Question

How do you write an equation of a line perpendicular to `y=-1/2x+2/3` and passes through (2,3)?

Answer 1

`y=2x-1`

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"`

`y=-1/2x+2/3" is in this form"`

`"with slope m "=-1/2`

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

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

`rArrm_("perpendicular")=-1/(-1/2)=2`

`rArry=2x+blarrcolor(blue)"is the partial equation"`

`"to find b substitute "(2,3)" into the partial equation"`

`3=4+brArrb=3-4=-1`

`rArry=2x-1larrcolor(red)"equation of perpendicular line"`

Answer 2

See below

Explanation:

If `y=mx+b` is the line equation, we call Slope to value `m` and y-intercept to value `b`

We know then that `m´=-1/m` is the slope of perpendiclar line to first. Thus we have

`m'=-1/(-1/2)=2`

Then, our perpendicular line is `y=2x+b`. Lets calculate b.

However `P(color(blue)2,color(red)3)` is a point belonging to this new line, must be

`color(red)3=2·color(blue)2+b`, then `b=-1`

The perpendicular line requested passing thru `P` is `y=2x-1`