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Coordinate Geometry Test - 1

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Coordinate Geometry Test - 1
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  • Question 1
    3 / -1

    Q. 

    We divided the plane of the paper into four equal parts. by drawing two mutually perpendicular lines, X'OX and YOY'. These lines are called the axes. Here X'OX is called x-axis and YOY' is called y-axis. There axes divide the plane of the paper into four parts, called quadrants.

    The position of a point in a plane is denoted by an ordered pair (a,b), where a is called the x co-ordinate and y is called y co-ordinate.

    In which quadrant does the point(-4, -7) lie? 

    Solution

    The point (-4, -7) lies in 3rd quadrant. 

  • Question 2
    3 / -1

    In which quadrant does the point(1, 5) lie? 

    Solution

    The point (1, 5) lies in 1st quadrant.

  • Question 3
    3 / -1

    In which quadrant does the point(9, -2) lie?

    Solution

    The point (9, -2) lies in 4th quadrant.

  • Question 4
    3 / -1

    In which quadrant does the point(-7, 6) lie?

    Solution

    The point (-7, 6) lies in 2nd quadrant.

  • Question 5
    3 / -1

    In which quadrant does the point(0, 9) lie? 

    Solution

    The point (0, 9) lies in y-axis.

  • Question 6
    3 / -1

    In which quadrant does the point(9, 0) lie? 

    Solution

    The point (9, 0) lies in x-axis.

  • Question 7
    3 / -1

    Find the coordinates of the point equidistant from the points A(1, 2), B(3, –4) and C(5, –6).       

    Solution

    Given three points A(1,2) B(3,-4) and C(5,-6).

    To find the perpendicular bisectors of AB:-

    • The slope of AB:
      = -4-2 / 3-1
      = -3
    • The slope of line perpendicular to AB:
      = 1 / 3
    • The mid-point of AB is:
      = (1 + 3 / 2, 2 - 4 / 2)
      = (2, -1)
    • The equation of perpendicular line is:
      y = 1 / 3(x - 2) - 1

    To find the perpendicular bisectors of AC:-

    • The slope of AC:
      = -6 - 2 / 5 - 1
      = -2
    • The slope of line perpendicular to AC:
      = 1 /2
    • The mid-point of AC is:
      = (1 + 5 / 2, 2-6 / 2)
       (3, -2)
    • The equation of perpendicular line is:
      y = 1 / 2(x - 3) - 2

    Now, solve the above two equations:-

    ⇒ 1 / 3(x - 2) - 1 = 1 / 2(x - 3) - 2
    ⇒ 2(x - 2) - 6 = 3(x - 3) - 12
    ⇒ x = 11
    ⇒ y = 1 / 2(x - 3) - 2 = 1 / 2(11 - 3) - 2 = 2

    The coordinates of the points equidistant from the point A(1, 2), B(3, -4) and C(5, -6) are (11, 2)

  • Question 8
    3 / -1

    Find the distance of the point A(3, -3) from the origin. 

    Solution

    OA = √32+(-3)2 = √9+9 = √18 = 3√2

  • Question 9
    3 / -1

    P is a point on x-axis at a distance of 4 units from y-axis to its right. The co-ordinates of P are: 

    Solution

    The co-ordinates of P are A(4, 0)

  • Question 10
    3 / -1

    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:

    Solution

    The co-ordinates of A are A(0, -5)

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