Question

(x^2+y^2)(a^2+b^2)=(ax+by)2 prove that x/a=y/b

Answer 1

$\underline{\mathfrak{\huge{The\:Question:}}}$

Given that :-

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

Prove that :-

$\tt{\frac{x}{a} = \frac{y}{b}}$

$\underline{\mathfrak{\huge{Your\:Answer:}}}$

Given :-

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

To Prove :-

$\tt{\frac{x}{a} = \frac{y}{b}}$

Proof :-

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

Multiply and open the brackets at the L.H.S. ( Left Hand Side ) :-

=》 $\tt{x^{2}a^{2} + x^{2}b^{2} + y^{2}a^{2} + y^{2}b^{2} = (ax + by)^{2}}$

Now, use the identity:-

$\tt{(a+b)^{2} = a^{2} + b^{2} + 2ab}$

in the R.H.S. ( Right Hand Side ) of the formed equation :-

=》 $\tt{x^{2}a^{2} + x^{2}b^{2} + y^{2}a^{2} + y^{2}b^{2} = a^{2}x^{2} + b^{2}y^{2} + 2axby}$

Now, cancel the common terms at R.H.S. and L.H.S. :-

=》 $\tt{b^{2}x^{2} + y^{2}a^{2} = 2axby}$

Now, take (2axby) to the L.H.S. and then solve :-

=》 $\tt{b^{2}x^{2} + y^{2}a^{2} - 2axby = 0}$

Now, if we see, the identity :-

$\tt{a^{2} + b^{2} - 2ab = (a-b)^{2}}$

can be used here to simplify the equation :-

=》 $\tt{(bx - ay)^{2} = 0}$

Now, take square root at both the sides :-

=》 $\tt{bx- ay = 0}$

Take (ay) to the R.H.S. :-

=》 $\tt{bx = ay}$

Take (b) to the R.H.S. and (a) to the L.H.S. and then, you'll get your answer :-

=》 $\tt{\frac{x}{a} = \frac{y}{b}}$

$\mathfrak{Hence\:Proved!}$