How do you write the equation of line passes through (4, -5), and is perpendicular to 2x-5y= -10?

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Answer 1 · 2016-09-02T20:51:20

#5x+2y=10#

Let us write the equation of line ##2x-5y=-10# in slope intercept form i.e.

#-5y=-2x-10# or #y=2/5x+2#

i.e. its slope is #2/5#. Hence the slope of a line perpendicular to it is #(-1)/(2/5)=-5/2#.

Now equation of a line passing through appoint #(x_1,y_1)# and having a slope #m# is given by

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

so the equationn of line passing through #(4,-5)# and having a slope of #-5/2# is

#(y-(-5))=-5/2(x-4)# or

#y+5=-5/2x+(5×4)/2# or

#y+5=-5/2x+10# or

#2y+10=-5x+20# or

#5x+2y=10#