Two lines `2x - 3y = 1 and x + 2y + 3 = 0` divide the xy plane in four compartments which are named as shown in the figure. Consider the locations of the points`(2, -1), (3, 2) and (-1, - 2)`.

Options

A. `(2,-1) in IV`
B. `(3,2) in III`
C. `(-1,-2) in II`
D. None of these

Answer

To solve the problem of determining the locations of the points (2, -1), (3, 2), and (-1, -2) relative to the lines given by the equations \(2x - 3y = 1\) and \(x + 2y + 3 = 0\), we will follow these steps:

### Step 1: Find the intersection point of the two lines.

1. **Write the equations of the lines:**
- Line 1: \(2x - 3y = 1\) (Equation 1)
- Line 2: \(x + 2y + 3 = 0\) (Equation 2)

2. **Rearrange Equation 2 to express y in terms of x:**
\[ x + 2y + 3 = 0 \implies 2y = -x - 3 \implies y = -\frac{1}{2}x - \frac{3}{2} \]

3. **Substitute \(y\) from Equation 2 into Equation 1:**
\[ 2x - 3\left(-\frac{1}{2}x - \frac{3}{2}\right) = 1 \]
\[ 2x + \frac{3}{2}x + \frac{9}{2} = 1 \]
\[ \frac{4x + 3x + 9}{2} = 1 \implies 7x + 9 = 2 \implies 7x = -7 \implies x = -1 \]

4. **Substitute \(x = -1\) back into Equation 2 to find \(y\):**
\[ -1 + 2y + 3 = 0 \implies 2y + 2 = 0 \implies 2y = -2 \implies y = -1 \]

5. **Thus, the intersection point is \((-1, -1)\).**

### Step 2: Determine the locations of the points.

1. **Point (2, -1):**
- Substitute \(x = 2\) and \(y = -1\) into both line equations:
- For Line 1: \(2(2) - 3(-1) = 4 + 3 = 7 \neq 1\) (above the line)
- For Line 2: \(2 + 2(-1) + 3 = 2 - 2 + 3 = 3 \neq 0\) (above the line)
- Since both conditions indicate that the point is above both lines, it lies in the **fourth quadrant**.

2. **Point (3, 2):**
- Substitute \(x = 3\) and \(y = 2\):
- For Line 1: \(2(3) - 3(2) = 6 - 6 = 0\) (on the line)
- For Line 2: \(3 + 2(2) + 3 = 3 + 4 + 3 = 10 \neq 0\) (above the line)
- Since it lies on Line 1 and above Line 2, it lies in the **first quadrant**.

3. **Point (-1, -2):**
- Substitute \(x = -1\) and \(y = -2\):
- For Line 1: \(2(-1) - 3(-2) = -2 + 6 = 4 \neq 1\) (above the line)
- For Line 2: \(-1 + 2(-2) + 3 = -1 - 4 + 3 = -2 \neq 0\) (below the line)
- Since it is above Line 1 and below Line 2, it lies in the **third quadrant**.

### Summary of Locations:
- Point (2, -1) is in the **fourth quadrant**.
- Point (3, 2) is in the **first quadrant**.
- Point (-1, -2) is in the **third quadrant**.