SOLUTION: if x÷b+c-a=y÷c+a-b=z÷a+b-c, then show that x(b-c)+y(c-a)+(a-b)=0

Algebra ->  Proportions  -> Lessons -> SOLUTION: if x÷b+c-a=y÷c+a-b=z÷a+b-c, then show that x(b-c)+y(c-a)+(a-b)=0      Log On


   

Question 1026678: if x÷b+c-a=y÷c+a-b=z÷a+b-c, then show that x(b-c)+y(c-a)+(a-b)=0
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
x%2F%28b%2Bc-a%29=y%2F%28c%2Ba-b%29=z%2F%28a%2Bb-c%29 is the symmetric equation of a straight line in R%5E3 with direction vector {b+c-a, c+a-b, a+b-c}.
One point on the line is (b+c-a, c+a-b, a+b-c). Another point is the origin (0,0,0). (Why?) Incidentally, the point (b+c-a, c+a-b, a+b-c) is also contained in the plane
x(b-c)+y(c-a)+z(a-b) = 0,
since (b+c-a)(b-c) + (c+a - b)(c-a) + (a+b-c)(a-b)=0. (VERIFY!!)
Hence the line x%2F%28b%2Bc-a%29=y%2F%28c%2Ba-b%29=z%2F%28a%2Bb-c%29 and the plane x(b-c)+y(c-a)+z(a-b) = 0 share two common points, namely (0,0,0) and (b+c-a,c+a-b,a+b-c), and thus the line x%2F%28b%2Bc-a%29=y%2F%28c%2Ba-b%29=z%2F%28a%2Bb-c%29 is contained completely in the plane x(b-c)+y(c-a)+z(a-b) = 0.
Therefore if (x,y,z) satisfy x%2F%28b%2Bc-a%29=y%2F%28c%2Ba-b%29=z%2F%28a%2Bb-c%29, then (x,y,z) should also satisfy x(b-c)+y(c-a)+z(a-b) = 0, and the statement is proved.