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