The reflection of the point (4,-13) in the line `5x+y+6=0` is

Options

A. `(-1,-14)`
B. `(3,4)`
C. `(1,2)`
D. `(-4,13)`

Answer

To find the reflection of the point \( P(4, -13) \) in the line \( 5x + y + 6 = 0 \), we will follow these steps:

### Step 1: Identify the line and point
The line is given by the equation \( 5x + y + 6 = 0 \) and the point is \( P(4, -13) \).

### Step 2: Find the slope of the line
The equation of the line can be rewritten in slope-intercept form \( y = -5x - 6 \). The slope of the line is \( -5 \).

### Step 3: Find the slope of the perpendicular line
The slope of the line perpendicular to this line will be the negative reciprocal of \( -5 \), which is \( \frac{1}{5} \).

### Step 4: Write the equation of the perpendicular line
Using the point-slope form of the line equation, the equation of the line passing through point \( P(4, -13) \) with slope \( \frac{1}{5} \) is:
\[ y + 13 = \frac{1}{5}(x - 4) \]
Simplifying this, we get:
\[ y + 13 = \frac{1}{5}x - \frac{4}{5} \]
\[ y = \frac{1}{5}x - \frac{4}{5} - 13 \]
\[ y = \frac{1}{5}x - \frac{4 + 65}{5} \]
\[ y = \frac{1}{5}x - \frac{69}{5} \]

### Step 5: Find the intersection of the two lines
Now we need to find the intersection of the lines \( 5x + y + 6 = 0 \) and \( y = \frac{1}{5}x - \frac{69}{5} \).

Substituting \( y \) from the second equation into the first:
\[ 5x + \left(\frac{1}{5}x - \frac{69}{5}\right) + 6 = 0 \]
Multiplying through by 5 to eliminate the fraction:
\[ 25x + x - 69 + 30 = 0 \]
\[ 26x - 39 = 0 \]
\[ 26x = 39 \]
\[ x = \frac{39}{26} = \frac{3}{2} \]

### Step 6: Find the y-coordinate of the intersection point
Substituting \( x = \frac{3}{2} \) back into the equation of the line:
\[ y = \frac{1}{5}\left(\frac{3}{2}\right) - \frac{69}{5} \]
\[ y = \frac{3}{10} - \frac{69}{5} \]
Converting \( \frac{69}{5} \) to a common denominator:
\[ y = \frac{3}{10} - \frac{138}{10} = \frac{3 - 138}{10} = \frac{-135}{10} = -13.5 \]

### Step 7: Find the reflection point
Let the intersection point be \( I\left(\frac{3}{2}, -13.5\right) \). The reflection point \( R(x', y') \) can be found using the midpoint formula:
\[ \left(\frac{4 + x'}{2}, \frac{-13 + y'}{2}\right) = \left(\frac{3}{2}, -13.5\right) \]

From the x-coordinates:
\[ \frac{4 + x'}{2} = \frac{3}{2} \]
\[ 4 + x' = 3 \implies x' = 3 - 4 = -1 \]

From the y-coordinates:
\[ \frac{-13 + y'}{2} = -13.5 \]
\[ -13 + y' = -27 \implies y' = -27 + 13 = -14 \]

### Final Answer
Thus, the reflection of the point \( (4, -13) \) in the line \( 5x + y + 6 = 0 \) is:
\[ (-1, -14) \]