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. none of these

Answer

To determine the nature of the triangle formed by 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**:
- Line 1: \(x + y - 4 = 0\) (Equation 1)
- Line 2: \(3x + y - 4 = 0\) (Equation 2)

Subtract Equation 1 from Equation 2:
\[ (3x + y - 4) - (x + y - 4) = 0 \implies 2x = 0 \implies x = 0 \]
Substitute \(x = 0\) into Equation 1:
\[ 0 + y - 4 = 0 \implies y = 4 \]
So, the first intersection point is \(A(0, 4)\).

2. **Intersection of Line 1 and Line 3**:
- Line 3: \(x + 3y - 4 = 0\) (Equation 3)

Subtract Equation 1 from Equation 3:
\[ (x + 3y - 4) - (x + y - 4) = 0 \implies 2y = 0 \implies y = 0 \]
Substitute \(y = 0\) into Equation 1:
\[ x + 0 - 4 = 0 \implies x = 4 \]
So, the second intersection point is \(B(4, 0)\).

3. **Intersection of Line 2 and Line 3**:
Subtract Equation 2 from Equation 3:
\[ (x + 3y - 4) - (3x + y - 4) = 0 \implies -2x + 2y = 0 \implies y = x \]
Substitute \(y = x\) into Equation 2:
\[ 3x + x - 4 = 0 \implies 4x = 4 \implies x = 1 \]
Thus, \(y = 1\) and the third intersection point is \(C(1, 1)\).

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

Using the distance formula \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\):

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

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 nature of the triangle

We have:
- \(AB = 4\sqrt{2}\)
- \(BC = \sqrt{10}\)
- \(AC = \sqrt{10}\)

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

### Conclusion

The triangle formed by the lines is an **isosceles triangle**.

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