The straight lines `x+y-4=0, 3x+y-4=0` and `x+3y-4=0` form a triangle, which is

Options

A. isosceles
B. right-angled
C. equilateral
D. scalene

Explanation

To determine the type of triangle formed by the intersection of the lines \(x + y - 4 = 0\), \(3x + y - 4 = 0\), and \(x + 3y - 4 = 0\), we will follow these steps:

### Step 1: Find the intersection points of the lines

1. **Intersection of Line 1 and Line 2**:
- Equations:
\[ x + y - 4 = 0 \quad \text{(1)} \]
\[ 3x + y - 4 = 0 \quad \text{(2)} \]
- Subtract equation (1) from (2):
\[ (3x + y - 4) - (x + y - 4) = 0 \implies 2x = 0 \implies x = 0 \]
- Substitute \(x = 0\) into equation (1):
\[ 0 + y - 4 = 0 \implies y = 4 \]
- **Point A**: \((0, 4)\)

2. **Intersection of Line 1 and Line 3**:
- Equations:
\[ x + y - 4 = 0 \quad \text{(1)} \]
\[ x + 3y - 4 = 0 \quad \text{(3)} \]
- Subtract equation (1) from (3):
\[ (x + 3y - 4) - (x + y - 4) = 0 \implies 2y = 0 \implies y = 0 \]
- Substitute \(y = 0\) into equation (1):
\[ x + 0 - 4 = 0 \implies x = 4 \]
- **Point B**: \((4, 0)\)

3. **Intersection of Line 2 and Line 3**:
- Equations:
\[ 3x + y - 4 = 0 \quad \text{(2)} \]
\[ x + 3y - 4 = 0 \quad \text{(3)} \]
- Multiply equation (3) by 3:
\[ 3x + 9y - 12 = 0 \]
- Subtract equation (2) from the modified equation (3):
\[ (3x + 9y - 12) - (3x + y - 4) = 0 \implies 8y - 8 = 0 \implies y = 1 \]
- Substitute \(y = 1\) into equation (2):
\[ 3x + 1 - 4 = 0 \implies 3x = 3 \implies x = 1 \]
- **Point C**: \((1, 1)\)

### Step 2: Calculate the lengths of the sides of the triangle

1. **Length of side AB**:
\[ AB = \sqrt{(4 - 0)^2 + (0 - 4)^2} = \sqrt{4^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \]

2. **Length of side BC**:
\[ BC = \sqrt{(1 - 4)^2 + (1 - 0)^2} = \sqrt{(-3)^2 + (1)^2} = \sqrt{9 + 1} = \sqrt{10} \]

3. **Length of side AC**:
\[ AC = \sqrt{(1 - 0)^2 + (1 - 4)^2} = \sqrt{(1)^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10} \]

### Step 3: Determine the type of triangle

- The lengths of the sides are:
- \(AB = 4\sqrt{2}\)
- \(BC = \sqrt{10}\)
- \(AC = \sqrt{10}\)

Since \(BC = AC\) and \(AB \neq AC\), the triangle is **isosceles**.

### Final Answer:
The triangle formed by the lines is an **isosceles triangle**.