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Algebra

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Still need help on this one

Drawing a blank on this one, not sure where to start.

Find the equation, in standard form, with all integer coefficients, of the line perpendicular to x + 3y = 12 and passing through (9, -5).

  • Algebra - ,

    Very easy
    Recall that the slopes of perpendicular lines have slopes that are negative reciprocals of each other, or in other words, the product of their slopes is -1

    Your first line is x + 3y = 12

    So the perpendicular line must be

    3x - y = c , the only thing we don't know is the constant
    But (9,-5) lies on this new line
    so 3(9) - (-5) = c
    c = 32

    The equation must be 3x - y = 32

  • Algebra - ,

    I got something totally different -
    x +3y = 12
    x - x + 3y = 12 - x
    3y = -x + 12
    3y/3 = -x/3 + 12/3
    y = -3x + 4
    y - -5 = 1/3(x - 9)
    y + 5 = 1/3x - 3
    y + 5 - 5 = 1/3x - 3 - 5
    y = 1/3x - 8

  • Algebra - ,

    no!
    the slope of the given line was -1/3, the slope of your new line is +1/3
    So they are not even perpendicular!!

    from your line:
    3y/3 = -x/3 + 12/3 so far so good
    now
    y = (-1/3)x + 4
    so the new slope must be +3

    y + 5 = 3(x - 9)
    y + 5 = 3x - 27
    y = 3x - 32, which when rearranged is the same as 3x - y = 32

    let me show you another approach

    the slope of the given line was -1/3,

  • Algebra - ,

    (continued)

    so the slope of the new line must be +3

    then the new equation is y = 3x + b
    but (9,-5) lies on it
    so -5 = (3)(9) + b
    b = -32

    for y = 3x - 32

  • Algebra - ,

    Thank you so much.

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