If the point `M(h, k)` lie on the line `2x+3y=5` such that `|MA-MB|` is maximum where (1, 2) and B(2, 3), then the value of `(h+k)` is

Answer

To solve the problem, we need to find the point \( M(h, k) \) that lies on the line \( 2x + 3y = 5 \) such that \( |MA - MB| \) is maximized, where \( A(1, 2) \) and \( B(2, 3) \).

### Step 1: Understand the condition for maximum \( |MA - MB| \)

The expression \( |MA - MB| \) is maximized when point \( M \) lies on the line segment joining points \( A \) and \( B \). This is because, geometrically, the maximum difference in distances occurs when the point is collinear with \( A \) and \( B \).

### Step 2: Find the equation of line \( AB \)

We can find the equation of the line passing through points \( A(1, 2) \) and \( B(2, 3) \) using the two-point form of the line equation:

\[ y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1) \]

Substituting \( (x_1, y_1) = (1, 2) \) and \( (x_2, y_2) = (2, 3) \):

\[ y - 2 = \frac{3 - 2}{2 - 1}(x - 1) \]

This simplifies to:

\[ y - 2 = (x - 1) \]
\[ y = x + 1 \]

### Step 3: Set up the system of equations

Now we have two equations:
1. The line \( 2x + 3y = 5 \)
2. The line \( y = x + 1 \)

We can substitute the second equation into the first to find the intersection point.

### Step 4: Substitute and solve

Substituting \( y = x + 1 \) into \( 2x + 3y = 5 \):

\[ 2x + 3(x + 1) = 5 \]
\[ 2x + 3x + 3 = 5 \]
\[ 5x + 3 = 5 \]
\[ 5x = 2 \]
\[ x = \frac{2}{5} \]

Now, substitute \( x \) back to find \( y \):

\[ y = x + 1 = \frac{2}{5} + 1 = \frac{2}{5} + \frac{5}{5} = \frac{7}{5} \]

Thus, the coordinates of point \( M \) are \( \left( \frac{2}{5}, \frac{7}{5} \right) \).

### Step 5: Find \( h + k \)

Now we can find \( h + k \):

\[ h + k = \frac{2}{5} + \frac{7}{5} = \frac{9}{5} \]

### Final Answer

The value of \( h + k \) is \( \frac{9}{5} \).

---