In an isosceles triangle ABC , the vertex A is (6, 1) and the...

In $△ ABC, AB = AC$

Solving $2x + y = 4 \text{ and } x + 3y = 7$, we get

$B ≡ \left(1, 2\right)$

Let $C ≡ \left(h, k\right)$ and as it lies on $2x + y = 4$

so $2h + k = 4$

Now, $A{B}^{2} = A{C}^{2}$

$26 = ( h - 6{)}^{2} + {\left(k - 1\right)}^{2}$$⇒ 26 = {\left(h - 6\right)}^{2} + {\left(3 - 2h\right)}^{2}$

$⇒ 26 = 5{h}^{2} - 24h + 45$ $⇒ \left(h - 1\right)\left(5h - 19\right) = 0$

$⇒ h = \dfrac{19}{5}$ (as $h = 1$rejected)

so $k = - \dfrac{18}{5}$

Hence, centroid $= \left(\dfrac{6 + 1 + \dfrac{19}{5}}{3}, \dfrac{1 + 2 - \dfrac{18}{5}}{3}\right) ≡ \left(\dfrac{18}{5}, - \dfrac{1}{5}\right)$

i.e. $15\left(\alpha + \beta\right) = 15 \times \dfrac{17}{5} = 51$