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

`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")`