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

6x+3y=16x+3y=1

Explanation:

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