Question

Find the third vertex of a triangle, if two of its vertices are at (-3, 1) and (0, -2) and the centroid is at the origin.

Answer 1

$\text{Concept used:}$

$\text{Centroid of a triangle having vertices (x_1,y_1), (x_2,y_2)\; and (x_3,y_3) is}$

$\bf(\frac{x_1+x_2+x_3 }{3},\frac{y_1+y_2+y_3 }{3})$

$\textbf{Given:}$

$(x_1,y_1)=(-3,1)$

$(x_2,y_2)=(0,-2)$

$\text{and Centroid is (0,0)}$

$(\frac{x_1+x_2+x_3 }{3},\frac{y_1+y_2+y_3 }{3})=(0,0)$

$(\frac{-3+0+x_3 }{3},\frac{1+(-2)+y_3 }{3})=(0,0)$

$\implies\frac{-3+0+x_3 }{3}=0,\;\frac{1+(-2)+y_3 }{3})=0$

$\implies\,-3+x_3 =0,\;-1+y_3=0$

$\implies\,x_3 =3,\;y_3=1$

$\therefore\textbf{The third vertex is (3,1)}$

Find more:

A straight line has non -zero intercepts a and b on the X axis and Y axis. find the centroid of the triangle formed with the axes and the origin

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