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The problem asks for the point \((a, b)\) on the line \( 2x - 3y + 6 = 0 \) that is closest to the point \((3, 1)\), and then to find the sum \(a + b\).

This type of problem involves finding the perpendicular distance from a point to a line and determining the coordinates where the perpendicular line intersects the given line. Here's how we can solve this:

### Step 1: Equation of the Line
The given line equation is:
\[
2x - 3y + 6 = 0
\]
This can be rewritten in the form of:
\[
2x - 3y = -6
\]

### Step 2: Slope of the Given Line
The slope \(m_1\) of the given line \( 2x - 3y + 6 = 0 \) can be found by rearranging it into slope-intercept form \(y = mx + c\):
\[
y = \frac{2}{3}x + 2
\]
So, the slope \(m_1 = \frac{2}{3}\).

### Step 3: Perpendicular Line Slope
The slope of the line perpendicular to this one is the negative reciprocal of \(m_1\). Therefore, the perpendicular slope \(m_2 = -\frac{3}{2}\).

### Step 4: Equation of the Perpendicular Line
We now find the equation of the line that passes through the point \((3, 1)\) with the slope \(m_2 = -\frac{3}{2}\). The equation of a line is given by \(y - y_1 = m(x - x_1)\). Substituting \((x_1, y_1) = (3, 1)\) and \(m = -\frac{3}{2}\), we get:
\[
y - 1 = -\frac{3}{2}(x - 3)
\]
\[
y - 1 = -\frac{3}{2}x + \frac{9}{2}
\]
\[
y = -\frac{3}{2}x + \frac{9}{2} + 1 = -\frac{3}{2}x + \frac{11}{2}
\]

### Step 5: Find the Intersection Point
To find the point \((a, b)\) where these two lines intersect, we need to solve the system of equations:
1. \( y = \frac{2}{3}x + 2 \) (the original line)
2. \( y = -\frac{3}{2}x + \frac{11}{2} \) (the perpendicular line)

Set the two equations equal to each other:
\[
\frac{2}{3}x + 2 = -\frac{3}{2}x + \frac{11}{2}
\]
To eliminate fractions, multiply the entire equation by 6:
\[
4x + 12 = -9x + 33
\]
Solve for \(x\):
\[
4x + 9x = 33 - 12
\]
\[
13x = 21
\]
\[
x = \frac{21}{13}
\]

Now substitute \(x = \frac{21}{13}\) back into one of the line equations, say \(y = \frac{2}{3}x + 2\):
\[
y = \frac{2}{3} \times \frac{21}{13} + 2 = \frac{42}{39} + 2 = \frac{42}{39} + \frac{78}{39} = \frac{120}{39} = \frac{40}{13}
\]

Thus, the point of intersection is \(\left(\frac{21}{13}, \frac{40}{13}\right)\).

### Step 6: Find \(a + b\)
The sum \(a + b = \frac{21}{13} + \frac{40}{13} = \frac{61}{13}\).

### Final Answer:
\[
a + b = \frac{61}{13}
\]

Would you like further details on any of the steps, or clarification?

Here are some related questions for deeper understanding:

1. How do we derive the formula for the distance from a point to a line?
2. Can you explain why the perpendicular line gives the closest distance?
3. How would this problem change if the point were on the line instead of being outside it?
4. What is the geometric significance of finding the negative reciprocal slope?
5. Could you solve the problem using vector projections?

**Tip:** Whenever you work with perpendicular distances, finding the slope of the line and its negative reciprocal is key to identifying the shortest path.

Find the sum point (a, b) on the line 2x - 3y + 6 = 0 that is closest to the point (3, 1). Find a + b.

Learn how to find the point (a, b) on the line 2x - 3y + 6 = 0 that is closest to the point (3, 1) and calculate the sum a + b. This problem involves key concepts of geometry, including the perpendicular distance and intersection of lines. Ideal for students in Grades 10-12.

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