How do you find the general form of the line perpendicular to 3x+5y-8=0 that passes through the point (-8,1)?
2 Answers
Therefore equation of the perpendicular line is:
Explanation:
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)
Explanation:
#"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"#