Q. 8 A4.8( 4 Votes )
The coordinates of the middle points of the sides of a triangle are (1,1), (2,3) and (4,1), find the coordinates of its vertices.
Answer :
Consider a ΔABC with A(x1, y1), B(x2, y2) and C(x3, y3). If P(1, 1), Q(2, 3) and R(4, 1) are the midpoints of AB, BC, and CA. Then,
…(i)
…(ii)
…(iii)
…(iv)
…(v)
…(vi)
Adding (i), (iii) and (v), we get
x1 + x2 + x2 + x3 + x1 + x3 = 2 + 4 + 8
⇒ 2(x1 + x2 + x3) = 14
⇒ x1 + x2 + x3 = 7 …(vii)
From (i) and (vii), we get
x3 = 7 – 2 = 5
From (iii) and (vii), we get
x1 = 7 – 4 = 3
From (v) and (vii), we get
x2 = 7 – 8 = -1
Now adding (ii), (iv) and (vi), we get
y1 + y2 + y2 + y3 + y1 + y3 = 2 + 6 + 2
⇒ 2(y1 + y2 + y3) = 10
⇒ y1 + y2 + y3 = 5 …(viii)
From (ii) and (viii), we get
y3 = 5 – 2 = 3
From (iv) and (vii), we get
y1 = 5 – 6 = -1
From (vi) and (vii), we get
y2 = 5 – 2 = 3
Hence, the vertices of ΔABC are A(3, -1), B(-1, 3) and C(5, 3)
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