
In △ABC,AB=AC
Solving 2x+y=4&x+3y=7, we get
B≡(1,2)
Let C≡(h,k) and as it lies on 2x+y=4
so 2h+k=4
Now, AB2=AC2
26=(h−6)2+(k−1)2⇒26=(h−6)2+(3−2h)2
⇒26=5h2−24h+45 ⇒(h−1)(5h−19)=0
⇒h=519 (as h=1rejected)
so k=−518
Hence, centroid =(36+1+519,31+2−518)≡(518,−51)
i.e. 15(α+β)=15×517=51