How do you write an equation of a line passing through (-8, -5), perpendicular to #x + 4y = 6#?

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Answer 1 · 2017-07-13T19:08:52

I got: #4x-y=-27#

We can write our line in the slope-intercept form #y=mx+c# so that we can "see" the slope #m#:

#y=-1/4x+6/4#

#y=-1/4x+3/2#

so that the slope will be #m=-1/4#

The perpendicular to this line will have a slope:

#m'=-1/m=4#

we then use the general form for a line through a point of coordinates #(x_0,y_0)# and slope #m'# as:

#y-y_0=m'(x-x_0)#

or:

#y-(-5)=4(x-(-8))#

#y=4x+32-5#

#y=4x+27#

or

#4x-y=-27#