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(x_{1}, y_{1}), B(x_{2}, y_{2}) and C(x_{3}, y_{3}). 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

x_{1} + x_{2} + x_{2} + x_{3} + x_{1} + x_{3} = 2 + 4 + 8

⇒ 2(x_{1} + x_{2} + x_{3}) = 14

⇒ x_{1} + x_{2} + x_{3} = 7 …(vii)

From (i) and (vii), we get

x_{3} = 7 – 2 = 5

From (iii) and (vii), we get

x_{1} = 7 – 4 = 3

From (v) and (vii), we get

x_{2} = 7 – 8 = -1

Now adding (ii), (iv) and (vi), we get

y_{1} + y_{2} + y_{2} + y_{3} + y_{1} + y_{3} = 2 + 6 + 2

⇒ 2(y_{1} + y_{2} + y_{3}) = 10

⇒ y_{1} + y_{2} + y_{3} = 5 …(viii)

From (ii) and (viii), we get

y_{3} = 5 – 2 = 3

From (iv) and (vii), we get

y_{1} = 5 – 6 = -1

From (vi) and (vii), we get

y_{2} = 5 – 2 = 3

Hence, the vertices of ΔABC are A(3, -1), B(-1, 3) and C(5, 3)

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