If the orthocentre of the triangle, whose vertices are (1, 2), (2,...

We have,

Here ${m}_{BH} \times {m}_{AC} = - 1$

$⇒ \left(\dfrac{\beta - 3}{\alpha - 2}\right)\left(\dfrac{1}{- 2}\right) = - 1$

$⇒ \beta - 3 = 2\alpha - 4$

$⇒ \beta = 2\alpha - 1$ ....(1)

${m}_{AH} \times {m}_{BC} = - 1$

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

$⇒ 2\beta - 4 = \alpha - 1$

Using (1), we can write:

$⇒ 2 ( 2\alpha - 1 ) - 4 = \alpha - 1$

$⇒ 2\left(2\alpha - 1\right) = \alpha + 3$

$⇒ 3\alpha = 5$

$\alpha = \dfrac{5}{3}, \beta = \dfrac{7}{3} ⇒ H\left(\dfrac{5}{3}, \dfrac{7}{3}\right)$

$\alpha + 4\beta = \dfrac{5}{3} + \dfrac{28}{3} = \dfrac{33}{3} = 11$

$\beta + 4\alpha = \dfrac{7}{3} + \dfrac{20}{3} = \dfrac{27}{3} = 9$

Required equation is

${x}^{2} - 20x + 99 = 0$