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Mathematics 23 Online
rvc (rvc):

the diagonals oy a parallelogram PQRS are along the lines x+3y=4 and 6x-2y=7 then PQRS must be a ?

rvc (rvc):

*of

rvc (rvc):

@rational please help

rvc (rvc):

@ikram002p may u please help me

OpenStudy (rational):

Hint : the given lines are perpendicular

rvc (rvc):

i know that

OpenStudy (rational):

Next use this : A parallelogram with perpendicular diagonals is called a "rhombus"

rvc (rvc):

oh so its not square?

rvc (rvc):

@rational

OpenStudy (rational):

it has to be a rhombus i think..

OpenStudy (ikram002p):

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OpenStudy (ikram002p):

hmm

OpenStudy (ikram002p):

yeah it can be square as well

OpenStudy (rational):

it has to be a rhombus

OpenStudy (jhannybean):

ikky took a ruler to her screen to draw out those lines :P Hahaha jk!

rvc (rvc):

so which is right

OpenStudy (ikram002p):

jhanny lol xD

OpenStudy (rational):

not all rhombuses are squares so saying "it has to be a square" is incorrect

OpenStudy (ikram002p):

i said it might be also square xD

OpenStudy (rational):

it can be a cube also if you stretch it and take it to 3D (some sarcasm)

OpenStudy (rational):

:P

rvc (rvc):

lol this souns interesting @rational

OpenStudy (ikram002p):

this is why the one who ask should be careful :P any imagination answer can be correct

OpenStudy (rational):

nope, the only possible answer is "rhombus"

OpenStudy (ikram002p):

but yeah according to the definition of rhombus its the parallelogram of perpendicular diagonals, and square belongs to rhombus family so hmmm

rvc (rvc):

well the ans is rhombus but i just need a proof for that :)

OpenStudy (rational):

consider a quadratic equation ax^2 + bx + c = 0 and it is given that b^2-4ac > 0 what can you say about the nature of roots ?

OpenStudy (ikram002p):

oh @rvc all u need is to find the slope of both lines to show they are perpendicular :)

OpenStudy (rational):

options : 1) roots must be integers 2) roots must be rationsl 3) roots must be perfect squares 4) roots must be mangoes

rvc (rvc):

okay then

rvc (rvc):

1)

OpenStudy (rational):

the answer is actually NONE OF THEM.

OpenStudy (rational):

because `b^2-4ac > 0` is only a sufficient condition for "real roots"

rvc (rvc):

ohh yes

OpenStudy (ikram002p):

hahaha ok got it xDDDDDDD if there is other options like they might be integer they might be rational they might be morons any thing fits :P

OpenStudy (rational):

ugh, the point im trying is to convince us that none of them fit

OpenStudy (ikram002p):

i know, i totally agree when the student still in school i just wanna see what @rvc is about to say

OpenStudy (rational):

OK \[b^2 - 4ac \ge 0\implies \text{roots are real}\] end of story.

OpenStudy (rational):

\[b^2 - 4ac \ge 0\implies \text{roots are integers}\] is incorrect as it doesn't hold always.

rvc (rvc):

rectangle,rhombus,square,cyclic quadrilateral

OpenStudy (rational):

\[\text{Sheela's dress became wet} \implies \text{It rained}\] the implication is wrong because there might be other reasons for Sheela to become wet

rvc (rvc):

sheela is my math teacher's name lol

OpenStudy (rational):

\[\text{perpendicular diagonals} \implies \text{parallelogram is square}\] the implication is wrong because there might be other possibilities for parallelogram other than square.

OpenStudy (ikram002p):

xD then dont tell her husbend were using her in question @rvc xD

OpenStudy (rational):

However below implication holds \[\text{perpendicular diagonals} \implies \text{parallelogram is rhombus}\]

OpenStudy (ikram002p):

ok i agree

rvc (rvc):

lol @ikram002p

rvc (rvc):

well im not convinced its a rhombus

OpenStudy (ikram002p):

what u are thinking it is ?

rvc (rvc):

im confused

OpenStudy (rational):

you may google for a proof of "A parallelogram with perpendicular diagonals is a rhombus"

OpenStudy (ikram002p):

ok so ur agree both lines are perpendicular diagonals ?

OpenStudy (ikram002p):

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