Question and Answers Forum

All Questions      Topic List

Coordinate Geometry Questions

Previous in All Question      Next in All Question      

Previous in Coordinate Geometry      Next in Coordinate Geometry      

Question Number 73919 by necxxx last updated on 16/Nov/19

Commented by necxxx last updated on 16/Nov/19

Good day sirs. This question was formed and solved by some of us here. I really do not remember the question or approaches applied. Please help. Thanks in advance.

$${Good}\:{day}\:{sirs}.\:{This}\:{question}\:{was}\:{formed} \\ $$$${and}\:{solved}\:{by}\:{some}\:{of}\:{us}\:{here}.\:{I}\:{really}\: \\ $$$${do}\:{not}\:{remember}\:{the}\:{question}\:{or} \\ $$$${approaches}\:{applied}.\:{Please}\:{help}. \\ $$$${Thanks}\:{in}\:{advance}. \\ $$

Commented by mr W last updated on 16/Nov/19

creature of ajfour sir

$${creature}\:{of}\:{ajfour}\:{sir} \\ $$

Commented by necxxx last updated on 16/Nov/19

yes sir

$${yes}\:{sir} \\ $$

Answered by mind is power last updated on 16/Nov/19

((ab)/2)=(a/2)+(b/2)+((√(a^2 +b^2 ))/2) ⇒ab=a+b+(√(a^2 +b^2 ))...E ⇒(ab−a−b)^2 =a^2 +b^2 ⇒a^2 b^2 +2ab−2a^2 b−2ab^2 =0 ⇒ab(ab+2−2a−2b)=0 ⇒ab+2−2a−2b=0 ⇒b=((2a−2)/(a−2))...G DT=DP=b−1 by thales⇒((b−1)/b)=((b−1)/((√(b^2 +a^2 ))−b+1)) ⇒b=(√(b^2 +a^2 ))−b+1⇒2b−1=(√(b^2 +a^2 )) E⇔ab=a+b+2b−1⇒b(a−3)=a−1⇒b=((a−1)/(a−3)) witheG⇒((a−1)/(a−3))=((2a−2)/(a−2))⇒a−2=2a−6⇒a=4 b=((4−1)/(4−3))=3

$$\frac{\mathrm{ab}}{\mathrm{2}}=\frac{\mathrm{a}}{\mathrm{2}}+\frac{\mathrm{b}}{\mathrm{2}}+\frac{\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{ab}=\mathrm{a}+\mathrm{b}+\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }...\mathrm{E} \\ $$$$\Rightarrow\left(\mathrm{ab}−\mathrm{a}−\mathrm{b}\right)^{\mathrm{2}} =\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{a}^{\mathrm{2}} \mathrm{b}^{\mathrm{2}} +\mathrm{2ab}−\mathrm{2a}^{\mathrm{2}} \mathrm{b}−\mathrm{2ab}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\mathrm{ab}\left(\mathrm{ab}+\mathrm{2}−\mathrm{2a}−\mathrm{2b}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{ab}+\mathrm{2}−\mathrm{2a}−\mathrm{2b}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{b}=\frac{\mathrm{2a}−\mathrm{2}}{\mathrm{a}−\mathrm{2}}...\mathrm{G} \\ $$$$ \\ $$$$\mathrm{DT}=\mathrm{DP}=\mathrm{b}−\mathrm{1} \\ $$$$\mathrm{by}\:\mathrm{thales}\Rightarrow\frac{\mathrm{b}−\mathrm{1}}{\mathrm{b}}=\frac{\mathrm{b}−\mathrm{1}}{\sqrt{\mathrm{b}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} }−\mathrm{b}+\mathrm{1}} \\ $$$$\Rightarrow\mathrm{b}=\sqrt{\mathrm{b}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} }−\mathrm{b}+\mathrm{1}\Rightarrow\mathrm{2b}−\mathrm{1}=\sqrt{\mathrm{b}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} } \\ $$$$\mathrm{E}\Leftrightarrow\mathrm{ab}=\mathrm{a}+\mathrm{b}+\mathrm{2b}−\mathrm{1}\Rightarrow\mathrm{b}\left(\mathrm{a}−\mathrm{3}\right)=\mathrm{a}−\mathrm{1}\Rightarrow\mathrm{b}=\frac{\mathrm{a}−\mathrm{1}}{\mathrm{a}−\mathrm{3}} \\ $$$$\mathrm{witheG}\Rightarrow\frac{\mathrm{a}−\mathrm{1}}{\mathrm{a}−\mathrm{3}}=\frac{\mathrm{2a}−\mathrm{2}}{\mathrm{a}−\mathrm{2}}\Rightarrow\mathrm{a}−\mathrm{2}=\mathrm{2a}−\mathrm{6}\Rightarrow\mathrm{a}=\mathrm{4} \\ $$$$\mathrm{b}=\frac{\mathrm{4}−\mathrm{1}}{\mathrm{4}−\mathrm{3}}=\mathrm{3} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by necxxx last updated on 17/Nov/19

Thank you so much but what do G and E represent?

$${Thank}\:{you}\:{so}\:{much}\:{but}\:{what}\:{do}\:{G}\:{and}\:{E} \\ $$$${represent}? \\ $$

Commented by mr W last updated on 17/Nov/19

he means equation G and equation E. other people may use eqn. (i) and (ii) in such cases.

$${he}\:{means}\:{equation}\:{G}\:{and}\:{equation}\:{E}. \\ $$$${other}\:{people}\:{may}\:{use}\:{eqn}.\:\left({i}\right)\:{and}\:\left({ii}\right) \\ $$$${in}\:{such}\:{cases}. \\ $$

Commented by necxxx last updated on 17/Nov/19

ok Thanks

$${ok}\:{Thanks} \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com