Question

If $A(\frac{2c}{a},\frac{c}{b}),B(\frac{c}{a},0)$ and $C(\frac{1+c}{a},\frac{1}{b})$ are three points, then

• A. $(AB{)}^2+(BC{)}^2-(CA{)}^2=2c$
• B. $\frac{(AB{)}^2+(BC{)}^2}{(CA{)}^2}=\frac{c^2+1}{(c-1{)}^2}$
• C. $(AB{)}^2+(BC{)}^2-(CA{)}^2=\frac{2c(a^2+b^2)}{a^2{b}^2}$
• D. $(AB{)}^2-(BC{)}^2-(CA{)}^2=2b$

Answer 1

Answer: (B), (C)

The correct options are
B $\frac{(AB{)}^2+(BC{)}^2}{(CA{)}^2}=\frac{c^2+1}{(c-1{)}^2}$
C $(AB{)}^2+(BC{)}^2-(CA{)}^2=\frac{2c(a^2+b^2)}{a^2{b}^2}$

As $A\equiv (\frac{2c}{a},\frac{c}{b}),B\equiv (\frac{c}{a},0)$ and $C\equiv (\frac{1+c}{a},\frac{1}{b})$
$AB=\sqrt{{(\frac{c}{a}-\frac{2c}{a})}^2+{(0-\frac{c}{b})}^2}=\sqrt{\frac{c^2}{a^2}+\frac{c^2}{b^2}}BC=\sqrt{{(\frac{1+c}{a}-\frac{c}{a})}^2+{(\frac{1}{b}-0)}^2}=\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}AC=\sqrt{{(\frac{1+c}{a}-\frac{2c}{a})}^2+{(\frac{1}{b}-\frac{c}{b})}^2}=\sqrt{{(\frac{1}{a}-\frac{c}{a})}^2+{(\frac{1}{b}-\frac{c}{b})}^2}$
Hence,
$\frac{{\left(AB\right)}^2+{\left(BC\right)}^2}{{\left(AC\right)}^2}=\frac{\frac{c^2}{a^2}+\frac{c^2}{b^2}+\frac{1}{a^2}+\frac{1}{b^2}}{{(\frac{1}{a}-\frac{c}{a})}^2+{(\frac{1}{b}-\frac{c}{b})}^2}=\frac{c^2+1}{{(c-1)}^2}{\left(AB\right)}^2+{\left(BC\right)}^2-{\left(AC\right)}^2=\frac{c^2}{a^2}+\frac{c^2}{b^2}+\frac{1}{a^2}+\frac{1}{b^2}-{(\frac{1}{a}-\frac{c}{a})}^2-{(\frac{1}{b}-\frac{c}{b})}^2=\frac{2c(a^2+b^2)}{a^2{b}^2}$