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प्रश्न
Find the lengths of the medians of a triangle whose vertices are A(–1, 1), B(5, –3) and C(3, 5).
उत्तर
Let the medians meet the lines BC, AC and AB at points be
\[P\left( x_1 , y_1 \right)\]
\[Q\left( x_2 , y_2 \right)\] and
\[R\left( x_3 , y_3 \right)\]respectively.
P is thus the mid point of line BC
\[P\left( x_1 , y_1 \right) = \left( \frac{5 + 3}{2}, \frac{5 - 3}{2} \right) = \left( 4, 1 \right)\]
\[AP = \sqrt{\left( -4 - 1 \right)^2 + \left( 1 - 1 \right)^2} = \sqrt{25} = 5\]
Q is the mid point of line AC.
\[Q\left( x_2 , y_2 \right) = \left( \frac{- 1 + 3}{2}, \frac{1 + 5}{2} \right) = \left( 1, 3 \right)\]
\[BQ = \sqrt{\left( 5 - 1 \right)^2 + \left( - 3 - 3 \right)^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}\]
R is the mid point of AB.
\[R\left( x_3 , y_3 \right) = \left( \frac{- 1 + 5}{2}, \frac{1 - 3}{2} \right) = \left( 2, - 1 \right)\]
\[RC = \sqrt{\left( 3 - 2 \right)^2 + \left( -1 - 5 \right)^2} = \sqrt{1 + 36} = \sqrt{37}\]
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The coordinates of the vertices of a triangle ABC are A (–7, 6), B(2, –2) and C(8, 5). Find coordinates of its centroid.
Solution: Suppose A(x1, y1) and B(x2, y2) and C(x3, y3)
x1 = –7, y1 = 6 and x2 = 2, y2 = –2 and x3 = 8, y3 = 5
Using Centroid formula
∴ Coordinates of the centroid of a traingle
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= `(square/3, square/3)`
∴ Coordinates of the centroid of a triangle ABC = `(3/3, square)`
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