Question

what is the centroid of a triangle with the coordinates (-5, 4) (4, -3) (7, 2)​

Answer 1

Answer:-

Given vertices of a triangle are ( - 5 , 4) , (4 , - 3) , (7 , 2).

We know that,

Centroid of a triangle with vertices (x₁ , y₁) , (x₂, y₂) , (x₃ , y₃) is :

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

Let,

• x₁ = - 5
• y₁ = 4
• x₂ = 4
• y₂ = - 3
• x₃ = 7
• y₃ = 2

Hence,

$\implies \: \sf \: G(x \: , \: y) = \bigg( \dfrac{ - 5 + 4+ 7}{3} \: \: ,\: \: \dfrac{4 - 3 + 2}{3} \bigg) \\ \\ \implies \sf \: G(x \:, \: y) = \bigg( \dfrac{6}{3} \: \: ,\: \: \dfrac{3}{3} \bigg) \\ \\ \\\implies \boxed{ \sf \: G(x \: , \: y) = ( 2 \: \: ,\: \: 1 ) }$

∴ Centroid of the given triangle is (2 , 1).

Some Important Formulae:-

• Distance between two points: $\sf \sqrt{(x_2 - x_1)^2 + (y_2-y_1)^2}$

• Area of triangle: $\sf \: \frac{1}{2} \begin{vmatrix} \sf \: x_1 - x_2& \sf x_1 - x_3 \\ \\ \sf y_1 - y_2& \sf \: y_1 - y_3 \end{vmatrix}$

• Area of Quadrilateral: $\sf \: \frac{1}{2} \begin{vmatrix} \sf \: x_1 - x_3& \sf x_2 - x_4 \\ \\ \sf y_1 - y_3& \sf \: y_2- y_4 \end{vmatrix}$

• Mid point of a line segment : p(x , y) = [ (x₁ + x₂)/2 , (y₁ + y₂)/2 ]

• Section formula: $\sf \bigg( \dfrac{mx_2 \pm nx_1}{m \pm \: n} \: \: , \: \: \dfrac{my_2 \pm \: ny_1}{m \pm \: n} \bigg) \: \: \: \binom{+ \longrightarrow \: internally}{ - \longrightarrow \: externally}$