A triangle ABC lying in the first quadrant has two vertices as A(1,...

${m}_{AC} = \dfrac{\beta - 2}{\alpha - 1}$

${m}_{AB} = \dfrac{2 - 1}{1 - 3} = - \dfrac{1}{2}$

$AB ⊥ AC$

$∴ \dfrac{\beta - 2}{\alpha - 1}\left(- \dfrac{1}{2}\right) = - 1$

$\beta = 2\alpha - 2 + 2$

$\beta = 2\alpha$

Now area of $\delta ABC = 5\sqrt{5} = \dfrac{1}{2}AB . AC$

$⇒ \dfrac{1}{2}\sqrt{{\left(3 - 1\right)}^{2} + {\left(1 - 2\right)}^{2}} . \sqrt{{\left(\alpha - 1\right)}^{2} + {\left(\beta - 2\right)}^{2}} = 5\sqrt{5}$

$⇒ \sqrt{{\left(\alpha - 1\right)}^{2} + {\left(2\alpha - 2\right)}^{2}} = 10$

$⇒ \sqrt{{\left(\alpha - 1\right)}^{2}}\sqrt{5} = 10$

$⇒ \left|\alpha - 1\right| = 2\sqrt{5}$

$⇒ \alpha = 1 \pm 2\sqrt{5}$.