Question

If $\dfrac{{\left( {{p^2} + {q^2}} \right)}}{{\left( {{r^2} + {s^2}} \right)}} = \dfrac{{pq}}{{rs}}$, then what is the value of $\dfrac{{p - q}}{{p + q}}$ in terms of $r$ and $s$?
1) $\dfrac{{r + s}}{{r - s}}$
2) $\dfrac{{r - s}}{{r + s}}$
3) $\dfrac{{r + s}}{{rs}}$
4) $\dfrac{{rs}}{{r - s}}$

Answer 1

Hint: We are given that $\dfrac{{\left( {{p^2} + {q^2}} \right)}}{{\left( {{r^2} + {s^2}} \right)}} = \dfrac{{pq}}{{rs}}$. We know that if $\dfrac{x}{y} = \dfrac{m}{n}$, then $\dfrac{{x + m}}{{y + n}} = \dfrac{m}{n}$ and $\dfrac{{x - m}}{{y - n}} = \dfrac{m}{n}$. We will use this properties to form the formula of ${a^2} + {b^2} + 2ab = {\left( {a + b} \right)^2}$ and ${a^2} + {b^2} - 2ab = {\left( {a - b} \right)^2}$ and simplify the equation. Divide the equations and rearrange it to find the value of $\dfrac{{p - q}}{{p + q}}$

Complete step by step answer:

We are given that $\dfrac{{\left( {{p^2} + {q^2}} \right)}}{{\left( {{r^2} + {s^2}} \right)}} = \dfrac{{pq}}{{rs}}$,
Simplify the expression using the rules of ratio
We will multiply the numerator and denominator of R.H.S as,
$\dfrac{{\left( {{p^2} + {q^2}} \right)}}{{\left( {{r^2} + {s^2}} \right)}} = \dfrac{{2pq}}{{2rs}}$
Now, if $\dfrac{x}{y} = \dfrac{m}{n}$, then $\dfrac{{x + m}}{{y + n}} = \dfrac{m}{n}$
Then, $\dfrac{{\left( {{p^2} + {q^2} + 2pq} \right)}}{{\left( {{r^2} + {s^2} + 2rs} \right)}} = \dfrac{{2pq}}{{2rs}}$ which is equal to
$\dfrac{{\left( {{p^2} + {q^2} + 2pq} \right)}}{{\left( {{r^2} + {s^2} + 2rs} \right)}} = \dfrac{{pq}}{{rs}}$
Now, we know that ${a^2} + {b^2} + 2ab = {\left( {a + b} \right)^2}$
Then,
$\dfrac{{{{\left( {p + q} \right)}^2}}}{{{{\left( {r + s} \right)}^2}}} = \dfrac{{pq}}{{rs}}$ eq(1)
Similarly, $\dfrac{{\left( {{p^2} + {q^2} - 2pq} \right)}}{{\left( {{r^2} + {s^2} - 2rs} \right)}} = \dfrac{{pq}}{{rs}}$
Also, ${a^2} + {b^2} - 2ab = {\left( {a - b} \right)^2}$
Then,
$\dfrac{{{{\left( {p - q} \right)}^2}}}{{{{\left( {r - s} \right)}^2}}} = \dfrac{{pq}}{{rs}}$ eq(2)
From equation (1) and (2), we will get,
$\dfrac{{{{\left( {p + q} \right)}^2}}}{{{{\left( {r + s} \right)}^2}}} = \dfrac{{{{\left( {p - q} \right)}^2}}}{{{{\left( {r - s} \right)}^2}}}$
Taking square roots on both sides,
$\dfrac{{p - q}}{{r - s}} = \dfrac{{p + q}}{{r + s}}$
But, we want to find the value of $\dfrac{{p - q}}{{p + q}}$, then we can write the above equation as
$\dfrac{{p - q}}{{p + q}} = \dfrac{{r - s}}{{r + s}}$
Therefore the value of $\dfrac{{p - q}}{{p + q}}$ is $\dfrac{{r - s}}{{r + s}}$
Hence, option (2) is correct.

Note: We can multiply or divide by the same number in ratio except 0 and the ratio will remain unchanged. If a constant number is added or subtracted to the ratio, then also the ratio remains the same. These properties will help to simplify the equations. There are various other properties of ratio like compenendo and divedendo.