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Deepak Bora
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If the line x – 4y – 6 = 0 is the perpendicular bisector of the line segment PQ and the co-ordinates of P are (1, 3), find the co-ordinates of Q.

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This question is Based on Equation of a Straight Line Chapter of M.L Aggarwal book for ICSE BOARD for class 10.
Here if a line is the perpendicular bisector of a line segment. and the co-ordinates of P are (1, 3), find the co-ordinates of Q.
This is the Question Number 41, Exercise 12.2 of M.L Aggarwal.

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  1. Given, line equation: x – 4y – 6 = 0 … (i)

    Co-ordinates of P are (1, 3)

    Let the co-ordinates of Q be (x , y)

    Now, the slope of the given line is

    4y = x – 6

    y = (1/4) x – 6/4

    slope (m) = ¼

    So, the slope of PQ will be (-1/m) [As the product of slopes of perpendicular lines is -1]

    Slope of PQ = -1/ (1/4) = -4

    Now, the equation of line PQ will be

    y – 3 = (-4) (x – 1)

    y – 3 = -4x + 4

    4x + y = 7 … (ii)

    On solving equations (i) and (ii), we get the coordinates of M

    ML Aggarwal Solutions for Class 10 Chapter 12 - 17Multiplying (ii) by 4 and adding with (i), we get

    x – 4y – 6 = 0

    16x + 4y = 28

    ——————

    17x = 34

    x = 34/17 = 2

    Putting the value of x in (i)

    2 – 4y – 6 = 0

    -4 – 4y = 0

    4y = -4

    y = -1

    So, the co-ordinates of M are (2, -1)

    But, M is the mid-point of line segment PQ

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

    (x + 1)/2 = 2

    x + 1 = 4

    x = 3

    And,

    (y + 3)/2 = -1

    y + 3 = -2

    y = -5

    Thus, the co-ordinates of Q are (3, -5).

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