Ask your own question, for FREE!
Mathematics 22 Online
OpenStudy (anonymous):

In rectangle ABCD, AB has a slope of 2. The slope of BC is by 4x-2. Solve for 'x'.

OpenStudy (anonymous):

Thanks a lot by the way! ♥

hartnn (hartnn):

the adjacent sides of rectangle are perpendicular, did u know this ?

OpenStudy (anonymous):

Uhh yeah I did. (:

hartnn (hartnn):

since sides are perpendicular, the product of slope = -1

hartnn (hartnn):

so you just need to solve 2 (4x-2) = -1 for x :)

OpenStudy (anonymous):

So what is -1 the slope of?

OpenStudy (anonymous):

you can understand it this way if line ab has a slope =2, then line bc perpendicular to line ab has a slope = -2

hartnn (hartnn):

-1 is not any slope. when lines are perpendicular, the product of their slopes is -1 like if i have line1 with slope = m and line2 with slope n, if these lines are perpendicular, then m*n =-1

OpenStudy (anonymous):

Is it always that? I never knew.

OpenStudy (anonymous):

therefore the slope of bc is -2 but we have an eqn for the slope, therefore, 4x-2= - 2

OpenStudy (anonymous):

Yeah and you get 4x=0

OpenStudy (anonymous):

So the answer would be 3/8?

hartnn (hartnn):

no... slope of BC is NOT -2

OpenStudy (anonymous):

its -1/2?

hartnn (hartnn):

you got this, right ??? "sides are perpendicular, the product of slope = -1"

hartnn (hartnn):

yes, its -1/2

hartnn (hartnn):

4x-2 = -1/2

OpenStudy (anonymous):

yeah I did. the answers 3/8 then?

hartnn (hartnn):

4x = 3/2 YES x = 3/8 is correct :)

OpenStudy (anonymous):

Thanks (:

OpenStudy (anonymous):

oh im sorry, i guess i messed up trying to explain how perpendicular slopes are related :(

hartnn (hartnn):

welcome ^_^

OpenStudy (anonymous):

Oh no its okay! It was just a small missup.

OpenStudy (anonymous):

Thanks for the help though! ♥

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!