If #(ay-bx)/c=(cx-az)/b=(bz-cy)/a#, then how to prove that #x/a=y/b=z/c#?

Question

22770 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-06T15:39:50

#(ay-bx)/c=(cx-az)/b=(bz-cy)/a#

#"each"= (acy-bcx)/c^2=(bcx-abz)/b^2=(abz-cay)/a^2#

By addendo we get

#"each"=( (acy-bcx)+(bcx-abz)+(abz-cay))/(a^2+b^2+c^2)=0#

So #(ay-bx)/c=0=>x/a=y/b#

#(cx-az)/b=0=>x/a=z/c#

Hence

#x/a=y/b=z/c#