The angle between the straight lines `2x-y+3=0 and x+2y+3=0` is-

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `90^(@)`

Answer

To find the angle between the two straight lines given by the equations \(2x - y + 3 = 0\) and \(x + 2y + 3 = 0\), we can follow these steps:

### Step 1: Convert the equations to slope-intercept form
The general form of a straight line is given by \(Ax + By + C = 0\). We can rearrange the equations to find their slopes.

1. For the first line \(2x - y + 3 = 0\):
\[ -y = -2x - 3 \implies y = 2x + 3 \]
The slope \(m_1\) of the first line is \(2\).

2. For the second line \(x + 2y + 3 = 0\):
\[ 2y = -x - 3 \implies y = -\frac{1}{2}x - \frac{3}{2} \]
The slope \(m_2\) of the second line is \(-\frac{1}{2}\).

### Step 2: Use the formula for the angle between two lines
The formula for the angle \(\theta\) between two lines with slopes \(m_1\) and \(m_2\) is given by:
\[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]

### Step 3: Substitute the slopes into the formula
Now, substituting \(m_1 = 2\) and \(m_2 = -\frac{1}{2}\):
\[ \tan \theta = \left| \frac{2 - \left(-\frac{1}{2}\right)}{1 + 2 \cdot \left(-\frac{1}{2}\right)} \right| \]
This simplifies to:
\[ \tan \theta = \left| \frac{2 + \frac{1}{2}}{1 - 1} \right| = \left| \frac{\frac{4}{2} + \frac{1}{2}}{0} \right| = \left| \frac{\frac{5}{2}}{0} \right| \]
Since the denominator is \(0\), \(\tan \theta\) approaches infinity.

### Step 4: Determine the angle
If \(\tan \theta\) is infinite, then \(\theta = 90^\circ\).

### Conclusion
The angle between the two given straight lines is \(90^\circ\).

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