Consider a ΔABC with A(x1,y1),B(x2,y2) and C(x3,y3). If P(10,5),Q(8,4) and R(6,6) are the midpoints of AB, BC, and CA. Then,
10=x1+x22⇒x1+x2=20 ....(i)
5=y1+y22⇒y1+y2=10......(ii)
8=x2+x32⇒x2+x3=16 ...(iii)
4=y2+y32⇒y2+y3=8 .........(iv)
6=x1+x32⇒x1+x3=12 ......(v)
6=y1+y32⇒y1+y3=12 ......(vi)
Adding (i), (iii) and (v), we get
x1+x2+x2+x3+x1+x3=20+16+12
⇒2(x1+x2+x3)=48
⇒x1+x2+x3=24 .......(vii)
From (i) and (vii), we get
x3=24−20=4
From (iii) and (vii), we get
x1=24−16=8
From (v) and (vii), we get
x2=24−12=12
Now adding (ii), (iv) and (vi), we get
y1+y2+y2+y3+y1+y3=10+8+12
⇒2(y1+y2+y3)=30
⇒y1+y2+y3=15 .......(viii)
From (ii) and (viii), we get
y1=15−8=7
From (vi) and (vii), we get
y2=15−12=3
Hence, the vertices of ΔABC are A(8,7),B(12,3) and C(4,5).