A rectangle R with end points of the one of its sides as (1, 2) and...

Equation of line $AB$ will be

$y - 2 = 2\left(x - 1\right)$⇒ 2x - y = 0$</p><p>Now the given diameter is parallel to$AB$</p><p>Then, the distance between diameter and line$AB$</p><p>$= \left(\dfrac{4}{\sqrt{{2}^{2} + {1}^{2}}}\right) = \dfrac{4}{\sqrt{5}}$</p><p>Thus,$BC = 2 \times \dfrac{4}{\sqrt{5}} = \dfrac{8}{\sqrt{5}}$</p><p>Also$AB = \sqrt{{\left(1 - 3\right)}^{2} + {\left(2 - 6\right)}^{2}} = \sqrt{20} = 2\sqrt{5}$</p><p>Area of the rectangle$= AB \times BC = \dfrac{8}{\sqrt{5}} \times 2\sqrt{5} = 16