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

Question

Eli Assouline · Accepted Answer

#3x-2y-15=0#

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)#

Abigail Allen · Answer

#y=3/2x-15/2# #"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"#

Eva Adamson · Answer

undefined First, we need to find the slope of the given line by rearranging it into slope-intercept form (y = mx + b), where m is the slope. Then, we find the negative reciprocal of this slope to get the slope of the perpendicular line. Finally, we use the point-slope form of a linear equation (y - y1 = m(x - x1)) and substitute the given point (3,-3) and the slope to find the equation of the perpendicular line.