iii) On eliminating 𝑡, from Eqs. (ii) and (iii), we get γ𝑥 − α𝑦 + αδ − βγ = 0 …(iv) For parallelism condition in Eqs. (i) and (iv) 𝑎 𝑏 = γ −α ⇒ 𝑎α + 𝑏γ = 0 89 (d) 21, Lakhanpur , Kanpur, Uttar Pradesh 208024 Page No. 15 8400072444, 7007286637 Download Aspire Study APP www.myaspirestudy.com  Aspire Study MCA Entrance Classes Visit www.aspirestudy.in Call Us: 8400072444, 7007286637 The point of intersection of the lines given by 𝑎𝑥 2 + 2 ℎ𝑥𝑦 + 𝑏𝑦 2 + 2 𝑔𝑥 + 2 𝑓𝑦 + 𝑐 = 0 is given by ℎ𝑓 − 𝑏𝑔 𝑔ℎ − 𝑎𝑓 ( , ) 𝑎𝑏 − ℎ2 𝑎𝑏 − ℎ2 Hence, the lines given by 2 𝑥 2 − 5 𝑥𝑦 + 2 𝑦 2 − 3 𝑥 + 3 𝑦 + 1 = 0 intersect at (1/3, −1/3) 91 (b) Equation belonging to both families will pass through two fixed points. First intersection point of lies 𝑥 + 2𝑦 = 0 and 1 1 3𝑥 + 2𝑦 + 1 = 0 is (− 2 , 4) and second interception point of lines 𝑥 − 2𝑦 = 0 and 𝑥 − 𝑦 + 1 = 0, is (−2, −1) 1 1 Line passing through (− 2 , 4) and (−2, −1) is 1 1 −1 − 4 1 𝑦− = (𝑥 + ) 4 −2 + 1 2 2 ⇒ 5𝑥 − 6𝑦 + 4 = 0 92 (d) Since, 𝑃(1, 2) is mid point of 𝐴𝐵. Therefore, coordinate of 𝐴 and 𝐵 are (2, 0) and (0, 4) respectively ∴ Equation of line 𝐴𝐵 is 4 𝑦−0= (𝑥 − 2) −2 ⇒ 2𝑥 + 𝑦 = 4 93 (d) Here, ℎ = √2, g = 2, 𝑎 = 1, 𝑐 = 1, 𝑏 = 2, 𝑓 = 2√2 g 2 − 𝑎𝑐 4−1 ∴ Distance = 2√ = 2√ = 2 units 𝑎(𝑎 + 𝑏) 1(1 + 2) 94 (b) Let 𝑃(ℎ, 𝑘) be the centroid of ∆ 𝑂𝐴𝐵. Let the coordinates of 𝐴 and 𝐵 be (𝑎, 0) and (0, 𝑏) respectively. then, 𝑎 𝑏 ℎ = ,𝑘 = 3 3 21, Lakhanpur , Kanpur, Uttar Pradesh 208024 Page No. 16 8400072444, 7007286637 Download Aspire Study APP www.myaspirestudy.com  Aspire Study MCA Entrance Classes Visit www.aspirestudy.in Call Us: 8400072444, 7007286637 in ∆ 𝑂𝐴𝐵, we have 𝑂𝐴2 + 𝑂𝐵2 = 𝐴𝐵2 ⇒ 𝑎2 + 𝑏 2 = 92 ⇒ 9ℎ2 + 9𝑘 2 = 92 ⇒ ℎ2 + 𝑘 2 = 9 Hence, the locus of (ℎ, 𝑘) is 𝑥 2 + 𝑦 2 = 9 95 (b) It is evident from the figure that 𝑃 moves on the line 𝑥 = 1. Clearly, 𝑦-coordinate of 𝑃 varies between 0 and 1 ∴ 0 ≤ 𝛼 ≤ ⇒ 𝛼 ∈ [0, 1] 96 (d) The given equation is 𝑥 2 (cos 2 θ − 1) − 𝑥𝑦sin2 θ + 𝑦 2 sin2 θ = 0 1 Here, 𝑎 = cos2 θ − 1, ℎ = − sin2 θ, 𝑏 = sin2 θ 2 𝑎 + 𝑏 = cos 2 θ + sin2 θ − 1 = 1 − 1 = 0 π ∴ The angle between the pair of straight lines is 2. 97 (a) We have, 1 4 3 21 1 3 −5 33 ∆1 = | |= , ∆2 = | |= 2 1 6 2 2 6 1 2 1 −5 −3 1 −3 −3 9 ∆3 = | | = 9, ∆4 = | |=− 2 1 −3 2 −3 0 2 1 −3 4 3 and, ∆5 = | |=− 2 0 1 2 ∴ Area of the pentagon = |∆1 + ∆2 + ∆3 + ∆4 + ∆5 | 21, Lakhanpur , Kanpur, Uttar Pradesh 208024 Page No. 17 8400072444, 7007286637 Download Aspire Study APP www.myaspirestudy.com  Aspire Study MCA Entrance Classes Visit www.aspirestudy.in Call Us: 8400072444, 7007286637 21 33 9 3 =| + + 9 − − | = 30 sq. units 2 2 2 2 98 (d) Since, 𝑆 is mid point of 𝑄𝑅 ∴ Coordinate of 𝑆 are 6 + 7 −1 + 3 13 ( , ) = ( , 1) 2 2 2 2−1 2 ∴ Slope of 𝑃𝑆 = =− 13 9 2− 2 The required equation which is 2 passing throught (1, −1) andslope − , is 9 2 𝑦 + 1 = − (𝑥 − 1) 9 ⇒ 9𝑦 + 9 = −2𝑥 + 2 ⇒ 2𝑥 + 9𝑦 + 7 = 0 99 (a) 𝑥 𝑦 Let the equation of the line be 𝑎 + 𝑏 = 1 It passes through (2,2) 2 2 ∴ + = 1 ⇒ 2(𝑎 + 𝑏) = 𝑎𝑏 … (i) 𝑎 𝑏 The line encloses a triangle of area 𝐴 square units with the coordinate axes 1 ∴ |𝑎||𝑏| = 𝐴 ⇒ |𝑎𝑏| = 2𝐴 ⇒ 𝑎𝑏 = ±2𝐴 … (ii) 2 From (i) and (ii), we get 𝑎 + 𝑏 = ±𝐴 The quadratic equation having 𝑎, 𝑏 as its roots is 𝑥 2 − 𝑥(𝑎 + 𝑏) + 𝑎𝑏 = 0 or, 𝑥 2 ∓ 𝐴𝑥 ± 2𝐴 = 0 21, Lakhanpur , Kanpur, Uttar Pradesh 208024 Page No. 18 8400072444, 7007286637 Download Aspire Study APP www.myaspirestudy.com  Aspire Study MCA Entrance Classes Visit www.aspirestudy.in Call Us: 8400072444, 7007286637 21, Lakhanpur , Kanpur, Uttar Pradesh 208024 Page No. 19 8400072444, 7007286637 Download Aspire Study APP www.myaspirestudy.com