Question

If bcx = cay = abz then show that $\frac{ax+by}{a^2+b^2}=\frac{by+cz}{b^2+c^2}=\frac{cz+ax}{c^2+a^2}$.

Answer 1

$$\begin{gathered} \mathrm{Let}: \\ bcx=cay=abz=k \\ \Rightarrow x=\frac{k}{bc},y=\frac{k}{ca},z=\frac{k}{ab}...\left(\mathrm{i}\right) \\ \therefore \frac{ax+by}{a^2+b^2}=\frac{a\times {\displaystyle \frac{k}{bc}}+b\times {\displaystyle \frac{k}{ca}}}{a^2+b^2}\left(\mathrm{Substituting}\mathrm{the}\mathrm{values}\mathrm{of}x\mathrm{and}y\right) \\ =\frac{k\left({\displaystyle \frac{a}{bc}}+{\displaystyle \frac{b}{ca}}\right)}{a^2+b^2} \\ =\frac{k\left(a^{2}c+b^{2}c\right)}{b{c}^{2}a\left(a^2+b^2\right)} \\ =\frac{kc\left(a^2+b^2\right)}{ab{c}^2\left(a^2+b^2\right)} \\ =\frac{k}{abc} \\ \therefore \frac{ax+by}{a^2+b^2}=\frac{k}{abc}...\left(\mathrm{ii}\right) \\ Also, \\ \frac{by+cz}{b^2+c^2}=\frac{b\times {\displaystyle \frac{k}{ca}}+c\times {\displaystyle \frac{k}{ab}}}{b^2+c^2} \\ =\frac{k\times \left({\displaystyle \frac{b}{ca}}+{\displaystyle \frac{c}{ab}}\right)}{b^2+c^2} \\ =\frac{k\left(a{b}^2+c^{2}a\right)}{c{a}^{2}b\left(b^2+c^2\right)} \\ =\frac{ka\left(b^2+c^2\right)}{bc{a}^2\left(b^2+c^2\right)} \\ =\frac{k}{abc} \\ \therefore \frac{by+cz}{b^2+c^2}=\frac{k}{abc}...\left(\mathrm{iii}\right) \\ And, \\ \frac{cz+ax}{c^2+a^2}=\frac{c\times {\displaystyle \frac{k}{ab}}+a\times {\displaystyle \frac{k}{bc}}}{c^2+a^2} \\ =\frac{k\left({\displaystyle \frac{c}{ab}}+{\displaystyle \frac{a}{bc}}\right)}{c^2+a^2} \\ =\frac{k\left(b{c}^2+a^{2}b\right)}{a{b}^{2}c\left(c^2+a^2\right)} \\ =\frac{kb\left(c^2+a^2\right)}{a{b}^{2}c\left(c^2+a^2\right)} \\ =\frac{k}{abc} \\ \therefore \frac{cz+ax}{c^2+a^2}=\frac{k}{abc}...\left(\mathrm{iv}\right) \\ \mathrm{From}\left(\mathrm{ii}\right),\left(\mathrm{iii}\right)\mathrm{and}\left(\mathrm{iv}\right),weget: \\ \frac{ax+by}{a^2+b^2}=\frac{by+cz}{b^2+c^2}=\frac{cz+ax}{c^2+a^2} \end{gathered}$$