The line given is `5x+y+6=0` and the point is (4,-13).

Let the reflection be (a,b), then the midpoint of (4,-13) and (a,b) lies on the given curve. Midpoint, (c,d) can be calculated as below.

`(c,d) = ((4+a)/2,(-13+b)/2)`

This midpoint is on the line, therefore it should satisfy the equation of the line.

`5((4+a)/2)+((-13+b)/2)+6=0`

`5(4+a)+(-13+b)=-12`

`20+5a-13+b=-12`

`5a+b = -19` ----------->` ` Equation 1

Also we know that, the line connecting the reflection point with the original point is perpendicular to the line given. Therefore if the gradient of given line is `m_1` and the gradient of the line connecting two points is `m_2` ,

`m_1*m_2 = -1`

Let's find `m_1` and `m_2` ,

`5x+y+6 = 0`

`y = -5x-6`

Therefore, `m_1` = -5

Let's find `m_2` in terms of a and b.

`m_2 = (b-(-13))/(a-4) = (b+13)/(a-4)`

But according to the above equation, `m_1m_2 = -1` .

`(-5)((b+13)/(a-4))=-1`

`(5(b+13))/(a-4) = 1`

`5b+65 = a-4`

`a - 5b = 69`    ----------> Equation 2

Now we have two simultaneous equations for a and b,

Solving them would give you,

a = -1   and b =-14

Therefore the reflection point of (4,-13) over the given line is (-1,-14).