# How do you find the general form of the line perpendicular to 3x+5y-8=0 that passes through the point (-8,1)?

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#"the general form of the equation of a line is"#

#color(red)(bar(ul(|color(white)(2/2)color(black)(Ax+By+C=0)color(white)(2/2)|)))#

#"where A, B and C are integers with A and B non-zero"#

#"obtain the equation in "color(blue)"slope-intercept form"#

#•color(white)(x)y=mx+b#

#"where m is the slope and b the y-intercept"#

#"rearrange "3x+5y-8=0" into this form"#

#"subtract "3x-8" from both sides"#

#rArr5y=-3x+8#

#"divide all terms by 5"#

#rArry=-3/5x+8/5larrcolor(blue)"in slope-intercept form"#

#"with slope m "=-3/5#

#"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/(-3/5)=5/3#

#"now find the equation of the perpendicular line"#

#rArry=5/3x+blarrcolor(blue)"is the partial equation"#

#"to find b substitute "(-8,1)" into the partial equation"#

#1=-40/3+brArrb=3/3+40/3=43/3#

#rArry=5/3x+43/3larrcolor(red)"in slope-intercept form"#

#"rearrange into general form by multiplying all terms by 3"#

#rArr3y=5x+43#

#"subtract "3y" from both sides"#

#rArr5x-3y+43=0larrcolor(red)"in general form"#

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Therefore equation of the perpendicular line is:

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Slope of perpendicular line is negative reciprocal of the original slope of the line.

So given linear equation is:

So the slope of the perpendicular line is

So the equation of the perpendicular line is:

Lets us find

Therefore equation of the perpendicular line is:

We can draw the graph and check as well (see the attached graphs)

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