If the vertices of a triangle $$ \Delta ABC $$ are $$ A(1, 3), B(3, 5), ext{and } C(x, y) $$ and the centroid is $$ \left(\frac{11}{3}, \frac{16}{3} ight) $$, then find the values of $$ (x, y) $$.

Question

If the vertices of a triangle $$ \Delta ABC $$ are $$ A(1, 3), B(3, 5), 	ext{and } C(x, y) $$ and the centroid is $$ \left(\frac{11}{3}, \frac{16}{3}ight) $$, then find the values of $$ (x, y) $$.

Accepted Answer

The centroid of a triangle is given by the formula:  
$$
	ext{Centroid} = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}ight),
$$  
where $$ (x_1, y_1), (x_2, y_2), (x_3, y_3) $$ are the coordinates of the vertices.

Given:  
- Vertex A = (1, 3), Vertex B = (3, 5), Vertex C = (x, y).  
- Centroid = $$ \left(\frac{11}{3}, \frac{16}{3}ight). $$

### Step 1: Solve for $$ x $$
Using the x-coordinate of the centroid:  
$$
\frac{x_1 + x_2 + x_3}{3} = \frac{11}{3}.
$$  
Substitute $$ x_1 = 1 $$ and $$ x_2 = 3 $$:  
$$
\frac{1 + 3 + x}{3} = \frac{11}{3}.
$$  
Multiply through by 3:  
$$
1 + 3 + x = 11.
$$  
Simplify:  
$$
x = 7.
$$

### Step 2: Solve for $$ y $$
Using the y-coordinate of the centroid:  
$$
\frac{y_1 + y_2 + y_3}{3} = \frac{16}{3}.
$$  
Substitute $$ y_1 = 3 $$ and $$ y_2 = 5 $$:  
$$
\frac{3 + 5 + y}{3} = \frac{16}{3}.
$$  
Multiply through by 3:  
$$
3 + 5 + y = 16.
$$  
Simplify:  
$$
y = 8.
$$

### Final Answer:
The coordinates of vertex C are **(7, 8)**. Hence, the correct answer is **(b)**.

Answer

Answer

Answer