Question

the coordinates of centroid of the triangles if points D(-7, 6), E(8, 5) and
F(2, -2) are the mid points of the sides of that triangle. Also find co-ordinates if Point A,B and C

Answer 1

Given :

Coordinates of the triangle are:

•D=(-7,6)

•E=(8,5)

•F=(2,-2)

To Find :

• The centroid of the triangle =?

Formula :

Centroid of the triangle :

$\\ \red\bigstar\boxed{\sf{\bigg(\dfrac{x_{1}+x_{2}+x_{3}}{3} , \dfrac{y_{1}+y_{2}+y_{3}}{3}\bigg) }}$

Solution :

$\red{\bf{Values}} \begin{cases} \sf{D=(-7,6) } \\ \sf{E=(8,5) } </p><p>\\ \sf{F=(2,-2)} \end{cases}$

$\\ \\ \implies\sf{Centroid \: of \: triangle \: \triangle DEF=\bigg(\dfrac{x_{1}+x_{2}+x_{3}}{3} , \dfrac{y_{1}+y_{2}+y_{3}}{3} \bigg) }$

Substituting the given values :

$\\ \\ \\ \implies\sf{Centroid \: of \: triangle \triangle DEF=\bigg(\dfrac{-7+8+2}{3}, \dfrac{6+5-2}{3}\bigg) }$

$\\ \\ \\ \implies\sf{Centroid \: of \: triangle \triangle DEF=\bigg(\dfrac{-7+10}{3},\dfrac{11-2}{3}\bigg) }$

$\\ \\ \\ \implies\sf{Centroid \: of \: triangle \triangle DEF=\bigg(\dfrac{3}{3} ,\dfrac{9}{3}\bigg) }$

$\\ \\ \\ \implies\sf{Centroid \: of \: triangle \triangle DEF=(1,3)}$

$\\ \\ \red\bigstar\green{\boxed{\sf{Centroid \: of \: triangle \triangle DEF=(1,3)}}}$

Answer 2

☆Given Question :-

Find the coordinates of centroid of the triangles if the points D(-7, 6), E(8, 5) and F(2, -2) are the mid points of the sides of that triangle. Also find co-ordinates if Point A,B and C.

$\huge \orange{AηsωeR}$

☆Given :-

• A triangle ABC, in which D(-7, 6), E(8, 5) and F(2, -2) are the mid points of the sides.

☆To Find :-

• Vertices of triangle ABC
• Centriod of triangle ABC.

☆Formula Used :-

☆ Midpoint Formula

Let us consider a line segment AB having coordinates

$\sf \: A(x_1,y_1) \: and \: B (x_2,y_2)$

and let C (x, y) be the midpoint of AB, then coordinates of C using midpoint Formula is given by

$\bf \:( x, y) = (\dfrac{x_1+x_2}{2} , \dfrac{y_1+y_2}{2} )$

☆Centriod of triangle

Let us consider a triangle ABC, having coordinates

$\sf \: A(x_1,y_1) \: and \: B(x_2,y_2) \: and \: C(x_3,y_3)$

and let G(x, y) be the centriod of triangle, then centriod is

$\bf \:( x, y) = (\dfrac{x_1+x_2 + x_3)}{3} , \dfrac{y_1+y_2 + y_3}{3} )$

☆Solution :-

⟼Let the coordinates of triangle ABC be A(a, b), B(c, d) and C(e, f) respectively.

⟼Let assume that D(-7, 6), E(8, 5) and F(2, -2) are the mid points of the sides AB, BC and CA respectively.

⟼Now, D(-7, 6) is the midpoint of AB.

⟼So, using midpoint Formula, we get

$\bf \:  ⟼ ( - 7, 6) = (\dfrac{a+c}{2} , \dfrac{b+d}{2} )$

On comparing, we get

$\bf \:  ⟼ \dfrac{a + c}{2} = - 7 \: and \: \dfrac{b + d}{2} = 6$

$\bf\implies \:a + c = - 14 \: ⟼ \: (1)$

$\bf\implies \:b + d = 12 \: ⟼ \: (2)$

⟼ Now, E (8, 5) is the midpoint of BC,

⟼ So, using midpoint Formula, we get

$\bf \:( 8 ,5 ) = (\dfrac{c+e}{2} , \dfrac{d+f}{2} )$

⟼ On comparing, we get

$\bf \:  ⟼ \dfrac{ c+ e}{2} = 8 \: and \: \dfrac{d+ f}{2} = 5$

$\bf\implies \:e + c = 16 \: ⟼ \: (3)$

$\bf\implies \:d + f = 10 \: ⟼ \: (4)$

⟼ Now, F(2, -2) is the midpoint of AC.

⟼ So, using midpoint Formula, we get

$\bf \:(2 , - 2) = (\dfrac{a+e}{2} , \dfrac{b+f}{2} )$

⟼ On comparing, we get

$\bf \:  ⟼ \dfrac{ a+e }{2} = 2 \: and \: \dfrac{b \: + f}{2} = - 2$

$\bf\implies \:a + e = 4 \: ⟼ \: (5)$

$\bf\implies \:b + f = - 4\: ⟼ \: (6)$

⟼ On Adding, equation (1), (3) and (5), we get

$\bf \:  ⟼ a + c + c + e + e + a = - 14 + 16 + 4$

$\bf \:  ⟼ 2(a + c + e) = 6$

$\bf\implies \:a + c + e = 3 \: ⟼ \: (7)$

⟼ On substituting, equation (1) in (7), we get

$\bf \:  ⟼ - 14 + e = 3$

$\bf\implies \:e = 17$

⟼ On substituting equation (3) in equation (7), we get

$\bf \:  ⟼ a + 16 = 3$

$\bf \:  ⟼ a = - 13$

⟼ On substituting equation (5) in equation (7), we get

$\bf \:  ⟼ 4 + c = 3$

$\bf\implies \:c = - 1$

⟼ Now, on Adding equation (2), (4) and (6), we get

$\bf \:  ⟼ b + d + d + f+ f+ b = 10 + 12 - 4$

$\bf \:  ⟼ 2( b+ d + f) = 18$

$\bf\implies \: b+ d + f = 9 \: ⟼ \: (8)$

⟼ On substituting equation (2) in equation (8), we get

$\bf \:  ⟼ 12 + f = 9$

$\bf\implies \:f = - 3$

⟼ On substituting equation (4) in equation (8), we get

$\bf \:  ⟼ b + 10 = 9$

$\bf\implies \:b = - 1$

⟼ On substituting equation (6) in equation (8), we get

$\bf \:  ⟼ d - 4 = 9$

$\bf\implies \:d = 13$

So coordinates of triangle ABC are

⟼ A (- 13, - 1)

⟼ B (- 1, 13)

⟼ C (17, - 3)

To find Centriod of triangle ABC

⟼ The coordinates of triangle ABC are A (- 13, - 1), B (- 1, 13), C (17, - 3), then Centriod G(x, y) is given by

$\bf \:  ⟼ G =( \dfrac{ - 13 - 1 + 17}{3} , \dfrac{ - 1 + 13 - 3}{3} )$

$\bf\implies \:G = (1, 3)$

$\begin{gathered}\begin{gathered}\bf Hence = \begin{cases} &\sf{Vertices \: are \: A (- 13, - 1),B (- 1, 13),C (17, - 3)} \\ &\sf{Centroid \: of \: triangle \: is \: G(1, 3)} \end{cases}\end{gathered}\end{gathered}$