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

2 Answers
Mar 31, 2018

3x-2y-15=0

Explanation:

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)

Mar 31, 2018

y=3/2x-15/2

Explanation:

"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"