Find the centroid of the triangle with these vertices: P (-1,7), Q (9,5), and R (4,3).
The coordinates of the centroid are?

[tex]\red{\boxed{ \sf{Centroid \: = \: \dfrac{x_1 \: + \: x_2 \: + \: x_3}{3} } \: , \: \dfrac{y_1 \: + \: y_2 \: + \: y_3}{3} }} \: \pink\bigstar [/tex]

We have :

x₁ = -1
x₂ = 9
x3 = 4
y₁ = 7
y₂ = 5
y3 = 3

Substituting the values :

★ For x co-ordinate:-

[tex]: \: \implies \: \sf{x\: = \: \dfrac{x_{1} \: + \: x_2 \: + \: x_3}{3} } \\ \\ : \: \implies \: \sf{x\: = \: \dfrac{ - 1 \: + \: 9 \: + \: 4}{3}} \\ \\ : \: \implies \: \sf{x\: = \: \dfrac{ 8 \: + \: 4}{3} } \\ \\ : \: \implies \: \sf{x\: = \: \dfrac{ 12}{3} } \\ \\ : \: \implies \: \sf{x\: = \: \cancel\dfrac{ 12}{3} } \\ \\ : \: \implies \: \sf{x\: = \: 4 }[/tex]

★ For y co-ordinate:-

[tex]: \: \implies \: \sf{y\: = \: \dfrac{y_{1} \: + \: y_2 \: + \: y_3}{3} } \\ \\ : \: \implies \: \sf{y\: = \: \dfrac{ 7 + 5 + 3}{3}} \\ \\ : \: \implies \: \sf{y\: = \: \dfrac{ 10 \: + \: 5}{3} } \\ \\ : \: \implies \: \sf{y\: = \: \dfrac{ 15}{3} } \\ \\ : \: \implies \: \sf{y\: = \: \cancel\dfrac{ 15}{3} } \\ \\ : \: \implies \: \sf{y\: = \: 5 }[/tex]

★ Henceforth,

Co-ordinates are (4 , 5)

Additional Information :

★ Midpoint of two points:-

[tex]\boxed{ \sf{M \: = \: \dfrac{x_1 \: + \: x_2 }{2} \: , \: \dfrac{y_1 \: + \: y_2 }{2}}} \: \pink\bigstar[/tex]

★ Distance Formula :-

[tex]\huge \large \boxed{\sf{{d \: = \: \sqrt{(x _{2} - x _{1}) {}^{2} \: + \: (y _{2} - y _{1}) {}^{2} }}}} \: \red\bigstar[/tex]

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Find the centroid of the triangle with these vertices: P (-1,7), Q (9,5), and R (4,3). The coordinates of the centroid are?​

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Find the centroid of the triangle with these vertices: P (-1,7), Q (9,5), and R (4,3).
The coordinates of the centroid are?