Question

If the coordinates of the mid-points of the sides of a triangle are ( 1 , 1), ( 2 , -3) and ( 3 , 4). Find its centroid.

Answer 1

Step-by-step explanation:

hope it helps u...........

Answer 2

$\overbrace{ \underbrace{ \fcolorbox{white}{pink}{ \blue\dag \tt Given \red\dag}}}$

$\tt The \: mid \: point \: of \: the \: sides \: of \\ \tt a \: triangle \: are (1, 5, -1), (0, 4,-2) and (2, 3, 4).$

$\overbrace{ \underbrace{ \fcolorbox{white}{blue}{ \blue\dag \tt \red{Find} \purple\dag}}}$

$\tt Coordinates \: of \: vertices$

$\overbrace{ \underbrace{ \fcolorbox{white}{red}{ \blue\dag \tt \green{Solution} \purple\dag}}}$

$\tt Let \: the \: vertices \: of \: triangle \: be \\ \tt A( x_1,y_1,z_1) , B( x_2,y_2,z_2) , C( x_3,y_3,z_3 )$

$\tt Mid - point \: of \: side \: AC \: is \: Q.$

$\tt \therefore( \frac{x_1+x_3}{2} , \frac{y_1+y_3}{2} , \frac{z_1+z_3}{2} ) = (0,4 , - 2)$

$\tt substituting \: values \: we \: get,$

$\tt So,C(x_3,y_3,z_3)=C(-x_1,8-y_1,-4-z_1).....(i)$

$\tt Mid - point \: of \: side \: AB \: is \: P.$

$\tt \therefore( \frac{x_1+x_2}{2} , \frac{y_1+y_2}{2} , \frac{z_1+z_2}{2} ) = (2,3 , 4)$

$\tt So,B(x_2,y_2,z_2)=B(4-x_1,6-y_1,8-z_1).....(ii)$

$\tt Mid - point \: of \: side \: BC \: is \: R.$

$\tt \therefore( \frac{x_2+x_3}{2} , \frac{y_2+y_3}{2} , \frac{z_2+z_3}{2} ) = (1,5,1)$

$\tt \therefore( \frac{ - x_1+4 - x_1}{2} = 1 , \frac{8 - y_1 + 6 - y_1}{2} = 5 , \frac{ - 4 - z_1 + 8 - z_1}{2} = 1)$

$\tt \implies x_1= 1,y_1 = 2,z_1= 3$

$\tt \therefore A = (1,2,3)$

$\tt by \: using \: eq(i) \ratio B = (3,4,5) \\ \tt by \: using \: eq(ii) \ratio C = ( - 1,6, - 7)$

$\tt So, \: centroid \: K \frac{(1 + 3 - 1)}{3} , \frac{(2 + 4 + 6)}{3} , \frac{(3 + 5 - 7)}{3} = (1 ,4 , \frac{1}{3})$

$\tt The \: vertices \: of \: triangle \: are \: A(1 ,2 ,3) \\ \tt B (3 ,4 ,5) and \: C( - 1 ,6 , - 7). \: Also, centroid \\ \tt of \: the \: triangle \: is \: K( 1 ,4 , \frac{1}{3} )$

$\tt Hence, vertices \: are\boxed{ \tt A(1 ,2 ,3) B (3 ,4 ,5) and \: C( - 1 ,6 , - 7).} \\ \tt and \: centroid \: are \boxed{ K( 1 ,4 , \frac{1}{3} )}$