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Question

If the points (10,5),(8,4) and (6,6) are the mid-points of the sides of a triangle, find its vertices.

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Solution

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+x22x1+x2=20 ....(i)
5=y1+y22y1+y2=10......(ii)
8=x2+x32x2+x3=16 ...(iii)
4=y2+y32y2+y3=8 .........(iv)
6=x1+x32x1+x3=12 ......(v)
6=y1+y32y1+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=2420=4
From (iii) and (vii), we get
x1=2416=8
From (v) and (vii), we get
x2=2412=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=158=7
From (vi) and (vii), we get
y2=1512=3
Hence, the vertices of ΔABC are A(8,7),B(12,3) and C(4,5).

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