Question

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

Answer 1

`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)`

Answer 2

`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"`