Question

find the coordinates of the centroid of the triangle having vertices at (-1,5),(1,-2) and (3,3).

Answer 1

Question 1.

Find the coordinates of the centroid of the triangle having vertices at $(-1,5)$ , $(1,-2)$ and $(3,3)$ .

Solution:

A(-1, 5), B(1, -2), C(3, 3).

Coordinates $(\mathrm{x},\mathrm{y})$ of the middle of the line segment with endpoints $\mathrm{A}_1(\mathrm{x}_1,\mathrm{y}_1)$ and $\mathrm{A}_2(\mathrm{x}_2,\mathrm{y}_2)$ are

$$x = \frac {x _ {1} + x _ {2}}{2}, \qquad y = \frac {y _ {1} + y _ {2}}{2}.$$

Let's find coordinates of the middle of the line segments AB, AC, BC.

$$x _ {E} = \frac {x _ {A} + x _ {B}}{2} = \frac {- 1 + 1}{2} = 0, \qquad y _ {E} = \frac {y _ {A} + y _ {B}}{2} = \frac {5 - 2}{2} = \frac {3}{2}.$$ $$E \left(x _ {E}, y _ {E}\right) = E \left(0, \frac {3}{2}\right).$$ $$x _ {F} = \frac {x _ {A} + x _ {C}}{2} = \frac {- 1 + 3}{2} = 1, \qquad y _ {F} = \frac {y _ {A} + y _ {C}}{2} = \frac {5 + 3}{2} = 4.$$ $$F \left(x _ {F}, y _ {F}\right) = F (1, 4).$$ $$x _ {D} = \frac {x _ {C} + x _ {B}}{2} = \frac {3 + 1}{2} = 2, \qquad y _ {D} = \frac {y _ {C} + y _ {B}}{2} = \frac {3 - 2}{2} = \frac {1}{2}.$$ $$D \left(x _ {D}, y _ {D}\right) = D \left(2, \frac {1}{2}\right).$$

AD, CE, BF are the medians.

Let's find the equation for lines AD, CE, BF:

The equation for non-vertical lines is often given in the slope-intercept form:

$$y = m x + c$$

Where:

m is the slope of the line.

c is the y-intercept of the line.

x is the independent variable of the function $y = f(x)$ .

Therefore,

AD:

$$y = m x + c$$ $$D \left(2, \frac {1}{2}\right), \quad A (- 1, 5)$$ $$\left\{ \begin{array}{l} \frac {1}{2} = 2 m + c \\ 5 = - m + c \end{array} \right. \quad \left\{ \begin{array}{l} 1 = 4 m + 2 c \\ 10 = - 2 m + 2 c \end{array} \right. = > m = - \frac {3}{2}, c = \frac {7}{2}$$ $$y _ {A D} = - \frac {3}{2} x + \frac {7}{2}.$$

CE:

$$y = m x + c$$ $$\mathrm {C} (3, 3), \quad E \left(0, \frac {3}{2}\right)$$ $$\left\{ \begin{array}{c} \frac {3}{2} = c \\ 3 = 3 m + c \end{array} \right. \quad \left\{ \begin{array}{c} \frac {3}{2} = c \\ m = \frac {1}{2} \end{array} \right.$$ $$y _ {C E} = \frac {1}{2} x + \frac {3}{2}.$$

BF:

$$y = m x + c$$ $$\mathrm {B} (1, - 2), \quad F (1, 4).$$ $$x _ {b} = 1 \text{ and } x _ {F} = 1 = > B F \text{ is a vertical line } x = 1.$$

The three medians intersect in a single point, the triangle's centroid.

$$\left\{ \begin{array}{c} y _ {A D} = - \frac {3}{2} x + \frac {7}{2} \\ y _ {C E} = \frac {1}{2} x + \frac {3}{2} \\ x = 1 \end{array} \right. \quad \left\{ \begin{array}{c} y _ {A D} = 2 \\ y _ {C E} = 2 \\ x = 1 \end{array} \right.$$

G is a centroid of the triangle. So, G(1, 2).

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