Question

If ax + hy + gz = 0, hx + by + fz = 0, prove that
$\left(1\right)\frac{x^2}{bc-f^2}=\frac{y^2}{ca-g^2}=\frac{z^2}{ab-h^2}.$
$\left(2\right)(bc-f^2)(ca-g^2)(ab-h^2)=(fg-ch)(gh-af)(hf-bg).$

Answer 1

$ax+hy+gz=0⟶1$

$hx+by+fz=0⟶2$

$gx+fy+cz=0⟶3$

Using cross multiplication $1$ and $2$

$\frac{x}{hf-bg}=\frac{y}{gh-af}=\frac{z}{ab-h^2}⟶4$

Using cross multiplication $2$ and $3$ equation

$\frac{x}{bc-f^2}=\frac{y}{fg-ch}=\frac{z}{hf-bg}⟶5$

Using cross multiplication $1$ and $3$ equation

$\frac{x}{fg-ch}=\frac{y}{ca-g^2}=\frac{z}{gh-af}⟶6$

From $5$ and $6$

$\frac{x}{bc-f^2}\times \frac{x}{fg-ch}=\frac{y}{fg-ch}\times \frac{y}{ca-g^2}$

$\therefore \frac{x^2}{bc-f^2}=\frac{y^2}{ca-g^2}$

Similarly from $4$ and $6$

$\frac{x^2}{bc-f^2}=\frac{z^2}{ab-h^2}$

$\therefore \frac{x^2}{bc-f^2}=\frac{y^2}{ca-g^2}=\frac{z^2}{ab-h^2}$

Now from equation $4$,$5$ and $6$

$\frac{x}{bc-f^2}\times \frac{y}{ca-g^2}\times \frac{z}{ab-h^2}=\frac{y}{fg-ch}\times \frac{z}{gh-af}\times \frac{x}{hf-bg}$

$\therefore (bc-f^2)\left({ca-g}^2\right)(ab-h^2)=(fg-ch)(gh-af)(hf-bg)$