What is the equation of the line that is perpendicular to #5y+3x=8# and passes through #(4, 6)#?

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Answer 1 · 2016-09-21T16:58:20

Equation of the line that is perpendicular to #5y+3x=8# and passing through #(4.6)# is #5x-3y-2=0#

Writing the equation of line #5y+3x=8#, in slope intercept form of #y=mx+c#

As #5y+3x=8#, #5y=-3x+8# or #y=-3/5x+8/5#

Hence slope of line #5y+3x=8# is #-3/5#

and slope of line perpendicular to it is #-1-:-3/5=-1xx-5/3=5/3#

Now equation of line passing through #(x_1,y_1)# and slope #m# is

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

and hence equation of line passing through #(4,6)# and slope #5/3# is

#(y-6)=5/3(x-4)# or

#3(y-6)=5(x-4)# or

#3y-18=5x-20# or

#5x-3y-2=0#