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Coordinate Geometry Test - 2

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Coordinate Geometry Test - 2
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  • Question 1
    3 / -1

    Distance of a point from the origin:

    The distance of a point A(x, y) from the origin O(0, 0) is given by OA = √x2 + y2

    Find the distance of the point A(4, -2) from the origin. 

    Solution

    OA = √4 - 02+(-2 - 0)2 = √16+4 = √20 = radic;4*5 = 2√5 units

  • Question 2
    3 / -1

    Distance between two points :

    If (x1, y1) and B(x2, y2) be two points, then AB = √(x2 - x1)2 + (y2 - y1)2

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

     

    Solution

    AB = √(2+4)2 + (-5-7)2

    = √62 + (-12)2

    = √36+144 = √180

    =√36*5 = 6√5 units.

  • Question 3
    3 / -1

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

    Solution

    AB = √(b-0)2-(0-a)2 

    = √b2+a2 

    = √a2+b2.

  • Question 4
    3 / -1

    If the distance of the point P(x, y) from A(a, 0) is a + x, then y2 = ? 

    Solution

    √(x-a)2+(y-0)2 = a + x 

    = (x-a)2+y2 

    = (a+x)2 => y2 = (x-a)2-(x-a)2-4ax => y2 = 4ax

  • Question 5
    3 / -1

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

    Solution

    AB2 = (2 - 5)2 + (3 + 7)2 

    => (-3)2 + (10)2 

    => 9 + 100 => √109

  • Question 6
    3 / -1

    Area of a triangle : 

    If A(x1,y1), B(x2,y2 and C(x3, y3) be three vertices of a ΔABC, then its area is given by:

    Δ = 1/2 [x1(y2 - y3 + x2(y3 - y1) + x3(y1 - y2)]

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

    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 

  • Question 7
    3 / -1

    Find the area of ΔABC whose vertices are A(2, -5), B(4, 9) and (6, -1). 

    Solution

    Here, x1 = 2, x2 = 4, x3 = 6 and y1 = -5, y2 = 9, y3 = -1

    = 1/2 [2(9+1) + 4(-1+5) + 6(-5-9)] 

    = 1/2 [2(10) + 4(4) + 6(-14)] 

    = 1/2 [20 + 16 - 84] 

    = 1/2 [-48] 

    = 24 sq.units

  • Question 8
    3 / -1

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

    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. ==> ΔABC is isosceles.

  • Question 9
    3 / -1

    The points A(-4, 0), B(1, -4), and C(5, 1) are the vertices of 

    Solution

    AB2 = (1 + 4)2 + (-4 - 0)2 

    = 25 + 16 = 41, 

    BC2 = (5 - 1)2 + (1 + 4)2 = 42 + 52

    = 16 + 25 = 41 

    AC2 = (5 + 4)2 + (1 - 0)2 

    = 81 + 1 = 82 

    AB = BC and AB2 = BC2 = AC2 

    ΔABC is an isosceles right angled triangle

  • Question 10
    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)

    we have to find the coordinates of the point equidistant from the points.

    The point that is equidistant from three points is called circumcenter which can be evaluated to find the perpendicular bisectors.

    To find the perpendicular bisectors of AB: 


    (11,2)

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