How do you write an equation of a line passing through (2, 6), perpendicular to #2x -3y = 12#?

1 Answer
Jan 21, 2017

Answer:

#3x+2y=18#

Explanation:

#color(red)"THINGS TO REMEMBER"#

#color(red)(bar(color(white)("XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX")#

#color(blue)("If a line has a slope of "m)#
#color(white)("XXXX")color(blue)("all lines perpendicular to it have a slope of "-1/m)#

#color(red)(bar(color(white)("XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX")#

#color(white)("XX")color(blue)("A line in standard form: "Ax+BY=C)color(white)("XX")#
#color(white)("XXXX")color(blue)("has a slope of "-A/B)#

#color(red)(bar(color(white)("XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX")#

#color(white)("XX")color(blue)("The point-slope form for a line with")#
#color(white)("XX")color(blue)("point "(a,b)" and slope "m" is"color(white)(*XXX"))#
#color(white)("XXXXXX")color(blue)(y-a=m(x-b))#

#color(red)(bar(color(white)("XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX")#

#color(green)("SOLVING THE GIVEN PROBLEM")#

#2x-3y=12# has a slope of #m=2/3#

All lines perpendicular to #2x-3y=12# have a slope of #-3/2#

The equation of a line with slope #(-3/2)# through the point #(2,6)#
in slope-point form is:
#color(white)("XXXXXX")y-6=-3/2(x-2)#

#color(green)("~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~")#

While the above slope-point form would be a valid answer,
we would normally convert this into standard form:

#y-6=-3/2(x-2)#

#color(white)("XXX")rarr 2y-12=-3x+6#

#color(white)("XXX")rarr 3x+2y=18#

#color(green)(bar(color(white)("XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX")#