find the equation of a line that is perpendicular to the line y=1/6x+8 and contains the point (-4,0)?

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  • Anonymous
    7 years ago
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    Two lines are said to be parallel if their slopes are equal i.e., m1=m2

    Two lines are said to be perpendicular if their slopes are negative reciprocals of each other i.e., m1= -1/m2

    Given equation is y=1/6x+8

    By comparing it with y=mx+c we get the slope 'm1'=1/6

    A line perpendicular to the above equation will have 'm2' as m2= -1/m1

    m2= -1 / (1/6)

    We get m2= -6

    Slope-intercept formula is y=mx+c

    To get the value of intercept, substitute m2= -6, x= -4 and y= 0- - -[x&y values are given as (-4,0)]

    - - -> 0= -6*-4 + c

    0= 24+c

    c= -24 (Intercept value)

    Now substitute m and c values in slope-intercept form

    y= -6x-24 ( Your Answer )

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