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

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thilina-g eNotes educator| Certified Educator

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.




`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`

`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).

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