Math, asked by kavana73, 1 year ago

solve it.please
if \: points \: \:  a(5.p) \:  \: b(1.5) \: c(6.2)and \: d(6.2) \: form \: a \: square \: abcd \: then \: p =  \\  \\  \\ (a)7 \\ (b)3 \\ (c)6 \\  (d)8
the question is from 10 th coordinate geometry.......

answer with step_by_step explaination..​


Anonymous: there is a mistake as it is not forming a square , it is forming a triangle as point c =point d
kavana73: how
Anonymous: point C (5,6) also point D(5,6)
Anonymous: sorry (6,2) both
Anonymous: C(6,2) and D(6,2) you provided

Answers

Answered by kajalsindhu252004
2

hey mate

here is your answer in the pic

Attachments:

kavana73: what
kajalsindhu252004: how
kavana73: so,gimme tge correct answer
Anonymous: you have written
kavana73: go n answer rohit
Anonymous: distance between b(1,5) and c(6,2) = √[(6-1)² + (2-5)² ]. = √[5² + 3² ] = √[ 25 + 9] = √(34)
Anonymous: but distance between c(6,2) and d(6,2) = √[(6-6)² + (2-2)²] = √[(0)² + (0)² ] = 0
Anonymous: so ig is not a square by the question provided
Anonymous: Well I was not able to answer that's why I have written here
kavana73: ok thanks
Answered by NishantMishra3
1

 \large \pink{  \textbf{side \: a \: and \: b \: should \: be \: }} \\  \large \pink{   \text{equal \: to \: side \: c \: and \: d \: now}} \\  \\ a(5,p) \:  \:  \: b(1,5) \:  \: c(6,2) \:  \: d(6,2) \\  \\  \sqrt{ ({y2}^{2}  -  {y1}^{2} )  +  ({x2}^{2}  -  {x1}^{2}) }  \\  \\  =  \sqrt{25 -  {p}^{2} + 1 - 25 }  =  \sqrt{1 -  {p}^{2} }  \\  \\ from \: c,d \: we \: have :  \\  \\ \sqrt{ ({y2}^{2}  -  {y1}^{2} )  +  ({x2}^{2}  -  {x1}^{2}) }  \\  \\  =  \sqrt{4 - 4  +   36 - 36 }  =  0 \\ \\ now.. \\  \\  \sqrt{1 -  {p}^{2} }  = 0 \\  \\ 1 -  {p}^{2}  = 0 \\  \\  {p}^{2}  = 1 \\  \\ p =  \sqrt{1}  = 1


NishantMishra3: p=1
kavana73: sry but its wrong
Anonymous: isn't distance formula is √[(x2 - x1)² + (y2-y1)²] ?
Anonymous: @NishantMishra22
NishantMishra3: hmm wait
NishantMishra3: someone reported this
NishantMishra3: fine if u got ans
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