How do you find the equation of the line passing through (3, -1) and perpendicular to 2x+7y=-1?
1 Answer
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"#
#y=7/2x-21/2-1#
#rArry=7/2x-23/2larrcolor(red)" in slope-intercept form"#
