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

Options

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

Answer

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

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

**1.1** Find the intersection of the first and second lines:
- The first line is \(x + y = 0\) (Equation 1).
- The second line is \(3x + y - 4 = 0\) (Equation 2).

To find the intersection, we can solve these equations simultaneously. From Equation 1, we can express \(y\) in terms of \(x\):
\[ y = -x \]
Substituting \(y = -x\) into Equation 2:
\[ 3x - x - 4 = 0 \implies 2x - 4 = 0 \implies 2x = 4 \implies x = 2 \]
Now substituting \(x = 2\) back into Equation 1 to find \(y\):
\[ y = -2 \]
Thus, the intersection point \(A\) is \((2, -2)\).

**1.2** Find the intersection of the first and third lines:
- The third line is \(x + 3y - 4 = 0\) (Equation 3).

Using Equation 1 again:
\[ y = -x \]
Substituting into Equation 3:
\[ x + 3(-x) - 4 = 0 \implies x - 3x - 4 = 0 \implies -2x - 4 = 0 \implies -2x = 4 \implies x = -2 \]
Substituting \(x = -2\) back into Equation 1:
\[ y = 2 \]
Thus, the intersection point \(B\) is \((-2, 2)\).

**1.3** Find the intersection of the second and third lines:
Using Equation 2 and Equation 3:
\[ 3x + y - 4 = 0 \quad (Equation 2) \]
\[ x + 3y - 4 = 0 \quad (Equation 3) \]
From Equation 2, express \(y\):
\[ y = 4 - 3x \]
Substituting \(y\) into Equation 3:
\[ x + 3(4 - 3x) - 4 = 0 \implies x + 12 - 9x - 4 = 0 \implies -8x + 8 = 0 \implies 8x = 8 \implies x = 1 \]
Substituting \(x = 1\) back into the expression for \(y\):
\[ y = 4 - 3(1) = 1 \]
Thus, the intersection point \(C\) is \((1, 1)\).

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

**2.1** Length of side \(AB\):
Using the distance formula:
\[ AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(2 - (-2))^2 + (-2 - 2)^2} = \sqrt{(2 + 2)^2 + (-4)^2} = \sqrt{4^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \]

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

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

### Step 3: Determine the type of triangle

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

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

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

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