Two point `(b+3,b+k)` and (2,b) are on the line `x-2y+9=0` find k

Answer

To solve the problem, we need to find the value of \( k \) such that the points \( (b+3, b+k) \) and \( (2, b) \) lie on the line given by the equation \( x - 2y + 9 = 0 \).

### Step 1: Substitute the coordinates of the point \( (2, b) \) into the line equation.

The point \( (2, b) \) has \( x = 2 \) and \( y = b \). Substitute these values into the line equation:

\[ 2 - 2b + 9 = 0 \]

### Step 2: Simplify the equation.

Combine the constants:

\[ 11 - 2b = 0 \]

### Step 3: Solve for \( b \).

Rearranging the equation gives:

\[ 2b = 11 \]
\[ b = \frac{11}{2} \]

### Step 4: Substitute \( b \) into the coordinates of the point \( (b+3, b+k) \).

Now that we have \( b \), we can find the coordinates of the point \( (b+3, b+k) \):

\[ b + 3 = \frac{11}{2} + 3 = \frac{11}{2} + \frac{6}{2} = \frac{17}{2} \]
\[ b + k = \frac{11}{2} + k \]

So the point \( P \) becomes:

\[ \left( \frac{17}{2}, \frac{11}{2} + k \right) \]

### Step 5: Substitute the coordinates of point \( P \) into the line equation.

Now substitute \( \left( \frac{17}{2}, \frac{11}{2} + k \right) \) into the line equation:

\[ \frac{17}{2} - 2\left(\frac{11}{2} + k\right) + 9 = 0 \]

### Step 6: Simplify the equation.

Distributing the \( -2 \):

\[ \frac{17}{2} - 2 \cdot \frac{11}{2} - 2k + 9 = 0 \]
\[ \frac{17}{2} - \frac{22}{2} - 2k + 9 = 0 \]
\[ -\frac{5}{2} - 2k + 9 = 0 \]

### Step 7: Combine the constants.

Convert \( 9 \) to a fraction:

\[ -\frac{5}{2} + \frac{18}{2} - 2k = 0 \]
\[ \frac{13}{2} - 2k = 0 \]

### Step 8: Solve for \( k \).

Rearranging gives:

\[ 2k = \frac{13}{2} \]
\[ k = \frac{13}{4} \]

### Final Answer:

Thus, the value of \( k \) is:

\[ \boxed{\frac{13}{4}} \]