Question

If $x^2+y^2+z^2-xy-yz-zx=0$ then the value of $\frac{x+y}{z}$

Answer 1

Solution:

Given $x^2+y^2+z^2-xy-yz-zx=0$

$2x^2+2y^2+2z^2-2xy-2yz-2zx=0$

$(x^2-2xy+y^2)+(y^2-2yz+z^2)+(z^2-2zx+x^2)=0$

$\Rightarrow (x+y{)}^2+(y-z{)}^2+(z-x{)}^2=0$

$\Rightarrow x=y=z=k$

$\Rightarrow \frac{x+y}{z}=\frac{2k}{k}=2$