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

Options

A. isosceles
B. right angled
C. equiateral
D. scalene

Answer

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

### Step 1: Rewrite the equations in slope-intercept form
1. For the first equation \(x + y - 4 = 0\):
\[ y = 4 - x \quad \text{(Equation 1)} \]

2. For the second equation \(3x + y - 4 = 0\):
\[ y = 4 - 3x \quad \text{(Equation 2)} \]

3. For the third equation \(x + 3y - 4 = 0\):
\[ 3y = 4 - x \implies y = \frac{4 - x}{3} \quad \text{(Equation 3)} \]

### Step 2: Find the intersection points of the lines
1. **Intersection of Equation 1 and Equation 2**:
\[ 4 - x = 4 - 3x \]
Simplifying gives:
\[ 2x = 0 \implies x = 0 \]
Substituting \(x = 0\) into Equation 1:
\[ y = 4 - 0 = 4 \]
Thus, the intersection point is \(A(0, 4)\).

2. **Intersection of Equation 2 and Equation 3**:
\[ 4 - 3x = \frac{4 - x}{3} \]
Multiplying through by 3 to eliminate the fraction:
\[ 12 - 9x = 4 - x \]
Rearranging gives:
\[ 8 = 8x \implies x = 1 \]
Substituting \(x = 1\) into Equation 2:
\[ y = 4 - 3(1) = 1 \]
Thus, the intersection point is \(B(1, 1)\).

3. **Intersection of Equation 1 and Equation 3**:
\[ 4 - x = \frac{4 - x}{3} \]
Multiplying through by 3:
\[ 12 - 3x = 4 - x \]
Rearranging gives:
\[ 8 = 2x \implies x = 4 \]
Substituting \(x = 4\) into Equation 1:
\[ y = 4 - 4 = 0 \]
Thus, the intersection point is \(C(4, 0)\).

### Step 3: Calculate the lengths of the sides of the triangle
1. **Length of side AB**:
\[ AB = \sqrt{(1 - 0)^2 + (1 - 4)^2} = \sqrt{1^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10} \]

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

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

### Step 4: Determine the type of triangle
We have the lengths:
- \(AB = \sqrt{10}\)
- \(BC = \sqrt{10}\)
- \(AC = 4\sqrt{2}\)

Since two sides are equal (\(AB\) and \(BC\)), the triangle formed is an **isosceles triangle**.

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