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"