Finding the Centroid of Triangle ABC using Coordinates

Let's find the centroid of triangle ABC by first determining the coordinates of each vertex A, B, and C. We are given some key pieces of information about the triangle.

Given Information about Triangle ABC

• Coordinates of vertex B: $(0, 0)$.
• Length of side AB: $AB = 2$.
• Angle at B: $\angle ABC = \pi/3$ radians, which is equal to 60 degrees.
• Coordinates of M, the midpoint of side BC: $(2, 0)$.

Determining the Vertex Coordinates

We already have the coordinates for vertex B: $B = (0, 0)$.

Next, let's find the coordinates of vertex C. We know that M $(2, 0)$ is the midpoint of the line segment BC. Using the midpoint formula, if $B = (x_B, y_B)$ and $C = (x_C, y_C)$, the midpoint $M = \left(\frac{x_B + x_C}{2}, \frac{y_B + y_C}{2}\right)$.

Substituting the known values:

$(2, 0) = \left(\frac{0 + x_C}{2}, \frac{0 + y_C}{2}\right)$

This gives us two equations:

• $\frac{x_C}{2} = 2 \implies x_C = 4$
• $\frac{y_C}{2} = 0 \implies y_C = 0$

So, the coordinates of vertex C are $C = (4, 0)$.

Now, let's find the coordinates of vertex A. Let $A = (x_A, y_A)$. We know that the distance from B to A is 2, so $AB = \sqrt{(x_A - 0)^2 + (y_A - 0)^2} = \sqrt{x_A^2 + y_A^2} = 2$. This means $x_A^2 + y_A^2 = 4$.

We are also given that the angle $\angle ABC = \pi/3$. We can use the dot product of the vectors $\vec{BA}$ and $\vec{BC}$ to relate the angle and the vertex coordinates. The vector $\vec{BA}$ is $(x_A - 0, y_A - 0) = (x_A, y_A)$. The vector $\vec{BC}$ is $(4 - 0, 0 - 0) = (4, 0)$.

The dot product is given by $\vec{BA} \cdot \vec{BC} = |\vec{BA}| |\vec{BC}| \cos(\angle ABC)$.

We know $|\vec{BA}| = AB = 2$ and $|\vec{BC}| = \sqrt{(4-0)^2 + (0-0)^2} = \sqrt{16} = 4$. Also, $\cos(\pi/3) = \frac{1}{2}$.

So, $x_A(4) + y_A(0) = (2)(4)\cos(\pi/3)$

$4x_A = 8 \times \frac{1}{2}$

$4x_A = 4$

$x_A = 1$

Now substitute $x_A = 1$ into the distance equation $x_A^2 + y_A^2 = 4$:

$1^2 + y_A^2 = 4$

$1 + y_A^2 = 4$

$y_A^2 = 3$

$y_A = \pm \sqrt{3}$.

Assuming the triangle is above the x-axis (based on standard representations and the options), we take the positive value, $y_A = \sqrt{3}$.

So, the coordinates of vertex A are $A = (1, \sqrt{3})$.

The coordinates of the vertices of $\Delta ABC$ are:

• A: $(1, \sqrt{3})$
• B: $(0, 0)$
• C: $(4, 0)$

Calculating the Centroid of Triangle ABC

The centroid G of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by the formula:

$G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)$

Substitute the coordinates of A, B, and C into the formula for the centroid of triangle ABC:

$G = \left( \frac{1 + 0 + 4}{3}, \frac{\sqrt{3} + 0 + 0}{3} \right)$

$G = \left( \frac{5}{3}, \frac{\sqrt{3}}{3} \right)$

We can rationalize the denominator of the y-coordinate:

$\frac{\sqrt{3}}{3} = \frac{\sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{1}{\sqrt{3}}$

So, the coordinates of the centroid of triangle ABC are $G = \left( \frac{5}{3}, \frac{1}{\sqrt{3}} \right)$.

Summary of Coordinate Geometry Steps

We used the midpoint formula to find vertex C, and a combination of the distance formula and the dot product (which relates the angle between vectors) to find vertex A. With the coordinates of all three vertices (A, B, and C) known, we applied the standard formula for the centroid of a triangle to get the final result. This systematic approach using coordinate geometry principles allowed us to determine the centroid.

The centroid of triangle ABC is $\left( \frac{5}{3}, \frac{1}{\sqrt{3}} \right)$.