find the equation of the line that is perpendicularto the line y=1/6x+4 contains the points (-8,0) write in slope intercept form y=

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sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You need to remember the equation that relates the slopes of two perpendicular lines such that:

`m_1*m_2 = -1`

Notice that you may find the slope of the line `y = (1/6)x + 4` , hence `m_1 = 1/6` .

Substituting `1/6`  in equation `m_1*m_2 = -1`  yields:

`(1/6)*m_2 = -1 => m_2 = -6`

You need to write the slope intercept form of equation of the line such that:

`y - 0 = m_2(x - (-8))`

`y = -6(x + 8) => y = -6x - 48`

Hence, evaluating the slope intercept form of equation of the line under the given conditions yields `y = -6x - 48.`

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dlccanada's profile pic

dlccanada | High School Teacher | (Level 1) eNoter

Posted on

given equation,

y=1/6x+4 and passes through the point(-8,0),

you know that, the equation of any line if you have given slope(m) and a point(x1,y1)

will be     y-y1=m(x-x1),

 as the finding equation is perpendicular to the given line y=1/6x+4 (whose slope m=1/6) so the slope of the perpendicular line will negetive reciprocal of the given line  i.e   m=-(6/1)=-6

therefore the equation will be

y-0=-6(x+8)

y=-6x-48

so the equation of the line will be  y=-6x-48 which is perpendicular th the y=1/6x+4  and passes through(-8,0)

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