Q.If \((x + y) ∝ (x - y)\), then \((x^2 + y^2) ∝ xy\).

The statement is true.
Given: \((x + y) ∝ (x - y)\) Or, \((x + y) = k(x - y)\) [where \(k\) is a non-zero constant of proportionality] So, \(\frac{x + y}{x - y} = k\) Squaring both sides: \(\frac{(x + y)^2}{(x - y)^2} = k^2\) Now, applying the sum and difference formula: \[ \frac{(x + y)^2 + (x - y)^2}{(x + y)^2 - (x - y)^2} = \frac{k^2 + 1}{k^2 - 1} \] Expanding both numerator and denominator: \[ \frac{2(x^2 + y^2)}{4xy} = \frac{k^2 + 1}{k^2 - 1} \] Simplifying: \[ \frac{x^2 + y^2}{xy} = \frac{2(k^2 + 1)}{k^2 - 1} = \text{constant} \] ∴ \(x^2 + y^2 ∝ xy\)
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