• ( 1 ) The distance between the points A(5, -7) and B(2, 3) is:

    • 1) 109
    • 2) 5√ 7
    • 3) √ 109
    • 4) 7√ 5
    • Discussion in forum
      Answer : 3) √ 109
      Solution : AB2 = (2 - 5)2 + (3 + 7)2
      => (-3)2 + (10)2
      => 9 + 100 => √ 109








      discussion


      Answer : 3) √ 109

    • ( 2 ) Find the value of k for which the points A(-2, 5), B(3, k) and C(6, 1) are collinear.

    • 1) 5
    • 2) 7
    • 3) 12
    • 4) 4
    • Discussion in forum
      Answer : 1) 5
      Solution : x1 = -2, x2 = 3, x3 = 6 and y1 = 5, y2 = k, y3 = -1
      Now Δ = 0 <=> -2(k + 1) + 3(-1 + 5) + 6(5 - k) = 0
      => -2 (k+1) + 3(4) + 6(5-k) = 0
      => -2k-2 + 12+30 -6k = 0
      => 40 - 8k = 0
      => -8k = -40
      => k = 5.








      discussion


      Answer : 1) 5

    • ( 3 ) The distance between the points A(b, 0) and B(0, a) is.

    • 1) √(a2-b2)
    • 2) √(a2+b2)
    • 3) √(a+b)
    • 4) a-b
    • Discussion in forum
      Answer : 2) √(a2+b2)
      Solution : AB = √((b-0)2 - (0 - a)2)
      = √(b2 + a2)
      = √(a2 + b2).








      discussion


      Answer : 2) √(a2+b2)

    • ( 4 ) The points A(0, 6), B(-5, 3) and C(3, 1) are the vertices of a triangle which is ?

    • 1) equilateral
    • 2) right angled
    • 3) isosceles
    • 4) scalene
    • Discussion in forum
      Answer : 3) isosceles
      Solution : AB2 = (-5 - 0)2 + (-3 - 0)2 = 25 + 9 = 34
      BC2 = (3 + 5)2 + (1-3)2 = 82 + (-2)2 = 64 + 4 = 68
      AC2 = (3 - 0)2 + (1 - 6)2 = 9 + 25 = 34.
      AB = AC. => Δ ABC is isosceles.








      discussion


      Answer : 3) isosceles

    • ( 5 ) The vertices of a quadrilateral ABCD are A(0, 0), B(3,3), C(3, 6) and D(0, 3). Then , ABCD is a

    • 1) square
    • 2) parallelogram
    • 3) rectangle
    • 4) rhombus
    • Discussion in forum
      Answer : 2) parallelogram
      Solution : AB2 = (3-0)2 + (3-0)2 = 18
      BC2 = (3-3)2 + (6-3)2 = 9 CD2 = (0-3)2 + (3-6)2 =18
      AD2 = (0-0)2 + (3-0)2 = 9
      AB = CD = √ 18 => 3√ 2,
      BC = AD = √ 9
      AC2 = (3-0)2 + (6-0)2 = 9 + 36 = 45
      BD2 = (0-3)2 + (3-3)2 = 9 + 0 = 9
      AC ≠ BD ABCD is a parallelogram.








      discussion


      Answer : 2) parallelogram

    • ( 6 ) Find the distance between the points A(-4, 7) and B(2, -5).

    • 1) 8√ 5 Units
    • 2) 6√ 4 Units
    • 3) 6√ 5 Units
    • 4) 7√ 5 Units
    • Discussion in forum
      Answer : 3) 6√ 5 Units
      Solution : AB = √((2+4)2 + (-5-7)2)
      = √(62 + (-12)2)
      = √(36 + 144) = √ 180
      =√ (36*5) = 6√ 5 units.








      discussion


      Answer : 3) 6√ 5 Units

    • ( 7 ) Find the distance of the point A(4, -4) from the origin.

    • 1) 3√ 2
    • 2) 2√ 8
    • 3) 6√ 2
    • 4) 8√ 2
    • Discussion in forum
      Answer : 4) 8√ 2
      Solution : OA = √(42 + (-4)2) = √(16 + 16) = √ 32 = 8√ 2








      discussion


      Answer : 4) 8√ 2

    • ( 8 ) A is a point on y-axis at a distance of 5 units from x-axis lying below x-axis. The co-ordinates of A are:

    • 1) (5, 0)
    • 2) (-5, 0)
    • 3) (0, 5)
    • 4) (0, -5)
    • Discussion in forum
      Answer : 4) (0, -5)
      Solution : The co-ordinates of A are A(0, -5)








      discussion


      Answer : 4) (0, -5)

    • ( 9 ) Find the area of ΔABC whose vertices are A(9, -5), B(3, 7) and (-2, 4).

    • 1) 29 units
    • 2) 35.9 sq.units
    • 3) 39 sq.units
    • 4) 39.5 sq.units
    • Discussion in forum
      Answer : 3) 39 sq.units
      Solution : Here, x1 = 9, x2 = 3, x3 = -2 and y1 = -5, y2 = 7, y3 = 4
      = 1/2 [9(7 - 4) + 3(4 + 5) + (-2)(-5 - 7)]
      = 1/2 [9(3) + 3(9) - 2(-12)]
      = 1/2 [27 + 27 + 24]
      = 1/2 [78]
      = 39 sq.units








      discussion


      Answer : 3) 39 sq.units

    • ( 10 ) If the sum of the dimensions of rectangular parallel piped is 24 cm and the length of the diagonal is 15 cm, then the total surface are of its is

    • 1) 420 cm2
    • 2) 275 cm2
    • 3) 351 cm2
    • 4) 378 cm2
    • Discussion in forum
      Answer : 3) 351 cm2
      Solution : Let the dimensions of the rectangular parallelepiped be x, y and z
      Then x + y + z = 24
      Length of the diagonal √(x2 + y2 + z2) = 15
      x2 + y2 + z2 = 225
      Now , (x+y+z)2 = x2 + y2 + z2 + 2(xy+yz+zx)
      (24)2 = 225+2(xy+yz+zx)
      2(xy+yz+zx) = 576-225 =351
      Total surface area = 2(xy+yz+zx) = 351 cm2








      discussion


      Answer : 3) 351 cm2





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