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.

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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