The distance of the point (1,2) from the line x+y+5=0 measured along the line parallel to 3x -y=7 is equal to

Options

A. `4sqrt10`
B. 40
C. `sqrt40`
D. `10sqrt2`

Explanation

To find the distance from the point (1, 2) to the line \(x + y + 5 = 0\) measured along the line parallel to \(3x - y = 7\), we will follow these steps:

### Step-by-Step Solution:

1. **Identify the given lines and point:**
- The first line is \(L_1: x + y + 5 = 0\).
- The second line is \(L_2: 3x - y - 7 = 0\).
- The point is \(P(1, 2)\).

2. **Find the slope of the line \(L_2\):**
- Rewrite \(L_2\) in slope-intercept form (y = mx + c):
\[ y = 3x - 7 \]
- The slope \(m\) of line \(L_2\) is \(3\).

3. **Determine the slope of the line \(PQ\):**
- Since \(PQ\) is parallel to \(L_2\), it will have the same slope:
\[ m_{PQ} = 3 \]

4. **Write the equation of the line \(PQ\) passing through point \(P(1, 2)\):**
- Using the point-slope form of the line:
\[ y - y_1 = m(x - x_1) \]
Substituting the values:
\[ y - 2 = 3(x - 1) \]
- Simplifying this:
\[ y - 2 = 3x - 3 \implies y = 3x - 1 \]

5. **Find the intersection point \(Q\) of lines \(L_1\) and \(PQ\):**
- Set the equations equal to each other:
\[ x + y + 5 = 0 \quad \text{and} \quad y = 3x - 1 \]
- Substitute \(y\) from the second equation into the first:
\[ x + (3x - 1) + 5 = 0 \]
Simplifying:
\[ 4x + 4 = 0 \implies 4x = -4 \implies x = -1 \]
- Substitute \(x = -1\) back into \(y = 3x - 1\):
\[ y = 3(-1) - 1 = -3 - 1 = -4 \]
- Thus, the coordinates of point \(Q\) are \((-1, -4)\).

6. **Calculate the distance \(PQ\) using the distance formula:**
- The distance formula is given by:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
- Here, \(P(1, 2)\) and \(Q(-1, -4)\):
\[ d = \sqrt{((-1) - 1)^2 + ((-4) - 2)^2} \]
Simplifying:
\[ d = \sqrt{(-2)^2 + (-6)^2} = \sqrt{4 + 36} = \sqrt{40} \]

7. **Final answer:**
- The distance from point \(P(1, 2)\) to the line \(x + y + 5 = 0\) measured along the line parallel to \(3x - y = 7\) is:
\[ \sqrt{40} = 2\sqrt{10} \]