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Deepak Bora
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Point P (3, – 5) is reflected to P’ in the x- axis. Also P on reflection in the y-axis is mapped as P”. (i) Find the co-ordinates of P’ and P”. (ii) Compute the distance P’ P”. (iii) Find the middle point of the line segment P’ P”. (iv) On which co-ordinate axis does the middle point of the line segment P P” lie ?

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ICSE Board Question Based on Section Formula Chapter of M.L Aggarwal for class10
In this question given a point is reflected to another point in the x-axis.Find the solution of the question
This is the Question Number 07, Exercise 11 of M.L Aggarwal.

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  1. (i) The image of P(3,-5) when reflected in X-axis will be (3,5).

    When you reflect a point across the X-axis, the x-coordinate remains the same,

    but the y-coordinate is transformed into its opposite (its sign is changed).

    Co-ordinates of P’ = (3,5)

    Image of P(3,-5) when reflected in Y axis will be (-3,-5).

    When you reflect a point across the Y-axis, the y-coordinate remains the same,

    but the x-coordinate is transformed into its opposite (its sign is changed)

    Co-ordinates of P’’ = (-3,-5)

    (ii)Let P’(x1, y1) and P’’(x2 , y2) be the given points

    By distance formula d(P’,P’’) = √[(x2-x1)2+(y2-y1)2]

    Co-ordinates of P’ = (3,5)

    Co-ordinates of P’’ = (-3,-5)

    Here x1 = 3, y1 = 5 , x2 = -3, y2 = -5

    d(P’,P’’) = √[(-3-3)2+(-5-5)2]

    = √[(-6)2+(-10)2]

    = √(36+100)

    = √136

    = √(4×34)

    = 2√34

    Hence the distance between P’ and P’’ is 2√34 units.

    (iii) Co-ordinates of P’ = (3,5)

    Co-ordinates of P’’ = (-3,-5)

    Here x1 = 3, y1 = 5 , x2 = -3, y2 = -5

    Let Q(x,y) be the midpoint of P’P’’

    By midpoint formula,

    x = (x1+x2)/2

    y = (y1+y2)/2

    x = (3+-3)/2 = 0/2 = 0

    y = (5+-5)/2 = 0/2 = 0

    Hence the co-ordinate of midpoint of P’P’’ is (0,0) .

    (iv) Co-ordinates of P = (3,-5)

    Co-ordinates of P’’ = (-3,-5)

    Here x1 = 3, y1 = -5 , x2 = -3, y2 = -5

    Let R(x,y) be the midpoint of PP’’

    By midpoint formula,

    x = (x1+x2)/2

    y = (y1+y2)/2

    x = (3+-3)/2 = 0/2 = 0

    y = (-5+-5)/2 = -10/2 = -5

    So the co-ordinate of midpoint of PP’’ is (0,-5) .

    Here x co-ordinate is zero.

    Hence the point lies on Y-axis.

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