How do you find the equation of a line that is perpendicular to #y=2/3x-4# and goes through point (6,-2)?

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Answer 1 · 2015-01-29T20:51:53

The answer is: #y=-3/2x+7#

The equation of a line, written in explicit form is:

#y=mx+q#

where

#m# is the slope of the line and
#q# is the y-intercept.

The formula that gives the equation of a line #r# if its slope, #m_r#, and a point #P# that #in r# are known is:

#y-y_P=m_r(x-x_P)#

Two lines #r and s# are perpendicular if #m_r=-1/m_s#.

So the slope of the requested line is: #m_r=-1/(2/3)=-3/2#.

The equation of the line is:

#y-(-2)=-3/2(x-6)rArry+2=-3/2x+9rArry=-3/2x+7#.