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Question-194695




Question Number 194695 by horsebrand11 last updated on 13/Jul/23
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$$\:\:\:\:\:\downharpoonleft\underline{\:} \\ $$
Answered by som(math1967) last updated on 13/Jul/23
let (x/(a+b−c))=(y/(b+c−a))=(z/(c+a−b))=k  x=k(a+b−c)  y=k(b+c−a)  z=k(c+a−b)  (a−b)x+(b−c)y+(c−a)z  =(a−b)(a+b−c)k+(b−c)(b+c−a)k  (c−a)(c+a−b)k  =k×(a^2 −b^2 −ca+bc+b^2 −c^2 −ab+ca  c^2 −a^2 −bc+ab)  =k×0=0
$$\boldsymbol{{let}}\:\frac{\boldsymbol{{x}}}{\boldsymbol{{a}}+\boldsymbol{{b}}−\boldsymbol{{c}}}=\frac{\boldsymbol{{y}}}{\boldsymbol{{b}}+\boldsymbol{{c}}−\boldsymbol{{a}}}=\frac{\boldsymbol{{z}}}{\boldsymbol{{c}}+\boldsymbol{{a}}−\boldsymbol{{b}}}=\boldsymbol{{k}} \\ $$$$\boldsymbol{{x}}=\boldsymbol{{k}}\left(\boldsymbol{{a}}+\boldsymbol{{b}}−\boldsymbol{{c}}\right) \\ $$$$\boldsymbol{{y}}=\boldsymbol{{k}}\left(\boldsymbol{{b}}+\boldsymbol{{c}}−\boldsymbol{{a}}\right) \\ $$$$\boldsymbol{{z}}=\boldsymbol{{k}}\left(\boldsymbol{{c}}+\boldsymbol{{a}}−\boldsymbol{{b}}\right) \\ $$$$\left(\boldsymbol{{a}}−\boldsymbol{{b}}\right)\boldsymbol{{x}}+\left(\boldsymbol{{b}}−\boldsymbol{{c}}\right)\boldsymbol{{y}}+\left(\boldsymbol{{c}}−\boldsymbol{{a}}\right)\boldsymbol{{z}} \\ $$$$=\left(\boldsymbol{{a}}−\boldsymbol{{b}}\right)\left(\boldsymbol{{a}}+\boldsymbol{{b}}−\boldsymbol{{c}}\right)\boldsymbol{{k}}+\left(\boldsymbol{{b}}−\boldsymbol{{c}}\right)\left(\boldsymbol{{b}}+\boldsymbol{{c}}−\boldsymbol{{a}}\right)\boldsymbol{{k}} \\ $$$$\left(\boldsymbol{{c}}−\boldsymbol{{a}}\right)\left(\boldsymbol{{c}}+\boldsymbol{{a}}−\boldsymbol{{b}}\right)\boldsymbol{{k}} \\ $$$$=\boldsymbol{{k}}×\left(\boldsymbol{{a}}^{\mathrm{2}} −\boldsymbol{{b}}^{\mathrm{2}} −\boldsymbol{{ca}}+\boldsymbol{{bc}}+\boldsymbol{{b}}^{\mathrm{2}} −\boldsymbol{{c}}^{\mathrm{2}} −\boldsymbol{{ab}}+\boldsymbol{{ca}}\right. \\ $$$$\left.\boldsymbol{{c}}^{\mathrm{2}} −\boldsymbol{{a}}^{\mathrm{2}} −\boldsymbol{{bc}}+\boldsymbol{{ab}}\right) \\ $$$$=\boldsymbol{{k}}×\mathrm{0}=\mathrm{0} \\ $$

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