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

2 Answers
Mar 23, 2018

Answer:

#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"#

Mar 23, 2018

Answer:

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#