Question

If $\frac{x}{a}=\frac{y}{b}=\frac{z}{c},$ then $xy+yz+zx=?$

• A. $\frac{\left|x^2\right(a+b+c{)}^2-a^2(x^2+y^2+z^2)|}{2a^2}$
• B. $(a+b+c{)}^2-(x^2+y^2+z^2)$
• C. $\frac{(a+b+c{)}^2}{x^2+y^2+z^2}$
• D. $3abc-(x^2+y^2+z^2)$

Answer 1

The correct option is A $\frac{\left|x^2\right(a+b+c{)}^2-a^2(x^2+y^2+z^2)|}{2a^2}$

$\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k$
$x=ak$
$y=bk$
$z=ck$
$(a+b+c{)}^2=a^2+b^2+c^2+2(ab+bc+ca)$
$xy+yz+xz=(ab+bc+ca)k^2$
$=\frac{x^2}{a^2}(ab+bc+ca)$
$=\frac{x^2}{2a^2}\left[\right(a+b+c{)}^2-(a^2+b^2+c^2)]$
$=\frac{x^2}{2a^2}\left[\right(a+b+c{)}^2-\left(\frac{x^2+y^2+z^2}{k^2}\right)]$
$=\frac{x^2(a+b+c{)}^2-a^2(x^2+y^2+z^2)}{2a^2}$