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

### 1 Answer

- Anonymous7 years agoFavorite Answer
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 )