The distance of the point (3,5) from the line `2x+3y-14=0` measured parallel to the line `x-2y=1` is

Options

A. `7/(sqrt(5))`
B. `7/(sqrt(13))`
C. `sqrt(5)`
D. `sqrt(13)`

Answer

To find the distance of the point (3, 5) from the line \(2x + 3y - 14 = 0\) measured parallel to the line \(x - 2y = 1\), we will follow these steps:

### Step 1: Find the equation of the line parallel to \(x - 2y = 1\) that passes through the point (3, 5).

The equation of a line parallel to \(x - 2y = 1\) can be expressed in the form:
\[ x - 2y = k \]
To find the value of \(k\), we substitute the coordinates of the point (3, 5) into the equation:
\[ 3 - 2(5) = k \\ 3 - 10 = k \\ k = -7 \]
Thus, the equation of the line passing through (3, 5) and parallel to \(x - 2y = 1\) is:
\[ x - 2y = -7 \]

### Step 2: Solve the system of equations to find the intersection point of the two lines.

Now we have two equations:
1. \(2x + 3y - 14 = 0\) (Equation of the given line)
2. \(x - 2y = -7\) (Equation of the parallel line)

We can express the second equation in terms of \(x\):
\[ x = 2y - 7 \]

Now, substitute \(x\) into the first equation:
\[ 2(2y - 7) + 3y - 14 = 0 \\ 4y - 14 + 3y - 14 = 0 \\ 7y - 28 = 0 \\ 7y = 28 \\ y = 4 \]

### Step 3: Substitute \(y\) back to find \(x\).

Now substitute \(y = 4\) back into the equation \(x - 2y = -7\):
\[ x - 2(4) = -7 \\ x - 8 = -7 \\ x = 1 \]
Thus, the intersection point \(B\) is \((1, 4)\).

### Step 4: Calculate the distance from point \(A(3, 5)\) to point \(B(1, 4)\).

We can use the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates:
\[ d = \sqrt{(1 - 3)^2 + (4 - 5)^2} \\ d = \sqrt{(-2)^2 + (-1)^2} \\ d = \sqrt{4 + 1} \\ d = \sqrt{5} \]

### Final Answer:
The distance of the point (3, 5) from the line \(2x + 3y - 14 = 0\) measured parallel to the line \(x - 2y = 1\) is \(\sqrt{5}\).

---