If point `P(3,2)` divides the line segment AB internally in the ratio of `3:2` and point `Q(-2,3)` divides AB externally in the ratio `4:3` then find the coordinates of points A and B.

Explanation

To find the coordinates of points A and B given the conditions of points P and Q, we can follow these steps:

### Step 1: Set Up the Problem
Let the coordinates of point A be (A, B) and the coordinates of point B be (C, D).

### Step 2: Use the Internal Division Formula
Point P(3, 2) divides the line segment AB internally in the ratio 3:2. According to the section formula for internal division, the coordinates of point P can be expressed as:
\[ P_x = \frac{mC + nA}{m+n} \quad \text{and} \quad P_y = \frac{mD + nB}{m+n} \]
where \( m = 3 \) and \( n = 2 \).

For the x-coordinate:
\[ 3 = \frac{3C + 2A}{3 + 2} = \frac{3C + 2A}{5} \]
Multiplying both sides by 5:
\[ 15 = 3C + 2A \quad \text{(Equation 1)} \]

For the y-coordinate:
\[ 2 = \frac{3D + 2B}{5} \]
Multiplying both sides by 5:
\[ 10 = 3D + 2B \quad \text{(Equation 2)} \]

### Step 3: Use the External Division Formula
Point Q(-2, 3) divides the line segment AB externally in the ratio 4:3. According to the section formula for external division, the coordinates of point Q can be expressed as:
\[ Q_x = \frac{mC - nA}{m-n} \quad \text{and} \quad Q_y = \frac{mD - nB}{m-n} \]
where \( m = 4 \) and \( n = 3 \).

For the x-coordinate:
\[ -2 = \frac{4C - 3A}{4 - 3} = 4C - 3A \]
Rearranging gives:
\[ 4C - 3A = -2 \quad \text{(Equation 3)} \]

For the y-coordinate:
\[ 3 = \frac{4D - 3B}{1} \]
This simplifies to:
\[ 4D - 3B = 3 \quad \text{(Equation 4)} \]

### Step 4: Solve the Equations
Now we have four equations:
1. \( 3C + 2A = 15 \) (Equation 1)
2. \( 3D + 2B = 10 \) (Equation 2)
3. \( 4C - 3A = -2 \) (Equation 3)
4. \( 4D - 3B = 3 \) (Equation 4)

#### Solving Equations 1 and 3
From Equation 1:
\[ 2A = 15 - 3C \quad \Rightarrow \quad A = \frac{15 - 3C}{2} \]
Substituting \( A \) into Equation 3:
\[ 4C - 3\left(\frac{15 - 3C}{2}\right) = -2 \]
Multiplying through by 2 to eliminate the fraction:
\[ 8C - 3(15 - 3C) = -4 \]
Expanding:
\[ 8C - 45 + 9C = -4 \]
Combining like terms:
\[ 17C - 45 = -4 \quad \Rightarrow \quad 17C = 41 \quad \Rightarrow \quad C = \frac{41}{17} \]

Substituting \( C \) back into Equation 1:
\[ 3\left(\frac{41}{17}\right) + 2A = 15 \]
Calculating:
\[ \frac{123}{17} + 2A = 15 \quad \Rightarrow \quad 2A = 15 - \frac{123}{17} \]
Converting 15 to a fraction:
\[ 2A = \frac{255 - 123}{17} = \frac{132}{17} \quad \Rightarrow \quad A = \frac{66}{17} \]

#### Solving Equations 2 and 4
From Equation 2:
\[ 2B = 10 - 3D \quad \Rightarrow \quad B = \frac{10 - 3D}{2} \]
Substituting \( B \) into Equation 4:
\[ 4D - 3\left(\frac{10 - 3D}{2}\right) = 3 \]
Multiplying through by 2:
\[ 8D - 3(10 - 3D) = 6 \]
Expanding:
\[ 8D - 30 + 9D = 6 \]
Combining like terms:
\[ 17D - 30 = 6 \quad \Rightarrow \quad 17D = 36 \quad \Rightarrow \quad D = \frac{36}{17} \]

Substituting \( D \) back into Equation 2:
\[ 3D + 2B = 10 \quad \Rightarrow \quad 3\left(\frac{36}{17}\right) + 2B = 10 \]
Calculating:
\[ \frac{108}{17} + 2B = 10 \quad \Rightarrow \quad 2B = 10 - \frac{108}{17} \]
Converting 10 to a fraction:
\[ 2B = \frac{170 - 108}{17} = \frac{62}{17} \quad \Rightarrow \quad B = \frac{31}{17} \]

### Step 5: Final Coordinates
The coordinates of points A and B are:
- A = \( \left( \frac{66}{17}, \frac{31}{17} \right) \)
- B = \( \left( \frac{41}{17}, \frac{36}{17} \right) \)