Answer:
[tex]3x+3y-5=0[/tex]
Step-by-step explanation:
The vertices of a triangle ABC are A(3, 2), B(1, -1) and C(5, -5).
Parallel lines share the same slope; hence, the slope of the straight line passing through the centroid of triangle ABC and parallel to side BC is identical to the slope of line BC. To determine the equation of this line, we first calculate the slope of side BC by substituting the coordinates of points B and C into the slope formula:
[tex]\textsf{Slope of $\overline{BC}$}=\dfrac{y_C-y_B}{x_C-x_B}=\dfrac{-5-(-1)}{5-1}=\dfrac{-4}{4}=-1[/tex]
The centroid of a triangle is the point of intersection of its medians, each of which connects a vertex to the midpoint of the opposite side.
The formula for the coordinates of the centroid of a triangle is:
[tex]\boxed{\begin{array}{c}\underline{\textsf{Centroid of a triangle}}\\\\\left(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}\right)\\\\\textsf{where $A(x_1, y_1), B(x_2, y_2)$, and $C(x_3, y_3)$ are the vertices of the triangle.}\end{array}}[/tex]
Substitute the coordinates of vertices A, B and C into the centroid formula:
[tex]\textsf{Centroid}=\left(\dfrac{3 + 1 + 5}{3}, \dfrac{2 + (-1) +(-5)}{3}\right)[/tex]
[tex]\textsf{Centroid}=\left(\dfrac{9}{3}, \dfrac{-4}{3}\right)[/tex]
[tex]\textsf{Centroid}=\left(3, -\dfrac{4}{3}\right)[/tex]
Therefore, the coordinates of the centroid are:
[tex]\left(3, -\dfrac{4}{3}\right)[/tex]
Now we have the slope m = -1 and the coordinates of the point it passes through (3, -4/3), we can substitute them into the point-slope formula to create an equation of the line:
[tex]\begin{aligned}y-y_1&=m(x-x_1)\\\\y-\left(-\dfrac{4}{3}\right)&=-1(x-3)\\\\y+\dfrac{4}{3}\right)&=-x+3\\\\3y+4&=-3x+9\\\\3x+3y+4-9&=0\\\\3x+3y-5&=0\end{aligned}[/tex]
Therefore, the equation of a straight line passing through the centroid of triangle ABC and parallel to the side BC is:
[tex]\Large\boxed{\boxed{3x+3y-5=0}}[/tex]
