Find the angle between the line joining the points (1,-2), (3,2) and the line `x+2y-7=0`

Explanation

To find the angle between the line joining the points (1, -2) and (3, 2) and the line given by the equation \(x + 2y - 7 = 0\), we can follow these steps:

### Step 1: Find the slope of the line joining the points (1, -2) and (3, 2).

The formula for the slope \(m\) of a line passing through two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:

\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]

Here, we have:
- \((x_1, y_1) = (1, -2)\)
- \((x_2, y_2) = (3, 2)\)

Substituting the values into the formula:

\[ m_1 = \frac{2 - (-2)}{3 - 1} = \frac{2 + 2}{2} = \frac{4}{2} = 2 \]

### Step 2: Find the slope of the line \(x + 2y - 7 = 0\).

To find the slope of the line in the form \(Ax + By + C = 0\), we can rearrange the equation into slope-intercept form \(y = mx + b\).

Starting with the equation:

\[ x + 2y - 7 = 0 \]

Rearranging gives:

\[ 2y = -x + 7 \implies y = -\frac{1}{2}x + \frac{7}{2} \]

From this, we can see that the slope \(m_2\) of the line is:

\[ m_2 = -\frac{1}{2} \]

### Step 3: Use the slopes to find the angle between the two lines.

The angle \(\theta\) between two lines with slopes \(m_1\) and \(m_2\) can be found using the formula:

\[ \tan(\theta) = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]

Substituting the values of \(m_1\) and \(m_2\):

\[ \tan(\theta) = \left| \frac{2 - \left(-\frac{1}{2}\right)}{1 + 2 \cdot \left(-\frac{1}{2}\right)} \right| = \left| \frac{2 + \frac{1}{2}}{1 - 1} \right| \]

Calculating the numerator:

\[ 2 + \frac{1}{2} = \frac{4}{2} + \frac{1}{2} = \frac{5}{2} \]

And the denominator:

\[ 1 - 1 = 0 \]

Since the denominator is zero, it indicates that the lines are perpendicular. Therefore, the angle \(\theta\) is:

\[ \theta = 90^\circ \text{ or } \frac{\pi}{2} \text{ radians.} \]

### Final Answer:
The angle between the line joining the points (1, -2) and (3, 2) and the line \(x + 2y - 7 = 0\) is \(90^\circ\) or \(\frac{\pi}{2}\) radians.

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