Which one of the following points on the line 2x - 3y = 5 is equidistant from (1, 2) and (3, 4) ?

Options

A. (7, 3)
B. (2, 4)
C. (1, - 1)
D. (4, 1)

Answer

To solve the question of finding which point on the line \(2x - 3y = 5\) is equidistant from the points \((1, 2)\) and \((3, 4)\), we will follow these steps:

### Step 1: Identify the points on the line
We need to find points that lie on the line given by the equation \(2x - 3y = 5\). We can rearrange this equation to express \(y\) in terms of \(x\):
\[ 3y = 2x - 5 \implies y = \frac{2}{3}x - \frac{5}{3} \]

### Step 2: Check the given points
Assuming we have a list of points to test, we will check each point to see if it satisfies the line equation \(2x - 3y = 5\).

### Step 3: Calculate distances
For each point that lies on the line, we will calculate the distance from this point to both \((1, 2)\) and \((3, 4)\) using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

### Step 4: Set distances equal
We will find the point where the distance to \((1, 2)\) is equal to the distance to \((3, 4)\).

### Example Calculation
Let’s assume we have the following points on the line to check: \((7, 3)\), \((2, 4)\), \((1, -1)\), and \((4, 1)\).

1. **Check Point (7, 3)**:
- Check if it satisfies the line equation:
\[ 2(7) - 3(3) = 14 - 9 = 5 \quad \text{(satisfies)} \]
- Calculate distances:
- Distance to \((1, 2)\):
\[ d_1 = \sqrt{(7 - 1)^2 + (3 - 2)^2} = \sqrt{6^2 + 1^2} = \sqrt{36 + 1} = \sqrt{37} \]
- Distance to \((3, 4)\):
\[ d_2 = \sqrt{(7 - 3)^2 + (3 - 4)^2} = \sqrt{4^2 + (-1)^2} = \sqrt{16 + 1} = \sqrt{17} \]
- Since \(\sqrt{37} \neq \sqrt{17}\), this point is not equidistant.

2. **Check Point (2, 4)**:
- Check if it satisfies the line equation:
\[ 2(2) - 3(4) = 4 - 12 = -8 \quad \text{(does not satisfy)} \]

3. **Check Point (1, -1)**:
- Check if it satisfies the line equation:
\[ 2(1) - 3(-1) = 2 + 3 = 5 \quad \text{(satisfies)} \]
- Calculate distances:
- Distance to \((1, 2)\):
\[ d_1 = \sqrt{(1 - 1)^2 + (-1 - 2)^2} = \sqrt{0 + (-3)^2} = \sqrt{9} = 3 \]
- Distance to \((3, 4)\):
\[ d_2 = \sqrt{(1 - 3)^2 + (-1 - 4)^2} = \sqrt{(-2)^2 + (-5)^2} = \sqrt{4 + 25} = \sqrt{29} \]
- Since \(3 \neq \sqrt{29}\), this point is not equidistant.

4. **Check Point (4, 1)**:
- Check if it satisfies the line equation:
\[ 2(4) - 3(1) = 8 - 3 = 5 \quad \text{(satisfies)} \]
- Calculate distances:
- Distance to \((1, 2)\):
\[ d_1 = \sqrt{(4 - 1)^2 + (1 - 2)^2} = \sqrt{3^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10} \]
- Distance to \((3, 4)\):
\[ d_2 = \sqrt{(4 - 3)^2 + (1 - 4)^2} = \sqrt{1^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10} \]
- Since \(d_1 = d_2\), this point is equidistant.

### Conclusion
The point \((4, 1)\) on the line \(2x - 3y = 5\) is equidistant from the points \((1, 2)\) and \((3, 4)\).