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Mathematics 12 Online
OpenStudy (anonymous):

WILL FAN AND MEDAL Line l is represented by the equation 2x+y=9 What it the Slope of a line parallel to line l? What is the slope of the line perpendicular to line l?

OpenStudy (anonymous):

first find the slope of \[2x+y=9\] by solving for \(y\) (in one step)

OpenStudy (anonymous):

the slope of the parallel line will be the same the slope of the perpendicular line will be the "negative reciprocal" i.e. flip it and change the sign

OpenStudy (anonymous):

did you get that or no?

OpenStudy (anonymous):

not really, sorry

OpenStudy (anonymous):

you have \[2x+y=9\] and you want the slope the slope of \[y=\color{red}mx+b\] is \(\color{red}m\) so you want to write the equation to look like that so you can see the slope with your eyeballs that is why i said "solve for x"

OpenStudy (anonymous):

i mean "solve for y"

OpenStudy (anonymous):

\[2x+y=9\] subtract \(2x\) from both sides, what do you get?

OpenStudy (anonymous):

you get y = 9 -2x

OpenStudy (anonymous):

right, or \[y=\color{red}{-2}x+9\] making the slope ???

OpenStudy (anonymous):

negative ?

OpenStudy (anonymous):

lol yes it is negative for sure more specifically it is \(-2\)

OpenStudy (anonymous):

got that? it is the number in front of the x

OpenStudy (anonymous):

yeah I got that

OpenStudy (anonymous):

ok then the parallel line has the same slope

OpenStudy (anonymous):

and the perpendicular is -1/2x?

OpenStudy (anonymous):

no

OpenStudy (anonymous):

the slope of the perpendicular line is what you are asked for, a number, no variable it is the "negative reciprocal" you flipped it, but did not change the sign

OpenStudy (anonymous):

i.e. the slope with be the number \(\frac{1}{2}\) the minus sign goes

OpenStudy (anonymous):

1/-2?

OpenStudy (anonymous):

\[-\frac{1}{2}=\frac{-1}{2}=\frac{1}{-2}\] but the negative reciprocal of \(-2\) is just \[\frac{1}{2}\] with no minus sign

OpenStudy (anonymous):

thank you. Mind I tag you in 1 more?

OpenStudy (anonymous):

no not at all

OpenStudy (anonymous):

or you can post it here if you like

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