How do you write an equation of a line passing through (0, 2), perpendicular to #y = 4x – 3#?

2 Answers

Answer:

#x+4y-8=0#

Explanation:

The slope of line: #y=4x-3\equiv y=mx+c# is #4#.

Hence the slope #m# of the line perpendicular to the given line: #y=4x-3#

#m=-1/{4}#

Now, the equation of line passing through the point #(x_1, y_1)\equiv(0, 2)# & having a slope #m=-1/4# is given as

#y-y_1-m(x-x_1)#

#y-2=-1/4(x-0)#

#4y-8=-x#

#x+4y-8=0#

Jul 21, 2018

Answer:

#y=-(1/4)x+2#

Explanation:

First, I'd start with the slope of the new equation. We know that this unknown equation is perpendicular to #y=4x-3#. Since it is perpendicular, the slope of the equation has to be the reciprocal and negative, giving us the new slope of #-1/4#.

So to recap, so far we have #y=-(1/4)x+b#, with #b# representing the #y#-intercept.

We know the #y#-intercept is #2# because #(0,2)# is a coordinate on the #y#-axis, but I will also solve for it, so it is more clear.

We can find the #y#-intercept by plugging in the coordinates that we are given #(0,2)# into this new equation and solving for #b#.

#2=-(1/4)(0)+b#

#2=b#

Now that we have the #y#-intercept, we can put together an equation.
#y=-(1/4)x+2#