How do you find a standard form equation for the line with (5,4) perpendicular to the line 3x+2y=7?

1 Answer
Sep 15, 2017

Answer:

#2x-3y=-2#

Explanation:

#3x+2y=7# has a slope of #-3/2color(white)("xxxx")# see Note 1

All lines perpendicular to #3x+2y=7# have a slope of #2/3color(white)("xxxx")#see Note 2

If such a perpendicular line goes through #(5,4)#
then we can write it equation in slope point form as:
#y-4=2/3(x-5)color(white)("xxxxxxxxxxxxx")#see Note 3

This can be converted into standard form as:
#2x-3y=-2color(white)("xxxxxxxxxxxxxxxx"#see Note 4

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Note 1
A relation in the form #Ax+By=C# has a slope of #-A/B#;
in this case #A=3# and #B=2#

If you are not familiar with this rule, you can convert given relation #3x+2y=7# into slope-intercept form:

#2y=-3x+7#

#y=(-3/2)x+7/2# with slope #(-3/2)# and y-intercept #7/2#

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Note 2
If a line has a slope of #m# then all line perpendicular to it have a slope of #(-1/m)#

In this case #-1/(-3/2)=2/3#

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Note 3

A line with slope #m# through a point #(x_0,y_0)#
has a slope-point form:
#color(white)("XXX")y-y_0=m(x-x_0)#

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Note 4
"standard form" for a linear equation is
#color(white)("XXX")Ax+By=C# with #A, B, C in ZZ, A>=0#

converting #y-4=2/3(x-5)# into this form:
#color(white)("XXX")3(y-4)=2(x-5)#

#color(white)("XXX")3y-12=2x-10#

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

#color(white)("XXX")2x-3y=-2#

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