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Deepak Bora
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The equation of a line is 3x + 4y – 7 = 0. Find (i) the slope of the line. (ii) the equation of a line perpendicular to the given line and passing through the intersection of the lines x – y + 2 = 0 and 3x + y – 10 = 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 given an equation of a line.
So , Find the the slope of the line and the equation of a line perpendicular to the given line and passing through the intersection of the given lines.
This is the Question Number 22, Exercise 12.2 of M.L Aggarwal.

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  1. Given line equation: 3x + 4y – 7 = 0

    (i) Slope of the line is given by,

    4y = -3x + 7

    y = (-3/4) x + 7

    Hence, slope (m1) = -3/4

    (ii) Let the slope of the perpendicular to the given line be m2

    Then, m1 x m2 = -1

    (-3/4) x m2 = -1

    m2 = 4/3

    Now, to find the point of intersection of

    x – y + 2 = 0 … (i)

    3x + y – 10 = 0 … (ii)

    On adding (i) and (ii), we get

    4x – 8 = 0

    4x = 8

    x = 8/4 = 2

    Putting x = 2 in (i), we get

    2 – y + 2 = 0

    y = 4

    Hence, the point of intersection of the lines is (2, 4)

    The equation of the line having slope m2 and passing through (2, 4) will be

    y – 4 = (4/3) (x – 2)

    3y – 12 = 4x – 8

    4x – 3y + 4 = 0

    Thus, the required line equation is 4x – 3y + 4 = 0.

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