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20 August, 05:30

What is the equation of the line that is perpendicular to y=2/3x-5 and passes through (6,-1) ?

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  1. 20 August, 09:12
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    Answer: 2y + 3x - 16 = 0

    Step-by-step explanation:

    Equation of the line is y = 2x/3 - 5

    From the equation, the slope m₁ = 2/3. therefore recall, from condition for perpendicularity, m₁m₂ = - 1

    The product of their gradient must be (-1).

    Now since m₁ = 2/3 and m₁m₂ = -1

    2m₂/3 = - 1

    Therefore, m₂ = - 3/2.

    Since the equation passes through the coordinate of (6,-1)

    we now substitute for x and y in the equation of a line to get the y intercept (c)

    y = mx + c

    -1 = - 3x/2 + c

    -1 = - 3/2 x 6 + c

    -1 = - 9 + c

    -1 + 9 = c

    Therefore c = 8

    Now to get the equation of line that is perpendicular to y = 2x/3 - 5

    y = - 3x/2 + 8

    making it a linear equation,

    2y = - 3x + 16

    2y + 3x - 16 = 0
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