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

Options

A. isosceles
B. right angled
C. equilateral
D. None

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 points of intersection of the lines

1. **Find the intersection of the lines \(x + y = 4\) and \(3x + y = 4\)**:
- From \(x + y = 4\), we can express \(y\) as \(y = 4 - x\).
- Substitute \(y\) in \(3x + y = 4\):
\[ 3x + (4 - x) = 4 \implies 3x - x + 4 = 4 \implies 2x = 0 \implies x = 0 \]
- Substitute \(x = 0\) back into \(y = 4 - x\):
\[ y = 4 - 0 = 4 \]
- Thus, the intersection point \(A\) is \((0, 4)\).

2. **Find the intersection of the lines \(x + y = 4\) and \(x + 3y = 4\)**:
- From \(x + y = 4\), express \(y\) as \(y = 4 - x\).
- Substitute \(y\) in \(x + 3y = 4\):
\[ x + 3(4 - x) = 4 \implies x + 12 - 3x = 4 \implies -2x + 12 = 4 \implies -2x = -8 \implies x = 4 \]
- Substitute \(x = 4\) back into \(y = 4 - x\):
\[ y = 4 - 4 = 0 \]
- Thus, the intersection point \(B\) is \((4, 0)\).

3. **Find the intersection of the lines \(3x + y = 4\) and \(x + 3y = 4\)**:
- From \(3x + y = 4\), express \(y\) as \(y = 4 - 3x\).
- Substitute \(y\) in \(x + 3y = 4\):
\[ x + 3(4 - 3x) = 4 \implies x + 12 - 9x = 4 \implies -8x + 12 = 4 \implies -8x = -8 \implies x = 1 \]
- Substitute \(x = 1\) back into \(y = 4 - 3x\):
\[ y = 4 - 3(1) = 4 - 3 = 1 \]
- Thus, the intersection point \(C\) is \((1, 1)\).

### 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 \(CA\)**:
\[ CA = \sqrt{(1 - 0)^2 + (1 - 4)^2} = \sqrt{(1)^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10} \]

### Step 3: Determine the nature of the triangle

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

- To check if it is a right triangle, we can use the Pythagorean theorem:
\[ AB^2 = BC^2 + CA^2 \]
\[ (4\sqrt{2})^2 = (\sqrt{10})^2 + (\sqrt{10})^2 \]
\[ 32 = 10 + 10 \implies 32 = 20 \quad \text{(not true)} \]

- Since \(AB^2\) does not equal \(BC^2 + CA^2\), we check if \(BC = CA\):
- Since \(BC = CA\), the triangle is isosceles.

### Conclusion

The triangle formed by the lines \(x + y - 4 = 0\), \(3x + y - 4 = 0\), and \(x + 3y - 4 = 0\) is an isosceles triangle.