Question

If (the above equation) then prove that x:a=y:b=z:c​

Answer 1

Given :-

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

To Prove

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf \:\dfrac{x}{a} = \dfrac{y}{b} = \dfrac{z}{c}$

$\large\underline{\sf{Solution-}}$

Basic Concept Used :-

Addendo Property of Proportions :-

$\sf \: Let \: \: \: \dfrac{a}{b} = \dfrac{c}{d} = \dfrac{e}{f} \: \: \: then \:$

$\rm :\longmapsto\: \: \dfrac{a}{b} = \dfrac{c}{d} = \dfrac{e}{f} \: = \: \dfrac{a + c + e}{b + d + f}$

Let's solve the problem now!!

Given that

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

can be rewritten as

$\rm\dfrac{acy - bxc}{ {c}^{2} } = \dfrac{cbx - abz}{ {b}^{2} } = \dfrac{abz - acy}{ {a}^{2} }$

and

Now,

By using Addendo Property, we have

$\rm\dfrac{acy - bxc}{ {c}^{2} } = \dfrac{cbx - abz}{ {b}^{2} } = \dfrac{abz - acy}{ {a}^{2}} \\ \rm \: = \: \dfrac{ \cancel{acy} - \cancel{bxc} + \cancel{cbx} - \cancel{abz} + \cancel{abz} - \cancel{acy}}{ {a}^{2} + {b}^{2} + {c}^{2}}$

$\rm\dfrac{acy - bxc}{ {c}^{2}}=\dfrac{cbx - abz}{{b}^{2}}=\dfrac{abz - acy}{{a}^{2}}=\dfrac{0}{{a}^{2}+{b}^{2}+{c}^{2}}$

$\rm\dfrac{acy - bxc}{ {c}^{2}}=\dfrac{cbx - abz}{{b}^{2}}=\dfrac{abz - acy}{{a}^{2}}=0$

$\rm :\implies\:\rm\dfrac{ay - bx}{c} = \dfrac{cx - az}{b} = \dfrac{bz - cy}{a} = 0$

$\rm :\implies\:ay - bx = 0 \: and \: cx - az = 0 \: and \: bz - cy = 0$

$\rm :\implies\:ay = bx \: and \: cx = az\: and \: bz = cy$

$\rm :\implies\:\dfrac{x}{a} = \dfrac{y}{b} \: and \: \dfrac{x}{a} = \dfrac{z}{c} \: and \: \dfrac{y}{b} = \dfrac{z}{c}$

$\bf\implies \:\dfrac{x}{a} = \dfrac{y}{b} = \dfrac{z}{c}$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information :-

Properties of Proportions :-

$\sf \: If \: \: \dfrac{a}{b} = \dfrac{c}{d} \: \: \: then$

$\sf \:\: \: \dfrac{a}{c} = \dfrac{b}{d} \: \: \: \: \: \: \: \: \: \{alternendo \: property \}$

$\sf \:\: \: \dfrac{b}{a} = \dfrac{d}{c} \: \: \: \: \: \: \: \: \: \{invertendo \: property \}$

$\sf \:\: \: \dfrac{a + b}{b} = \dfrac{c + d}{d} \: \: \: \: \: \: \: \: \: \{componendo \: property \}$

$\sf \:\: \: \dfrac{a - b}{b} = \dfrac{c - d}{d} \: \: \: \: \: \: \: \: \: \{dividendo \: property \}$

$\sf \:\: \: \dfrac{a}{b} = \dfrac{c}{d} = \dfrac{a + c}{b + a} \: \: \: \: \: \: \: \: \: \{addendo \: property \}$