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Re: In the xy-coordinate plane shown above, [#permalink]
1
CozmoP wrote:
Ks1859 wrote:
Solution:

Slope of line l =\(\frac{y2-y1}{x2-x1}\)=\(\frac{1-0}{2-0}\)=\(\frac{1}{2}\)

A Slope perpendicular to line l will be negative reciprocal of line l =-2

Thus \(\frac{1}{2}\)>-2

Qty A>Qty B

IMO A



Hope this helps!


Nice.

Conceptually, since the slope of line l is positive, we don't necessarily need to know the value. Just need to know that any parallel line will have a negative slope. Since a negative is always less than positive, A is correct.


Hi there!

Great explanation! Kudos!

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Re: In the xy-coordinate plane shown above, [#permalink]
Ks1859 wrote:
Solution:

Slope of line l =\(\frac{y2-y1}{x2-x1}\)=\(\frac{1-0}{2-0}\)=\(\frac{1}{2}\)

A Slope perpendicular to line l will be negative reciprocal of line l =-2

Thus \(\frac{1}{2}\)>-2

Qty A>Qty B

IMO A



Hope this helps!




the perpendicular slope would be: ( 1 / − 1 ) = -1 ..Not -2...If i am Wrong Correct me??
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Re: In the xy-coordinate plane shown above, [#permalink]
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