How do you write the equation that represents the line perpendicular to #y=-3x +4# and passing through the point (-1,1)?

2 Answers
Dec 12, 2016

#y = 1/3x + 4/3#

Explanation:

Step 1) Because this is already solved for #y# we can get the slope of the perpendicular line by taking the slope of the given line (-3) and "flipping" it and reversing the sign:

#-3# becomes #- -1/3 -> + 1/3 -> 1/3#

Therefore the slope of the perpendicular line is #1/3#

Step 2) Use the point slope for to find the perpendicular equation:

#y - 1 = 1/3(x - -1)#

#y - 1 = 1/3(x + 1)#

#y - 1 = 1/3x + 1/3#

#y - 1 + 1 = 1/3x + 1/3 + 1#

#y - 0 = 1/3x + 1/3 + 3/3#

#y = 1/3x + 4/3#

Dec 12, 2016

#y=(1/3)x+(4/3)#

Explanation:

Any linear equation of the form
#color(white)("XXX")y=color(green)mx+color(blue)b#
has a slope of #color(green)m#

Any line perpendicular to a line with slope #color(green)m#
has a slope #color(green)(""(-1/m))#

#y=color(green)(-3)x+4# has a slope of #color(green)(-3)#
#rarr# any line perpendicular to it has a slope #color(green)(1/3)#

A line with slope #color(magenta)k# through the point #(color(red)a,color(blue)b)#
can be written in slope-point form as:
#color(white)("XXX")(y-color(blue)b)=color(magenta)k(x-color(red)a)#

Therefore the line perpendicular to #y=-3x+4# and through the point #(color(red)(-1),color(blue)1)#
can be written as:
#color(white)("XXX")y-color(red)1=color(green)(1/3)(x-color(red)(""(-1)))#

While this is a perfectly valid answer to the given question, we would normally convert this into standard form as:
#color(white)("XXX")x-3y=-4#
or slope-vertex form (as was the initial equation):
#color(white)("XXX")y=1/3x+4/3#