The three straight lines `2x+11y-5=0, 24x+7y=20` and `4x-3y-2=0` are such that

Options

A. they form a triangle
B. they are all concurrent
C. one of them is a bisector of the other two
D. none

Answer

To determine the relationship between the three lines given by the equations \(2x + 11y - 5 = 0\), \(24x + 7y - 20 = 0\), and \(4x - 3y - 2 = 0\), we will analyze the conditions for concurrency and the angle bisector.

### Step 1: Identify the equations of the lines
We have three lines:
1. \(L_1: 2x + 11y - 5 = 0\)
2. \(L_2: 24x + 7y - 20 = 0\)
3. \(L_3: 4x - 3y - 2 = 0\)

### Step 2: Find the slopes of the lines
To determine if the lines are concurrent, we can find the slopes of each line.

- For \(L_1\):
\[ 11y = -2x + 5 \implies y = -\frac{2}{11}x + \frac{5}{11} \]
The slope \(m_1 = -\frac{2}{11}\).

- For \(L_2\):
\[ 7y = -24x + 20 \implies y = -\frac{24}{7}x + \frac{20}{7} \]
The slope \(m_2 = -\frac{24}{7}\).

- For \(L_3\):
\[ 3y = 4x - 2 \implies y = \frac{4}{3}x - \frac{2}{3} \]
The slope \(m_3 = \frac{4}{3}\).

### Step 3: Check for concurrency
Three lines are concurrent if the determinant of their coefficients is zero. We can write the equations in the form of a matrix:

\[ \begin{vmatrix} 2 & 11 & -5 \\ 24 & 7 & -20 \\ 4 & -3 & -2 \end{vmatrix} = 0 \]

Calculating the determinant:
\[ = 2 \begin{vmatrix} 7 & -20 \\ -3 & -2 \end{vmatrix} - 11 \begin{vmatrix} 24 & -20 \\ 4 & -2 \end{vmatrix} - 5 \begin{vmatrix} 24 & 7 \\ 4 & -3 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix} 7 & -20 \\ -3 & -2 \end{vmatrix} = (7)(-2) - (-20)(-3) = -14 - 60 = -74\)

2. \( \begin{vmatrix} 24 & -20 \\ 4 & -2 \end{vmatrix} = (24)(-2) - (-20)(4) = -48 + 80 = 32\)

3. \( \begin{vmatrix} 24 & 7 \\ 4 & -3 \end{vmatrix} = (24)(-3) - (7)(4) = -72 - 28 = -100\)

Now substituting back:
\[ = 2(-74) - 11(32) - 5(-100) = -148 - 352 + 500 = 0 \]

Since the determinant is zero, the lines are concurrent.

### Step 4: Check if one line is the angle bisector of the other two
To check if \(L_1\) is the angle bisector of \(L_2\) and \(L_3\), we can use the angle bisector theorem for lines:

The angle bisector condition states:
\[ \frac{a_1x + b_1y + c_1}{\sqrt{a_1^2 + b_1^2}} = \pm \frac{a_2x + b_2y + c_2}{\sqrt{a_2^2 + b_2^2}} \]

For \(L_2\) and \(L_3\):
\[ \frac{24x + 7y - 20}{\sqrt{24^2 + 7^2}} = \pm \frac{4x - 3y - 2}{\sqrt{4^2 + 3^2}} \]

Calculating the magnitudes:
- For \(L_2\): \(\sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25\)
- For \(L_3\): \(\sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5\)

Thus, we can write:
\[ \frac{24x + 7y - 20}{25} = \pm \frac{4x - 3y - 2}{5} \]

Cross-multiplying gives:
\[ 24x + 7y - 20 = \pm 5(4x - 3y - 2) \]

### Conclusion
After simplifying, we find that the line \(L_1\) is indeed the angle bisector of the lines \(L_2\) and \(L_3\), confirming that all three lines are concurrent and that one is the angle bisector of the others.