To find the coordinates of points A and B given the points P(3,2) and Q(-2,3) that divide the line segment AB in specific ratios, we can use the section formula. Let's break down the solution step by step.
### Step 1: Use the Section Formula for Internal Division
Point P(3, 2) divides the line segment AB internally in the ratio 3:2. According to the section formula, if a point divides a line segment internally in the ratio m:n, the coordinates (x, y) of the point can be calculated as follows:
\[
x = \frac{mx_2 + nx_1}{m+n}
\]
\[
y = \frac{my_2 + ny_1}{m+n}
\]
Here, let A be (x1, y1) and B be (x2, y2). For point P:
- m = 3
- n = 2
- P(3, 2) = (x, y)
Substituting the values, we get:
\[
3 = \frac{3x_2 + 2x_1}{3 + 2} \quad \text{(1)}
\]
\[
2 = \frac{3y_2 + 2y_1}{3 + 2} \quad \text{(2)}
\]
### Step 2: Simplify the Equations
From equation (1):
\[
3(3 + 2) = 3x_2 + 2x_1
\]
\[
15 = 3x_2 + 2x_1 \quad \text{(Equation 1)}
\]
From equation (2):
\[
2(3 + 2) = 3y_2 + 2y_1
\]
\[
10 = 3y_2 + 2y_1 \quad \text{(Equation 2)}
\]
### Step 3: Use the Section Formula for External Division
Point Q(-2, 3) divides the line segment AB externally in the ratio 4:3. For external division, the formula is slightly different:
\[
x = \frac{mx_2 - nx_1}{m-n}
\]
\[
y = \frac{my_2 - ny_1}{m-n}
\]
For point Q:
- m = 4
- n = 3
- Q(-2, 3) = (x, y)
Substituting the values, we get:
\[
-2 = \frac{4x_2 - 3x_1}{4 - 3} \quad \text{(3)}
\]
\[
3 = \frac{4y_2 - 3y_1}{4 - 3} \quad \text{(4)}
\]
### Step 4: Simplify the External Division Equations
From equation (3):
\[
-2 = 4x_2 - 3x_1
\]
\[
4x_2 - 3x_1 = -2 \quad \text{(Equation 3)}
\]
From equation (4):
\[
3 = 4y_2 - 3y_1
\]
\[
4y_2 - 3y_1 = 3 \quad \text{(Equation 4)}
\]
### Step 5: Solve the System of Equations
Now we have four equations:
1. \(3x_2 + 2x_1 = 15\) (Equation 1)
2. \(3y_2 + 2y_1 = 10\) (Equation 2)
3. \(4x_2 - 3x_1 = -2\) (Equation 3)
4. \(4y_2 - 3y_1 = 3\) (Equation 4)
#### Solving for x-coordinates:
From Equation 1:
\[
2x_1 = 15 - 3x_2 \quad \Rightarrow \quad x_1 = \frac{15 - 3x_2}{2}
\]
Substituting \(x_1\) into Equation 3:
\[
4x_2 - 3\left(\frac{15 - 3x_2}{2}\right) = -2
\]
Multiply through by 2 to eliminate the fraction:
\[
8x_2 - 3(15 - 3x_2) = -4
\]
\[
8x_2 - 45 + 9x_2 = -4
\]
\[
17x_2 = 41 \quad \Rightarrow \quad x_2 = \frac{41}{17}
\]
Now substitute \(x_2\) back to find \(x_1\):
\[
x_1 = \frac{15 - 3\left(\frac{41}{17}\right)}{2} = \frac{15 - \frac{123}{17}}{2} = \frac{\frac{255 - 123}{17}}{2} = \frac{\frac{132}{17}}{2} = \frac{66}{17}
\]
#### Solving for y-coordinates:
From Equation 2:
\[
2y_1 = 10 - 3y_2 \quad \Rightarrow \quad y_1 = \frac{10 - 3y_2}{2}
\]
Substituting \(y_1\) into Equation 4:
\[
4y_2 - 3\left(\frac{10 - 3y_2}{2}\right) = 3
\]
Multiply through by 2:
\[
8y_2 - 3(10 - 3y_2) = 6
\]
\[
8y_2 - 30 + 9y_2 = 6
\]
\[
17y_2 = 36 \quad \Rightarrow \quad y_2 = \frac{36}{17}
\]
Now substitute \(y_2\) back to find \(y_1\):
\[
y_1 = \frac{10 - 3\left(\frac{36}{17}\right)}{2} = \frac{10 - \frac{108}{17}}{2} = \frac{\frac{170 - 108}{17}}{2} = \frac{\frac{62}{17}}{2} = \frac{31}{17}
\]
### Final Coordinates
Thus, the coordinates of points A and B are:
- A = \(\left(\frac{66}{17}, \frac{31}{17}\right)\)
- B = \(\left(\frac{41}{17}, \frac{36}{17}\right)\)
