Find the centroid of the triangle whose vertices are (−1,0), (1,0), 𝑎𝑛𝑑 (0,3
We know that the centroid of the triangle (x1+x2+x33;y1+y2+y33)(\frac{x_1+x_2+x_3}{3};\frac{y_1+y_2+y_3}{3})(3x1+x2+x3;3y1+y2+y3)
Where (x1;y1);(x2;y2);(x3;y3)(x_1;y_1);(x_2;y_2);(x_3;y_3)(x1;y1);(x2;y2);(x3;y3) are the vertices of the
triangle SO, centroid= ((−1)+1+03;0+0+33)=(03;33)=(0;1)(\frac{(-1)+1+0}{3};\frac{0+0+3}{3})=(\frac{0}{3};\frac{3}{3})=(0;1)(3(−1)+1+0;30+0+3)=(30;33)=(0;1)
Hence, this is the answer.
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments