What is the equation of the line that is perpendicular to #2y=-6x+8# if its #y#-intercept is #5#?

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Answer 1 · 2016-07-26T09:49:36

#y=1/3x+5#

Given -

#2y=-6x+8#

#y=(-6)/2 x+8/2#
#y=-3x+4#

The slope of this line is #m_1=-3#

Another line is passing through #(0, 5)#
This line is perpendicular to the line #y=-3x+4#

Find the slope of the other line -

#m_2# is the slope of the other line.

For two lines to be perpendicular -

#m_1 xx m_2 = -1#

Then

#m_2=(-1)/(-3)=1/3#

The equation is

#y=mx+c#
#y=1/3x+5#

Look at the diagram