The centre of the ellipse `(x+y-3)^(2)/9+(x-y+1)^(2)/16=1` is

Options

A. (-1,2)
B. (1,-2)
C. (-1,-2)
D. (1,2)

Answer

To find the center of the ellipse given by the equation

\[ \frac{(x+y-3)^2}{9} + \frac{(x-y+1)^2}{16} = 1, \]

we need to analyze the equation and identify the center.

### Step-by-Step Solution:

1. **Identify the standard form of the ellipse**:
The given equation is in the form of an ellipse centered at some point \((h, k)\). The general form of an ellipse is

\[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1. \]

However, our equation is not in this standard form yet.

2. **Rewrite the equation**:
We can rewrite the equation to identify the center. The terms \((x+y-3)\) and \((x-y+1)\) suggest a transformation of variables.

Let:
\[ u = x + y \quad \text{and} \quad v = x - y. \]

Then, we can express the ellipse in terms of \(u\) and \(v\):
\[ \frac{(u-3)^2}{9} + \frac{(v+1)^2}{16} = 1. \]

3. **Find the center in terms of \(u\) and \(v\)**:
The center of this ellipse in the \(u-v\) coordinate system is at \((3, -1)\).

4. **Convert back to \(x\) and \(y\)**:
We need to convert the center back to the \(x-y\) coordinate system. We have:
\[ u = x + y = 3 \quad \text{and} \quad v = x - y = -1. \]

We can solve these two equations simultaneously to find \(x\) and \(y\).

5. **Solve the equations**:
From \(u = 3\):
\[ x + y = 3 \quad \text{(1)} \]
From \(v = -1\):
\[ x - y = -1 \quad \text{(2)} \]

Now, we can add equations (1) and (2):
\[ (x + y) + (x - y) = 3 - 1 \implies 2x = 2 \implies x = 1. \]

Substitute \(x = 1\) back into equation (1):
\[ 1 + y = 3 \implies y = 2. \]

6. **Conclusion**:
The center of the ellipse is at the point \((1, 2)\).

### Final Answer:
The center of the ellipse is \((1, 2)\).

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