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If the lines 3x + by + 5 = 0 and ax – 5y + 7 = 0 are perpendicular to each other, find the relation connecting a and b.

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ICSE Board Question Based on Equation of a Straight Line Chapter of M.L Aggarwal for class10
In this question given lines are perpendicular to each other so you have to find the relation connecting a and b.
This is the Question Number 07, Exercise 12.2 of M.L Aggarwal.

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  1. Given that the lines 3x + by + 5 = 0 and ax – 5y + 7 = 0 are perpendicular to each other

    Then the product of their slopes must be -1.

    Slope of line 3x + by + 5 = 0 is,

    by = -3x – 5

    y = (-3/b) – 5/b

    So, slope (m1) = -3/b

    And,

    The slope of line ax – 5y + 7 = 0 is

    5y = ax + 7

    y = (a/5) x + 7/5

    So, slope (m2) = a/5

    As the lines are perpendicular, we have

    m1 x m2 = -1

    -3/b x a/5 = -1

    -3a/5b = -1

    -3a = – 5b

    3a = 5b

    Hence, the relation connecting a and b is 3a = 5b.

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