Question

if (x^2+y^2)(a^2+b^2)=(ax+by)^2. Prove that x/a=y/b

Answer 1

prove that this question

Answer 2

$\dfrac{x}{a}=\dfrac{y}{b},$ proved.

Step-by-step explanation:

We have,

$(x^2+y^2)(a^2+b^2)=(ax+by)^2$

Prove that, $\dfrac{x}{a}=\dfrac{y}{b}$.

∴ $(x^2+y^2)(a^2+b^2)=(ax+by)^2$

⇒ $x^2(a^2+b^2)+y^2(a^2+b^2)=(ax)^2+(by)^{2} +2(ax)(by)$

Using algebraic identity,

$(x+y)^{2}=x^{2}+y^{2}+2xy$

⇒ $x^2a^2+x^2b^2+y^2a^2+y^2b^2=a^2x^2+b^2y^{2} +2abxy$

⇒ $x^2b^2+y^2a^2=2abxy$

⇒ $x^2b^2+y^2a^2-2abxy=0$

⇒ $(xb)^2+(ya)^2-2(xb)(ya)=0$

Using algebraic identity,

$(x-y)^{2}=x^{2}+y^{2}-2xy$

⇒ $(xb-ya)^2=0$

⇒ xb - ya = 0

⇒ xb = ya

⇒ $\dfrac{x}{a}=\dfrac{y}{b}$, proved.

Hence, $\dfrac{x}{a}=\dfrac{y}{b},$ proved.