How do you write the standard form of a line given (1/2 , -2/3 ); perpendicular to the line 4x − 8y = 1?

1 Answer
Feb 16, 2018

Answer:

#6x+3y=1#

Explanation:

#"the equation of a line in "color(blue)"standard form"# is.

#color(red)(bar(ul(|color(white)(2/2)color(black)(Ax+By=C)color(white)(2/2)|)))#

#"where A is a positive integer and B, C are integers"#

#"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 "4x-8y=1" into this form"#

#-8y=-4x+1rArry=1/2x-1/8tom=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_(color(red)"perpendicular")=-1/(1/2)=-2#

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

#"to find b substitute "(1/2,-2/3)" into the partial"#
#"equation"#

#-2/3=-1+brArrb=-2/3+1=1/3#

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

#"multiply through by 3 to eliminate fraction"#

#rArr3y=-6x+1#

#rArr6x+3y=1larrcolor(red)"in standard form"#