The equations of the sided of a triangle are `x+y-5=0,x-y+1=0,` and `x+y-sqrt(2)=0` is `(-oo,-4/3)uu(4/3,+oo)` `(-4/3,4/3)` (c) `(-3/4,4/3)` none of these

Options

A. `(-oo, -4//3) uu (4//3, +oo)`
B. (-4/3, 4/3)
C. (-3/4,4/3)
D. none of these

Explanation

To solve the problem, we need to find the area of the triangle formed by the given lines and determine the circumcenter. The equations of the sides of the triangle are:

1. \( x + y - 5 = 0 \)
2. \( x - y + 1 = 0 \)
3. \( x + y - \sqrt{2} = 0 \)

### Step 1: Find the Points of Intersection

We will find the points of intersection of the lines to determine the vertices of the triangle.

**Intersection of Line 1 and Line 2:**

From Line 1:
\[ y = 5 - x \]

Substituting into Line 2:
\[ x - (5 - x) + 1 = 0 \]
\[ x - 5 + x + 1 = 0 \]
\[ 2x - 4 = 0 \]
\[ x = 2 \]

Now substitute \( x = 2 \) back into Line 1:
\[ y = 5 - 2 = 3 \]

So, the first vertex is \( A(2, 3) \).

---

**Intersection of Line 1 and Line 3:**

From Line 1:
\[ y = 5 - x \]

Substituting into Line 3:
\[ x + (5 - x) - \sqrt{2} = 0 \]
\[ 5 - \sqrt{2} = 0 \]

This does not yield a valid intersection since it does not depend on \( x \).

---

**Intersection of Line 2 and Line 3:**

From Line 2:
\[ y = x + 1 \]

Substituting into Line 3:
\[ x + (x + 1) - \sqrt{2} = 0 \]
\[ 2x + 1 - \sqrt{2} = 0 \]
\[ 2x = \sqrt{2} - 1 \]
\[ x = \frac{\sqrt{2} - 1}{2} \]

Now substitute \( x \) back into Line 2:
\[ y = \frac{\sqrt{2} - 1}{2} + 1 = \frac{\sqrt{2} + 1}{2} \]

So the second vertex is \( B\left(\frac{\sqrt{2} - 1}{2}, \frac{\sqrt{2} + 1}{2}\right) \).

---

**Intersection of Line 1 and Line 3:**

We already found that this does not yield a valid intersection.

### Step 2: Find the Circumcenter

For a right triangle, the circumcenter is the midpoint of the hypotenuse. We need to find the hypotenuse first.

The hypotenuse is between points \( A(2, 3) \) and \( B\left(\frac{\sqrt{2} - 1}{2}, \frac{\sqrt{2} + 1}{2}\right) \).

**Midpoint Calculation:**
The midpoint \( M \) of segment \( AB \) is given by:
\[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]

Substituting the coordinates:
\[ M = \left( \frac{2 + \frac{\sqrt{2} - 1}{2}}{2}, \frac{3 + \frac{\sqrt{2} + 1}{2}}{2} \right) \]

Calculating the x-coordinate:
\[ M_x = \frac{4 + \sqrt{2} - 1}{4} = \frac{3 + \sqrt{2}}{4} \]

Calculating the y-coordinate:
\[ M_y = \frac{6 + \sqrt{2} + 1}{4} = \frac{7 + \sqrt{2}}{4} \]

Thus, the circumcenter is:
\[ M\left(\frac{3 + \sqrt{2}}{4}, \frac{7 + \sqrt{2}}{4}\right) \]

### Conclusion

The circumcenter of the triangle formed by the given lines is \( M\left(\frac{3 + \sqrt{2}}{4}, \frac{7 + \sqrt{2}}{4}\right) \).