Question

Please solve this question

Answer 1

HELLO DEAR,

x(b-c)+y(c-a)+z(a-b)=0 

=> bx - cx + cy - ay + az - bz = 0 ---------(1)
[multiply by c ]

=> bcx - c²x + c²y - acy + acz - bcz = 0 

=> bcx - c²x + acz = bcz - c²y + acy 

[subtract (abz) both sides]

=> bcx - abz - c²x + acz = bcz - abz - c²y + acy 

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

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

then, 

(bz - cy) / (b - c) = (cx - az) / (c - a)---------(2)

-------------and --------------
from--(1)
multiply by "a"in Equation

bx - cx + cy - ay + az - bz = 0 

we get,

abx - acx + acy - a²y +a²z-abz = 0

=> abx - acx + acy - a²y +a²z-abz = 0

=> abx - a²y + acy = acx -a²z +abz
[subtract Both side (bcx)]

=> abx - a²y - bcx + acy = acx - a²z - bcx +abz

=> acy - bcx -a²y + abx = acx - a²z - bcx + abz

=> c(ay - bx) - a(ay - bx) = a(cx - az) - b(cx-az)

=> (ay - bx)(c-a) = (a-b)(cx-az)

=> (ay-bx)/(a-b) = (cx-az)/(c-a)-------(3)



from ---(2) and--(3)

we get,

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




I HOPE ITS HELP YOU DEAR,
THANKS,☺️☺️☺️☺️

Answer 2

It is difficult to solve directly proving to this question ,
Let's try to make it easy ,
x( b - c) + y(c - a) + z(a - b) = 0
⇒bx - cx + cy - ay + az - bz = 0
⇒-(bz - cy) - (cx - az) - (ay - bx) = 0
⇒(bz - cy) + (cx - az) + (ay - bx) = 0 -------(1)

Now, (bz - cy)/(b - c) = (cx - az)/(c - a) = (ay - bx)/(a - b) = K ( let )------(2)
(bz - cy) = K(b - c) -----(3)
(cx - az ) = K(c - a)------(4)
(ay - bx) = K(a - b) -------(5)

If and only if equation (2) is correct when we get LHS = RHS of equation (1) by putting values of (3) , (4) and (5)
Just put now,
K(b - c) + K(c - a) + K(a - b) = 0
K{(b - c) + (c - a) + (a - b)} = 0
K { 0 } = 0
LHS = RHS

Hence, if x(b - c) + y(c - a) + z(a - b) then ,
(bz - cy)/(b - c) = (cx - az)/(c - a) = (ay - bx)/(a - b)