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Question
The points (2, -1), (-1, 4) and (-2, 2) are midpoints of the sides ofa triangle. Find its vertices.
Solution
Let A(x1,y1), B(x2, y2 ) and C(x3, y3) be the coordinates of the vertices of Δ ABC.
Midpoint of AB, i.e. D
D (2 , 1) = D `(("x"_1 + "x"_2)/2 , ("y"_1 + "y"_2)/2)`
`2 = ("x"_1 + "x"_2)/2 , ("y"_1 + "y"_2)/2` = -1
X1 + X2 = 4 .....(1) Y1 + Y2 = -2 ........(2)
Similarly,
X1 + X3 = -2 .....(3) y1 + y3 = 8 . ....(4)
X2 + X3 = -4 .....(5) Y2 + Y3 = 4 .....(6)
Adding (1), (3) and (5)
2(x1 + x2 + x3) = -2
x1 + x2 + x3 = -1
4 + x3 = -1 ... (from (1))
x3 = -5
From (3)
x1 - 5 = -2
x1 = 3
From (5)
x2 = -5 = -4
x2 = 1
Adding (2),(4) and (6)
2( Y1 + Y2 + Y3)= 10
Y1 +Y2 +Y3 = 5
-2 + Y3 = 5 [from(2)]
y3 = 7
From (4)
y1 + 7 = 8
y1 = 1
From (6)
y2 + 7 = 4
y2 = -3
The coordinates of the vertices of A.ABC are (3, 1), ( 1,-3) and ( -5,7)
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