Question

$If \: x(b - c) + y(c - a) + z(a - b) = 0 \\ \\ \star \: \underline {Then \: prove \: that : } \\ \\ \large \boxed{ \frac{(bz - cy)}{b - c} = \frac{(cx - az)}{c - a} = \frac{(ay - bx)}{a - b} }$


$NOTE :\\ \dag \underline{\: No \: spams \: or \: Copied \: answers}$

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Answer 1

Given :-

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

$where\:a,b,c\neq0$

To Prove :-

$\dfrac{bz-cy}{b-c}=\dfrac{cx-az}{c-a}=\dfrac{ay-bx}{a-b}$

Proof :-

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

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

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

Multiply with 'c' in whole equation,

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

Subtract 'abz' from both sides,

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

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

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

$\boxed{\dfrac{bz-cy}{b-c}=\dfrac{cx-az}{c-a}}\longrightarrow Equation\:(1)$

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

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

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

Multiply with 'a' in whole equation,

$acy-a^2y+abx=acx-a^2z+abz$

Subtract 'bcx' from both sides,

$acy-bcx-a^2y+abx=acx-a^2z-bcx+abz$

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

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

$\boxed{\dfrac{cx-az}{c-a}=\dfrac{ay-bx}{a-b}}\longrightarrow Equation\:(2)$

From Equation (1) and (2),

$\underline{\boxed{\dfrac{bz-cy}{b-c}=\dfrac{cx-az}{c-a}=\dfrac{ay-bx}{a-b}}}$

Hence, Proved !!