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Find the image of the point (1, 2) in the line x – 2y – 7 = 0.

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M.L Aggarwal book Important Question of class 10 chapter Based on Equation of a Straight Line for ICSE BOARD.
You have to find the image of the point (1, 2) in the line x – 2y – 7 = 0.
This is the Question Number 40, Exercise 12.2 of M.L Aggarwal.

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  1. This answer was edited.
    1. Given line equation: x – 2y – 7 = 0 … (i)

    ML Aggarwal Solutions for Class 10 Chapter 12 - 16Draw a perpendicular from point P (1, 2) on the line

    Let P’ be the image of P and let its co-ordinates be (x, y)

    The slope of the given line is given as,

    2y = x – 7

    y = (1/2) x – 7

    Slope (m1) = ½

    Let the slope of line segment PP’ be m2

    As PP’ is perpendicular to the given line, product of slopes: m1 x m2 = -1

    So, ½ x m2 = -1

    m2 = -2

    So, the equation of the line perpendicular to the given line and passing through P (1, 2) is

    y – 2 = (-2) (x – 1)

    y – 2 = -2x + 2

    2x + y – 4 = 0 … (ii)

    Let the intersection point of lines (i) and (ii) be taken as M.

    Solving both the line equations, we have

    Multiplying (ii) by 2 and adding with (i)

    x – 2y – 7 = 0

    4x + 2y – 8 = 0

    ———————

    5x – 15 = 0

    x = 15/5 = 3

    Putting value of x in (i), we get

    3 – 2y – 7 = 0

    2y = -4

    y = -4/2 = -2

    So, the co-ordinates of M are (3, -2)

    Hence, its seen that M should be the mid-point of the line segment PP’

    (3, -2) = ((x + 1)/2, (y + 2)/2)

    (x + 1)/2 = 3

    x + 1 = 6

    x = 6 – 1 = 5

    And,

    (y + 2)/2 = -2

    y + 2 = -4

    y = -4 – 2 = -6

    Therefore, the co-ordinates of P’ are (5, -6).

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