Find the equation of a line that is perpendicular to `3x+2y=8` and passes through the mid-point of the line segment joining :
`(i) (5,-2)` and `(2,2)`
`(ii) (2,7) ` and `(-4,1)`

Answer

To find the equation of a line that is perpendicular to \(3x + 2y = 8\) and passes through the midpoints of the given line segments, we will follow these steps:

### Step 1: Find the slope of the given line
The equation of the line is given as \(3x + 2y = 8\). To find its slope, we can rewrite it in the slope-intercept form \(y = mx + b\).

1. Rearranging the equation:
\[ 2y = -3x + 8 \]
\[ y = -\frac{3}{2}x + 4 \]
The slope \(m_1\) of the line is \(-\frac{3}{2}\).

**Hint:** The slope of a line in the form \(Ax + By = C\) can be found using the formula \(m = -\frac{A}{B}\).

### Step 2: Find the slope of the perpendicular line
The slope of a line that is perpendicular to another is the negative reciprocal of the original slope.

\[ m_2 = -\frac{1}{m_1} = -\frac{1}{-\frac{3}{2}} = \frac{2}{3} \]

**Hint:** To find the slope of the perpendicular line, take the negative reciprocal of the original slope.

### Step 3: Find the midpoints of the given line segments
#### (i) For the points (5, -2) and (2, 2):
The midpoint \(M_1\) is given by:
\[ M_1\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) = M_1\left(\frac{5 + 2}{2}, \frac{-2 + 2}{2}\right) = M_1\left(\frac{7}{2}, 0\right) \]

#### (ii) For the points (2, 7) and (-4, 1):
The midpoint \(M_2\) is given by:
\[ M_2\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) = M_2\left(\frac{2 - 4}{2}, \frac{7 + 1}{2}\right) = M_2\left(-1, 4\right) \]

**Hint:** The midpoint of two points \((x_1, y_1)\) and \((x_2, y_2)\) is calculated using the formula \(\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\).

### Step 4: Write the equation of the line using point-slope form
The point-slope form of a line is given by:
\[ y - y_1 = m(x - x_1) \]

#### For the midpoint \(M_1\left(\frac{7}{2}, 0\right)\):
Using the slope \(m_2 = \frac{2}{3}\):
\[ y - 0 = \frac{2}{3}\left(x - \frac{7}{2}\right) \]
Multiplying through by 3 to eliminate the fraction:
\[ 3y = 2x - 7 \]
Rearranging gives:
\[ 2x - 3y - 7 = 0 \]

#### For the midpoint \(M_2(-1, 4)\):
Using the same slope \(m_2 = \frac{2}{3}\):
\[ y - 4 = \frac{2}{3}(x + 1) \]
Multiplying through by 3:
\[ 3(y - 4) = 2(x + 1) \]
Expanding gives:
\[ 3y - 12 = 2x + 2 \]
Rearranging gives:
\[ 2x - 3y + 14 = 0 \]

### Final Answers:
1. The equation of the line through midpoint \(M_1\) is \(2x - 3y - 7 = 0\).
2. The equation of the line through midpoint \(M_2\) is \(2x - 3y + 14 = 0\).