Question

If $\frac{ay-bx}{c}=\frac{cx-az}{b}=\frac{bz-cy}{a}$ show that $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$

Answer 1

Let $\frac{ay-bx}{c}=\frac{cx-az}{b}=\frac{bz-cy}{a}=k$

$ay-bx=ck\Rightarrow \frac{ay-bx}{ab}=\frac{ck}{ab}\Rightarrow \frac{y}{b}-\frac{x}{a}=\frac{ck}{ab}....\left(1\right)cx-az=bk\Rightarrow \frac{cx-az}{ac}=\frac{bk}{ac}\Rightarrow \frac{x}{a}-\frac{z}{z}=\frac{bk}{ac}....\left(2\right)bz-cy=ak\Rightarrow \frac{bz-cy}{bc}=\frac{ak}{bc}\Rightarrow \frac{z}{c}-\frac{y}{b}=\frac{ak}{bc}....\left(3\right)$

adding $\left(1\right),\left(2\right),\left(3\right)$ we get

$\frac{y}{b}-\frac{x}{a}+\frac{x}{a}-\frac{z}{c}+\frac{z}{c}-\frac{y}{b}=\frac{ck}{ab}+\frac{bk}{ac}+\frac{ak}{bc}\Rightarrow \frac{(a^2+b^2+c^2)k}{abc}=0$

as $a^2+b^2+c^2\ne 0$ so $k=0$

so $\frac{ay-bx}{c}=0\Rightarrow ay-bx=0\Rightarrow ay=bx\Rightarrow \frac{x}{a}=\frac{y}{b}....\left(4\right)$

similarily $\frac{cx-az}{b}=0\Rightarrow cx-az=0\Rightarrow cx=az\Rightarrow \frac{x}{a}=\frac{z}{c}....\left(5\right)$

from $\left(4\right),\left(5\right)$ $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$

(Hence proved)