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Question
Two vertices of a triangle are ( -1, 4) and (5, 2). If the centroid is (0, 3), find the third vertex.
Solution
Let G be the centroid of Δ ABC whose coordinaes are (0 , -3) and let C (x , y) be the coordinates of thgird vertex
coordinates of G are ,
G (0 , -3) = G `((- 1 + 5 + "x")/3 , (4 + 2 + "y")/3)`
O = `(4 + "x")/3 , -3 = (6 + 4)/3`
x = -4 , y = - 15
Coordinates of third vertex are (-4 , - 15)
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Solution :
Suppose, (–4, 2) = (x1, y1) and (6, 2) = (x2, y2) and co-ordinates of P are (x, y).
∴ According to the midpoint theorem,
x = `(x_1 + x_2)/2 = (square + 6)/2 = square/2 = square`
y = `(y_1 + y_2)/2 = (2 + square)/2 = 4/2 = square`
∴ Co-ordinates of midpoint P are `square`.
