Question

If x(b-c)+y(c-a)+z(a-b)=0
prove that bz-cy)/b-c=cx-az)/c-a=ay-bx)/a-b​

Answer 1

Answer:

=> (1/2)*{x(0 - b) + a(b - y) + 0(y - 0)} = 0

=> -bx + ab - ay = 0

=> bx + ay = ab

=> bx/ab + ay/ab = 1

=> x/a + y/b = 1

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

x(b-c)+y(c-a)+z(a-b)=0

bx - cx + cy - ay + az - bz = 0

multiply by c

bcx - c^2x + c^2y - acy + acz - bcz = 0

transpose some .

bcx - c^2x + acz = bcz - c^2y + acy  

subtract abz both sides .

bcx - abz - c^2x + acz = bcz - abz - c^2y + acy

b(cx - az) - c(cx - az) = bz(c - a) - cy (c - a)

(b - c) (cx - az) = (c - a) (bz - cy)

thus  

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

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