Answer:
Step-by-step explanation:
Hello,
Let's develop both expressions and try to re factorise.
[tex]\forall (a,b,x,y) \in \mathbb{R}^4 \\ \\ (a^2+b^2)(x^2+y^2)=(ax+by)^2\\ \\<=>(ax)^2+(ay)^2+(bx)^2+(by)^2=(ax)^2+(by)^2+2(ax)(by)\\ \\<=> (ay)^2+(bx)^2-2(ay)(bx)=0\\ \\<=> (ay-bx)^2=0\\ \\<=> ay-bx=0\\ \\<=> \boxed{\sf \bf ay=bx}[/tex]
Thanks