Find the equation of line perpendicular to #y=-5/3x-8# and passing through #(-9,-3)#?

1 Answer
Oct 12, 2016

Answer:

#3x-5y+12=0#

Explanation:

The two lines are perpendicular to each other, if the product of their slopes is #-1#. Hence if slope of one line is given as #a/b#, slope of line perpendicular to it is #-b/a#.

As equation of one line is given in slope-intercept form as #y=-5/3x-8#, its slope is #-5/3#.

Hence slope of line perpendicular to it is #3/5#.

Now equation of line passing through #(x_1,y_1)# and having slope #m# is #(y-y_1)=m(x-x_1)#

Hence, equation of a line passing through #(-9,-3)# and having a slope #3/5# is

#(y-(-3))=3/5(x-(-9))#

or #5(y+3)=3(x+9)#

or #5y+15=3x+27#

or #3x-5y+12=0#
graph{(3x-5y+12)(y+5x/3+8)((x+9)^2+(y+3)^2-0.05)=0 [-20, 20, -10, 10]}