The triangle formed by the lines `x + y = 0,3x + y = 4,x+3y = 4` is

Options

A. Right angled
B. Equitateral
C. Scalene
D. Isosceles

Answer

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

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

1. **Intersection of Line 1 and Line 2:**
- Line 1: \( x + y = 0 \) (or \( y = -x \))
- Line 2: \( 3x + y = 4 \)
- Substitute \( y = -x \) into Line 2:
\[ 3x - x = 4 \implies 2x = 4 \implies x = 2 \]
Then, substitute \( x = 2 \) back into Line 1:
\[ y = -2 \]
- So, the intersection point is \( A(2, -2) \).

2. **Intersection of Line 2 and Line 3:**
- Line 2: \( 3x + y = 4 \)
- Line 3: \( x + 3y = 4 \)
- Rearranging Line 3 gives \( x = 4 - 3y \).
- Substitute \( x = 4 - 3y \) into Line 2:
\[ 3(4 - 3y) + y = 4 \implies 12 - 9y + y = 4 \implies -8y = -8 \implies y = 1 \]
Substitute \( y = 1 \) back into Line 2:
\[ 3x + 1 = 4 \implies 3x = 3 \implies x = 1 \]
- So, the intersection point is \( B(1, 1) \).

3. **Intersection of Line 1 and Line 3:**
- Line 1: \( x + y = 0 \) (or \( y = -x \))
- Line 3: \( x + 3y = 4 \)
- Substitute \( y = -x \) into Line 3:
\[ x + 3(-x) = 4 \implies x - 3x = 4 \implies -2x = 4 \implies x = -2 \]
Substitute \( x = -2 \) back into Line 1:
\[ y = 2 \]
- So, the intersection point is \( C(-2, 2) \).

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

1. **Length of side AB:**
- Points \( A(2, -2) \) and \( B(1, 1) \):
\[ AB = \sqrt{(1 - 2)^2 + (1 - (-2))^2} = \sqrt{(-1)^2 + (3)^2} = \sqrt{1 + 9} = \sqrt{10} \]

2. **Length of side BC:**
- Points \( B(1, 1) \) and \( C(-2, 2) \):
\[ BC = \sqrt{(-2 - 1)^2 + (2 - 1)^2} = \sqrt{(-3)^2 + (1)^2} = \sqrt{9 + 1} = \sqrt{10} \]

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

### Step 3: Determine the type of triangle.

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

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

### Final Answer:
The triangle formed by the lines \( x + y = 0 \), \( 3x + y = 4 \), and \( x + 3y = 4 \) is an **isosceles triangle**.

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