The straight lines `4x-3y-5=0, x-2y=0, 7x+y-40=0` and `x+3y+10=0` from

Options

A. a rectangle
B. a parallelogram
C. a cyclic quadrilateral
D. none of these

Answer

To determine the type of quadrilateral formed by the given straight lines, we will follow these steps:

### Step 1: Write down the equations of the lines
The equations of the lines are:
1. \( 4x - 3y - 5 = 0 \)
2. \( x - 2y = 0 \)
3. \( 7x + y - 40 = 0 \)
4. \( x + 3y + 10 = 0 \)

### Step 2: Find the slopes of each line
To find the slopes, we will rearrange each equation into the slope-intercept form \( y = mx + b \), where \( m \) is the slope.

1. For \( 4x - 3y - 5 = 0 \):
\[ 3y = 4x - 5 \implies y = \frac{4}{3}x - \frac{5}{3} \]
**Slope \( m_1 = \frac{4}{3} \)**

2. For \( x - 2y = 0 \):
\[ 2y = x \implies y = \frac{1}{2}x \]
**Slope \( m_2 = \frac{1}{2} \)**

3. For \( 7x + y - 40 = 0 \):
\[ y = -7x + 40 \]
**Slope \( m_3 = -7 \)**

4. For \( x + 3y + 10 = 0 \):
\[ 3y = -x - 10 \implies y = -\frac{1}{3}x - \frac{10}{3} \]
**Slope \( m_4 = -\frac{1}{3} \)**

### Step 3: Determine the angles between the lines
To check if the lines form a cyclic quadrilateral, we need to find the angles between opposite pairs of lines. The formula for the tangent of the angle \( \theta \) between two lines with slopes \( m_1 \) and \( m_2 \) is given by:
\[ \tan \theta = \frac{m_2 - m_1}{1 + m_1 m_2} \]

#### Angle between lines 1 and 3
Using slopes \( m_1 = \frac{4}{3} \) and \( m_3 = -7 \):
\[ \tan b = \frac{-7 - \frac{4}{3}}{1 + \left(\frac{4}{3}\right)(-7)} = \frac{-\frac{21}{3} - \frac{4}{3}}{1 - \frac{28}{3}} = \frac{-\frac{25}{3}}{-\frac{25}{3}} = 1 \]
Thus, \( b = 135^\circ \).

#### Angle between lines 2 and 4
Using slopes \( m_2 = \frac{1}{2} \) and \( m_4 = -\frac{1}{3} \):
\[ \tan d = \frac{-\frac{1}{3} - \frac{1}{2}}{1 + \left(\frac{1}{2}\right)(-\frac{1}{3})} = \frac{-\frac{2}{6} - \frac{3}{6}}{1 - \frac{1}{6}} = \frac{-\frac{5}{6}}{\frac{5}{6}} = -1 \]
Thus, \( d = 45^\circ \).

### Step 4: Check if the quadrilateral is cyclic
For a quadrilateral to be cyclic, the sum of the opposite angles must equal \( 180^\circ \):
\[ b + d = 135^\circ + 45^\circ = 180^\circ \]
Since the sum of the opposite angles is \( 180^\circ \), the quadrilateral formed by the lines is cyclic.

### Conclusion
The lines \( 4x - 3y - 5 = 0 \), \( x - 2y = 0 \), \( 7x + y - 40 = 0 \), and \( x + 3y + 10 = 0 \) form a cyclic quadrilateral.

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