Line K has a slope of 3. Line J is perpendicular to like K and passes through the point (3,8). Type the equation of the line in y=Mx+b

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Line K has a slope of 3. Line J is perpendicular to like K and passes through the point (3,8). Type the equation of the line in y=Mx+b

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Quinn 1 week 2022-01-08T19:16:32+00:00 1 Answer 0 views 0

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    2022-01-08T19:17:49+00:00

    I’m assuming you are finding the equation of the line of Line J

    Slope-intercept form:  y = mx + b

    [m is the slope, b is the y-intercept or the y value when x = 0 –> (0, y) or the point where the line crosses through the y-axis]

    For lines to be perpendicular, the slopes have to be negative reciprocals of each other. (basically changing the sign (+/-) and flipping the fraction/switching the numerator and the denominator)

    For example:

    slope = 2 or  \frac{2}{1}

    Perpendicular line’s slope = -\frac{1}{2}    [changed sign from + to -, and flipped the fraction]

    slope = -\frac{2}{5}

    Perpendicular line’s slope = \frac{5}{2}   [changed sign from – to +, and flipped the fraction]

    Since you know the slope is 3, the perpendicular line’s slope is -\frac{1}{3}.  Plug this into the equation

    y = mx + b

    y=-\frac{1}{3} x+b    To find b, plug in the point (3, 8)

    8=-\frac{1}{3}(3)+b

    8 = -1 + b    Add 1 on both sides to get “b” by itself

    8 + 1 = -1 + 1 + b

    9 = b

    y=-\frac{1}{3} x+9

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45:7+7-4:2-5:5*4+35:2 =? ( )