The acute angle between two straight lines passing through the point `M(-6,-8)` and the points in which the line segment `2x+y+10=0` enclosed between the co-ordinate axes is divided in the ratio 1:2:2 in the direction from the point of its intersection with the x-axis to the point of intersection with the y-axis is: `pi/3` (b) `pi/4` (c) `pi/6` (d) `pi/(12)`

Options

A. `pi//3`
B. `pi//4`
C. `pi//6`
D. `pi//12`

Explanation

To solve the problem, we need to find the acute angle between two straight lines that pass through the point \( M(-6, -8) \) and the points where the line segment defined by the equation \( 2x + y + 10 = 0 \) intersects the coordinate axes. The segment is divided in the ratio \( 1:2:2 \) from the x-axis intersection to the y-axis intersection.

### Step 1: Find the points of intersection with the axes.

1. **Find the x-intercept** by setting \( y = 0 \):
\[ 2x + 0 + 10 = 0 \implies 2x = -10 \implies x = -5 \]
So, the x-intercept is \( A(-5, 0) \).

2. **Find the y-intercept** by setting \( x = 0 \):
\[ 2(0) + y + 10 = 0 \implies y = -10 \]
So, the y-intercept is \( B(0, -10) \).

### Step 2: Divide the line segment \( AB \) in the ratio \( 1:2:2 \).

The total ratio is \( 1 + 2 + 2 = 5 \). We will find the coordinates of points \( P \) and \( Q \) that divide the segment \( AB \).

1. **Find point \( P \)** which divides \( AB \) in the ratio \( 1:4 \) (1 part towards A and 4 parts towards B):
\[ P = \left( \frac{1 \cdot 0 + 4 \cdot (-5)}{1 + 4}, \frac{1 \cdot (-10) + 4 \cdot 0}{1 + 4} \right) = \left( \frac{0 - 20}{5}, \frac{-10 + 0}{5} \right) = \left( -4, -2 \right) \]

2. **Find point \( Q \)** which divides \( AB \) in the ratio \( 3:2 \) (3 parts towards A and 2 parts towards B):
\[ Q = \left( \frac{3 \cdot 0 + 2 \cdot (-5)}{3 + 2}, \frac{3 \cdot (-10) + 2 \cdot 0}{3 + 2} \right) = \left( \frac{0 - 10}{5}, \frac{-30 + 0}{5} \right) = \left( -2, -6 \right) \]

### Step 3: Find the slopes of lines \( MP \) and \( MQ \).

1. **Slope of line \( MP \)**:
\[ m_1 = \frac{-2 - (-8)}{-4 - (-6)} = \frac{6}{2} = 3 \]

2. **Slope of line \( MQ \)**:
\[ m_2 = \frac{-6 - (-8)}{-2 - (-6)} = \frac{2}{4} = \frac{1}{2} \]

### Step 4: Find the angle \( \theta \) between the two lines.

Using the formula for the angle between two lines:
\[ \tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right| \]
Substituting \( m_1 = 3 \) and \( m_2 = \frac{1}{2} \):
\[ \tan \theta = \left| \frac{\frac{1}{2} - 3}{1 + 3 \cdot \frac{1}{2}} \right| = \left| \frac{\frac{1}{2} - \frac{6}{2}}{1 + \frac{3}{2}} \right| = \left| \frac{-\frac{5}{2}}{\frac{5}{2}} \right| = 1 \]

### Step 5: Find \( \theta \).

Since \( \tan \theta = 1 \), we have:
\[ \theta = \frac{\pi}{4} \]

### Conclusion

The acute angle between the two lines is \( \frac{\pi}{4} \).