How do you find the equation of the line passing through (3, -1) and perpendicular to 2x+7y=-1?

1 Answer
Apr 29, 2017

Answer:

#y=7/2x-23/2#

Explanation:

The equation of a line in #color(blue)"point-slope form"# is.

#color(red)(bar(ul(|color(white)(2/2)color(black)(y-y_1=m(x-x_1))color(white)(2/2)|)))#
where m represents the slope and # (x_1,y_1)" a point on the line"#

#"Require to know the following fact"#

#"For 2 perpendicular lines with slopes " m_1" and " m_2#

#"then " color(red)(bar(ul(|color(white)(2/2)color(black)(m_1xxm_2=-1)color(white)(2/2)|)))#

#"rearrange " 2x+7y=-1" into slope-intercept form"#

#"that is " y=mx+b" where m is the slope"#

#"subtract 2x from both sides"#

#cancel(2x)cancel(-2x)+7y=-2x-1#

#rArr7y=-2x-1#

#"divide ALL terms by 7"#

#(cancel(7) y)/cancel(7)=-2/7x-1/7#

#rArry=-2/7x-1/7larrcolor(red)" in form y=mx+b"#

#rArr"slope " =m=-2/7#

#rArrm_("perpendicular")=(-1)/(-2/7)=7/2#

#"using " m=7/2" and " (x_1,y_1)=(3,-1)#

#y-(-1)=7/2(x-3)#

#rArry+1=7/2(x-3)larrcolor(red)" in point-slope form"#

#"distribute and simplify"#

#y=7/2x-21/2-1#

#rArry=7/2x-23/2larrcolor(red)" in slope-intercept form"#