Question

Find the centroid of the triangle formed by the points (2,7),(3,-1),(-5,6)

Answer 1

ㅤㅤㅤㅤㅤ$\pmb {Given:-}$

✇ Vertices of triangle are (2, 7) , (3, -1) , (-5,6)

ㅤㅤ$\pmb{To\: find:-}$

✇ Centroid of the triangle .

ㅤㅤㅤ$\pmb{Diagram:-}$

$\setlength{\unitlength}{2.5mm}\begin{picture}(0,0)\linethickness{0.3mm}\qbezier(0,0)(0,0)(8,17)\qbezier(0,0)(0,0)(18,0)\qbezier(18,0)(18,0)(8,17)\put(8,17.8){\sf A=(2,7)}\put( - 1, - 1){\sf B=(3,-1)}\put(18,-1){\sf C=(-5,6)}\put(8, - 1.5){\sf }\put(15,8.1){\sf }\put( - 0,8.1){\sf }\end{picture}$

ㅤㅤ$\pmb{Required\: formula:-}$

If (x1, y1) , (x2,y2) , (x3,y3) are the vertices of triangle then, Centroid of triangle is given by ,

$\rm{\bigg(\dfrac{x_1+x_2+x_3}{3} ,\dfrac{y_1+y_2+y_3}{3}}\bigg)$ㅤㅤㅤ‏‏‎ ‎‏‏‎ ‎________________________

$\pmb{(x_1,y_1) =(2,7)}$

$\pmb{(x_2,y_2) = (3, -1)}$

$\pmb{(x_3,y_3) = (-5,6)}$

ㅤㅤ$\mathfrak{Substituting\: the \: values\: in\: formula}$

$\rm {G=}\rm{\bigg(\dfrac{x_1+x_2+x_3}{3} ,\dfrac{y_1+y_2+y_3}{3}}\bigg)$

$\rm G=\bigg(\rm \dfrac{2+3-5}{3} , \dfrac{7-1+6}{3}\bigg)$

$\rm G =\bigg(\rm \dfrac{5-5}{3} , \dfrac{13-1}{3}\bigg)$

$\rm G =\bigg(\rm \dfrac{0}{3} , \dfrac{12}{3}\bigg)$

$\rm G = (0,4)$

$\mathfrak {So,\: the\: centroid \: of \: triangle \: is (0,4)}$ㅤㅤㅤㅤ‎‎‎

ㅤㅤㅤ_______________________

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$\pmb {Know \: more\: formulae :-}$

$\pmb {Section \: formula (Internal \: division)}\\\rm \bigg(\dfrac{mx_2+nx_1}{m+n} , \dfrac{my_2+ny_1}{m+n} \bigg)$

$\pmb {Section \: formula\:External \: division)}\\\rm \bigg(\dfrac{mx_2-nx_1}{m-n} , \dfrac{my_2-ny_1}{m-n} \bigg)$

$\pmb{Distance\: formula:-}\\\rm \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

$\pmb {Midpoint\: formula:-}\\\rm \bigg(\dfrac{x_1+x_2}{2} , \dfrac{y_1+y_2}{2}\bigg)$

$\pmb{Area\: of \: triangle:-}\\\rm \dfrac{1}{2}\bigg|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\bigg|$

Answer 2

Answer:

0,4

Step-by-step explanation:

Centroid of triangle = (x1+x2+x3)/3, (y1+y2+y3) /3

So,

(2+3-5)/3, (7-1+6)/3

0,4

The centroid is 0,4