The lines `x+2y-5=0, 2x-3y+4=0,6x+4y-13=0`

Options

A. are concurrent
B. form a right angled triangle
C. form an isosceles triangle
D. form an equilateral triangle.

Answer

To determine the type of triangle formed by the lines given by the equations \(x + 2y - 5 = 0\), \(2x - 3y + 4 = 0\), and \(6x + 4y - 13 = 0\), we will follow these steps:

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

1. **Solve the first two equations**:
- \(x + 2y - 5 = 0\) (Equation 1)
- \(2x - 3y + 4 = 0\) (Equation 2)

We can express \(x\) from Equation 1:
\[ x = 5 - 2y \]
Substitute \(x\) in Equation 2:
\[ 2(5 - 2y) - 3y + 4 = 0 \]
Simplifying this:
\[ 10 - 4y - 3y + 4 = 0 \implies 14 - 7y = 0 \implies y = 2 \]
Now substitute \(y = 2\) back into \(x = 5 - 2y\):
\[ x = 5 - 2(2) = 1 \]
So, the intersection point of the first two lines is \(A(1, 2)\).

### Step 2: Check if the point \(A\) lies on the third line.

Substituting \(A(1, 2)\) into the third equation \(6x + 4y - 13 = 0\):
\[ 6(1) + 4(2) - 13 = 6 + 8 - 13 = 1 \neq 0 \]
Thus, point \(A\) does not lie on the third line, indicating that the three lines are not concurrent.

### Step 3: Find the slopes of the lines.

1. **Slope of the first line**:
From \(x + 2y - 5 = 0\):
\[ 2y = -x + 5 \implies y = -\frac{1}{2}x + \frac{5}{2} \quad \text{(slope \(m_1 = -\frac{1}{2}\))} \]

2. **Slope of the second line**:
From \(2x - 3y + 4 = 0\):
\[ -3y = -2x - 4 \implies y = \frac{2}{3}x + \frac{4}{3} \quad \text{(slope \(m_2 = \frac{2}{3}\))} \]

3. **Slope of the third line**:
From \(6x + 4y - 13 = 0\):
\[ 4y = -6x + 13 \implies y = -\frac{3}{2}x + \frac{13}{4} \quad \text{(slope \(m_3 = -\frac{3}{2}\))} \]

### Step 4: Check for perpendicularity.

To check if any two lines are perpendicular, we compute the product of their slopes:

- For lines 1 and 2:
\[ m_1 \cdot m_2 = -\frac{1}{2} \cdot \frac{2}{3} = -\frac{1}{3} \quad \text{(not perpendicular)} \]

- For lines 1 and 3:
\[ m_1 \cdot m_3 = -\frac{1}{2} \cdot -\frac{3}{2} = \frac{3}{4} \quad \text{(not perpendicular)} \]

- For lines 2 and 3:
\[ m_2 \cdot m_3 = \frac{2}{3} \cdot -\frac{3}{2} = -1 \quad \text{(perpendicular)} \]

### Conclusion

Since lines 2 and 3 are perpendicular, the triangle formed by the three lines is a right-angled triangle.

### Final Answer
The triangle formed by the lines is a right-angled triangle.

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