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Sagot :
Given:
Vertices of a triangle ABC are A(-3,0), B(3,0) and C(0,4).
To find:
The intersection point of all the medians of the triangle ABC.
Solution:
We know that, intersection point of all the medians of a triangle is called centroid.
The formula of centroid is
[tex]Centroid=\left(\dfrac{x_1+x_2+x_3}{3},\dfrac{y_1+y_2+y_3}{3}\right)[/tex]
Vertices of a triangle ABC are A(-3,0), B(3,0) and C(0,4).
[tex]Centroid=\left(\dfrac{-3+3+0}{3},\dfrac{0+0+4}{3}\right)[/tex]
[tex]Centroid=\left(\dfrac{0}{3},\dfrac{4}{3}\right)[/tex]
[tex]Centroid=\left(0,1.333...\right)[/tex]
Round each coordinate to the nearest tenth.
[tex]Centroid=\left(0,1.3\right)[/tex]
The coordinates of required point are (0,1.3). Therefore, the correct option is C.
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