Mathematics - Point Question with Solution | TestHub

In $△ABC, AB=AC$

Solving $2x+y=4 \& x+3y=7$, we get 

$B\equiv \left(1,2\right)$

Let $C\equiv \left(h,k\right)$ 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$$\Rightarrow 26={\left(h-6\right)}^2+{\left(3-2h\right)}^2$

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

$\Rightarrow h=\frac{19}{5}$    (as $h=1$rejected)

so $k=-\frac{18}{5}$

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

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