How do you find the equation of a line perpendicular to 2x+3y=9 and passing through (3,-3)?

We know that,

the slope of the line #ax+by+c=0# is #m=-a/b#

#:.# The slope of the line #2x+3y=9# is #m_1=-2/3#

#:.# The slope of the line perpendicular to #2x+3y=9#

is #m_2=-1/m_1=-1/(-2/3)=3/2#.

Hence,the equn.of line passing through #(3,-3) and m_2=3/2# is

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

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

#=>2y+6=3x-9#

#=>3x-2y-15=0#

Note:

The equn.of line passing through #A(x_1,y_1) and #with slope
#m# is

#y-y_1=m(x-x_1)#

#"the equation of a line in "color(blue)"slope-intercept form"# is.

#•color(white)(x)y=mx+b#

#"where m is the slope and b the y-intercept"#

#"rearrange "2x+3y=9" into this form"#

#rArr3y=-2x+9#

#rArry=-2/3x+3larrcolor(blue)"in slope-intercept form"#

#"with slope m "=-2/3#

#"Given a line with slope then the slope of a line"#
#"perpendicular to it is"#

#•color(white)(x)m_(color(red)"perpendicular")=-1/m#

#rArrm_("perpendicular")=-1/(-2/3)=3/2#

#rArry=3/2x+blarrcolor(blue)"is the partial equation"#

#"to find b substitute "(3,-3)" into the partial equation"#

#-3=9/2+brArrb=-6/2-9/2=-15/2#

#rArry=3/2x-15/2larrcolor(red)"equation of perpendicular line"#