The slope of a line joining P(6, k) and Q(1 - 3k, 3) is `1/(2)`. Find :
(ii) Midpoint of PQ, using the value of 'k' found in (i)

Answer

To solve the problem, we need to find the value of \( k \) first and then use it to find the midpoint of the line segment joining points \( P(6, k) \) and \( Q(1 - 3k, 3) \).

### Step 1: Find the slope of line segment PQ
The slope \( m \) of a line joining two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by the formula:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]
For points \( P(6, k) \) and \( Q(1 - 3k, 3) \):
- \( x_1 = 6 \), \( y_1 = k \)
- \( x_2 = 1 - 3k \), \( y_2 = 3 \)

Substituting these values into the slope formula, we get:
\[ m = \frac{3 - k}{(1 - 3k) - 6} \]
This simplifies to:
\[ m = \frac{3 - k}{1 - 3k - 6} = \frac{3 - k}{-3k - 5} \]

### Step 2: Set the slope equal to \( \frac{1}{2} \)
According to the problem, the slope is given as \( \frac{1}{2} \):
\[ \frac{3 - k}{-3k - 5} = \frac{1}{2} \]

### Step 3: Cross-multiply to solve for \( k \)
Cross-multiplying gives:
\[ 2(3 - k) = 1(-3k - 5) \]
Expanding both sides:
\[ 6 - 2k = -3k - 5 \]

### Step 4: Rearranging the equation
Rearranging the equation to isolate \( k \):
\[ 6 + 5 = -3k + 2k \]
\[ 11 = -k \]
Thus,
\[ k = -11 \]

### Step 5: Substitute \( k \) back into points P and Q
Now we substitute \( k = -11 \) back into the coordinates of points \( P \) and \( Q \):
- \( P(6, -11) \)
- \( Q(1 - 3(-11), 3) = Q(1 + 33, 3) = Q(34, 3) \)

### Step 6: Find the midpoint of PQ
The midpoint \( M \) of a line segment joining points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:
\[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]
Substituting the coordinates of points \( P \) and \( Q \):
\[ M = \left( \frac{6 + 34}{2}, \frac{-11 + 3}{2} \right) \]
Calculating the coordinates:
\[ M = \left( \frac{40}{2}, \frac{-8}{2} \right) = (20, -4) \]

### Final Answer
The midpoint of the line segment PQ is \( (20, -4) \).