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Coordinate Geometry Test - 6

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Coordinate Geometry Test - 6
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  • Question 1
    1 / -0
    The linear equation x = 6 + 2y cuts x-axis at
    Solution
    Since the line cuts x-axis,
    y = 0

    Given: x = 6 + 2y
    x = 6 + 2 × 0 = 6
    So, x = 6 + 2y cuts x-axis at (6, 0).
  • Question 2
    1 / -0
    If a linear equation has solutions (-2, 2), (0, 0) and (2, -2), then it is of the form
    Solution
    Consider the linear equation of the form y = mx + c. ... (1)
    Since (-2, 2) and (0, 0) are solutions of the equation (1);
    2 = -2m + c
    c - 2m = 2 ... (2)
    And
    0 = m(0) + c
    c = 0
    Using the value of c in equation (2), we get m = -1
    Then, the required equation is:
    y = -(1)x + 0 y + x = 0
  • Question 3
    1 / -0
    The ordered pair (1, 1) is a solution of the equation
    Solution
    We have coordinates (1, 1).
    Putting the given coordinates in y = 2x - 1,
    1 = 2 × 1 - 1
    1 = 2 -1
    1 = 1
    Hence, the coordinates satisfy the equation y = 2x - 1.
    So, option (1) is correct.
  • Question 4
    1 / -0
    If the sum of squares of abscissa and ordinate is 2 and product is 1, then the value of abscissa is
    Solution
    Let the abscissa and ordinate be x and y respectively.
    Then, x2 + y2 = 2 ... (1)
    And
    xy = 1 ... (2)
    Substituting the value of y from (2) in (1) we get,
    x2 + (1/x)2 = 2
    x4 + 1 = 2x2
    (x2 - 1)2 = 0
    x2 - 1 = 0
    x2 = 1
    x = 1 or x = -1
  • Question 5
    1 / -0
    Which of the following statements is correct?

    a) Coordinate axes divide the plane into four equal parts.
    b) Coordinate axes divide the plane into five equal parts.
    c) Coordinate axes divide the plane into two equal parts.
    d) Coordinate axes divide the plane into three equal parts.
    Solution
    Coordinate axes always divide the plane into 4 equal parts called the quadrants.

  • Question 6
    1 / -0
    What are the respective values of x and y, if (2x - 9, 4) = (3, 6 - 2y)?
    Solution
    Since (2x - 9, 4) = (3, 6 - 2y),
    2x - 9 = 3
    2x = 9 + 3
    x = = 6
    Similarly,
    6 - 2y = 4
    -2y = 4 - 6
    -2y = -2
    y = = 1
  • Question 7
    1 / -0
    What is the value of y in the ordered pair (6, y), if the abscissa of the pair is 2 more than the ordinate?
    Solution
    Abscissa = 2 + ordinate
    An ordered pair consists of two terms - the abscissa (horizontal, usually x) and the ordinate (vertical, usually y).
    6 = 2 + ordinate (y)
    y = 4
  • Question 8
    1 / -0
    The point (-2, 4) lies in
    Solution
    The point (-2, 4) lies in the 2nd quadrant as shown below.

  • Question 9
    1 / -0
    The distance from a point to the vertical or y-axis, measured parallel to the horizontal or x-axis is called
    Solution
    The distance from a point to the vertical or y-axis, measured parallel to the horizontal or x-axis is called abscissa.

  • Question 10
    1 / -0
    If the coordinates of two points are A(6, 7) and B(-3, 4), then (Ordinate of A) - (Ordinate of B) is
    Solution
    Point A(6,7) has ordinate as 7 and point B(-3, 4) has ordinate as 4.
    Thus, (Ordinate of A) - (Ordinate of B) = 7 - 4 = 3
  • Question 11
    1 / -0
    The signs of abscissa and ordinate of a point in quadrant III, respectively, are
    Solution
    The signs of abscissa and ordinate of a point in quadrant III are (-, -), respectively.
  • Question 12
    1 / -0
    Two points having different abscissa but same ordinates lie on
    Solution
    Two points having different abscissa but same ordinates lie on a line parallel to x axis which are indicated as points M and N in the picture given below.

  • Question 13
    1 / -0
    Study the graph and answer the question that follows.



    What are the coordinate of point C?
    Solution
    From the graph, the coordinates of point C are (-6, -4).
  • Question 14
    1 / -0
    Study the graph and answer the question that follows.



    (Abscissa of A) - (Abscissa of B) is
    Solution
    Abscissa of point A = 4
    Abscissa of point B = -6
    According to the question:
    Abscissa of A - Abscissa of B = 4 - (-6) = 4 + 6 = 10
  • Question 15
    1 / -0
    Study the graph and answer the question that follows.



    The point whose abscissa becomes zero when it is decreased by 2 is
    Solution
    The coordinates of D are (2, -6). When abscissa is decreased by 2 units, the new coordinates are (0, -6).
  • Question 16
    1 / -0
    Study the graph and answer the question that follows.


    What is the difference between the ordinates of A and D?


    Solution

    The coordinates of A are (4 , 4) and the coordinates of D are (2 , -6).
    So, the required difference is 4 - (-6) = 4 + 6 = 10
  • Question 17
    1 / -0
    What is the area of the triangle formed by the points A(0, 2), B(0, 8) and C(5, 5)?
    Solution


    Area of triangle ABC == = sq.units
  • Question 18
    1 / -0
    The perpendicular distance of (-4, 5) from x-axis is
    Solution

    Distance of point (-4, 5) from x-axis (-4,0) =

    =

    =

    = 5
  • Question 19
    1 / -0
    The point (0, 5) lies
    Solution
    The point (0, 5) lies on the y axis, as abscissa is 0 here and ordinate is 5.
  • Question 20
    1 / -0
    The signs of the abscissa and ordinate of a point in the second quadrant respectively are
    Solution
    The horizontal (x) value in a pair of coordinates is abscissa
    The vertical (y) value in a pair of coordinates is ordinate.



    So, in the quadrant 2nd: abscissa is negative and ordinate is positive.
  • Question 21
    1 / -0
    State 'T' for true and 'F' for false in the following statements.

    (i) The coordinates of a point lying on x axis at a distance of 3 units from y axis are (3, 0).
    (ii) The ordinate of a point is positive in quadrant 4 and 2.
    (iii) The point (-4, 4) would lie in the 4th quadrant.
    (iv) The vertical (y) value in a pair of coordinates is 'ordinate'.
    Solution
    (i) The coordinates of a point lying on x axis at a distance of 3 units from y axis are (3, 0).
    This is a true statement as abscissa should be 3 here because the point is on the x axis and ordinate would be 0. (T)

    (ii) The ordinate of a point is positive in quadrant 4 and 2.
    This is not a true statement as ordinates in quadrant 2 are positive but ordinates are negative in quadrant 4. (F)



    (iii) The point (-4, 4) would lie in the 4th quadrant.
    This is a false statement as abscissa is -4 here and ordinate is 4. So, it would lie in the 2ndquadrant, and not in the 4th. (F)



    (iv) The vertical (y) value in a pair of coordinates is 'ordinate'.
    This is a correct statement. (T)

    So, the answer will be: T, F, F, T.
  • Question 22
    1 / -0
    Fill in the blanks.

    (1) Point X is 4 spaces left and two spaces above the point (3, 2). So, point X lies in quadrant A .
    (2) Point X is 10 spaces right and 8 spaces below the point (-12, 11). So, coordinates of point X are B .
    (3) Point X is 8 spaces left and 4 spaces below the point (10, 1). So, point X lies in quadrant C .
    (4) Point X is 35 spaces right and 25 spaces above the point (-40, -19). So, coordinates of point X are D .
    Solution
    (1) Since point X is 4 spaces left and two spaces above the point (3, 2), coordinates of point X = (3 - 4, 2 + 2) = (-1, 4). Therefore, point X lies in quadrant II.
    (2) Since point X is 10 spaces right and 8 spaces below the point (-12, 11), coordinates of point X = (-12 + 10, 11 - 8) = (-2, 3). Therefore, coordinates of point X are (-2, 3).
    (3) Since point X is 8 spaces left and 4 spaces below the point (10, 1), coordinates of point X = (10 - 8, 1 - 4) = (2,-3). Therefore, Point X lies in quadrant IV.
    (4) Since point X is 35 spaces right and 25 spaces above the point (-40, -19), coordinates of point X = (-40 + 35, -19 + 25) = (-5, 6). Therefore, coordinates of point X are (-5, 6).
  • Question 23
    1 / -0
    In the given figure, ABCD is a rhombus whose diagonals AC and BD are along coordinate axis, and AC = 10 units and BD = 8 units. Now, if P is a point which is 4 spaces left and 1 space above D, find

    (1) sum of ordinates of B, C and P
    (2) sum of abscissa of A and P

    Solution

    From the given information,Coordinates of point P are (-4, 5).
    (1) Ordinate of B = -4
    Ordinate of C = 0
    Ordinate of P = 5
    Therefore, required sum = -4 + 0 + 5 = 1
    (2) Abscissa of A = -5
    Abscissa of P = -4
    Therefore, required sum = -5 - 4 = -9
  • Question 24
    1 / -0
    In the given figure, ABCD is a rectangle with its centre at origin, length AB = 10 units, and breadth BC = 6 units.
    Find coordinates of all of its vertices.

    Solution

  • Question 25
    1 / -0
    Match the following:

    Column - I Column - II
    (X) The area of △OPQ with O(0, 0), P(2, 0) and Q(0, 4) is (1) 15 sq. units
    (Y) The area of △PQR with P(4, 0), Q(6, 0) and R(2, 6) is (2) 4 sq. units
    (Z) The area of △OPQ with O(0, 0), P(5, 0) and Q(0, 6) is (3) 6 sq. units
    Solution
    (X) Area of △OPQ = (1/2) × OP × OQ = (1/2) × 2 × 4 = 4 sq. units


    (Y) Area of △PQR = × 2 × 6 = 6 sq. units

    (Z) Area of △OPQ = × 5 × 6 = 15 sq. units

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