How do you write an equation for the line in slope intercept form that is perpendicular to the given line and that passes through the given point 3x – 12y = –9; (1, –10)?

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Answer 1 · 2016-06-12T16:59:12

#y=-4x-6#.

The slope of the given line is (using its standard eqn. #ax+by+c=0#)
#-a/b=-(3/-12)=1/4.#

Hence, the slope of reqd. perpendicular line #=-1/(1/4)=-4.#

The rqd. line is thro. #(1,-10)#.

Therefore, using slope-point form of reqd. line, is, #y-(-10)=-4(x-1).#.
#:. y+10=-4x+4#, or, #y=-4x-6.# (slope-intercept form)

Otherwise

A sort-cut method to find the eqn. of a line perpendicular to a given line #ax+by+c=0# & passing thro. #(x_1,y_1)# is #b(x-x_1)-a(y-y_1)=0.#

In our case, it is given by #-12(x-1)-3{y-(-10)}=0.#
#:. y+10=-4(x-1)=-4x+4#, or, #y=-4x-6#.