Çözüldü Coordinate Geometry - Symmety Between Lines

If the symmetric line of 2x + y - 5 = 0 to 2x + y - 3 = 0 passes through the point (-4, p), what is p?

Solution - 1:
The slope and equation of the symmetric line is respectively -2 and y - p = -2[ x - (-4) ] which is 2x + y + 8 - p = 0.
The distance between the lines given in the problem is |-5 - (-3)| / √(2^2 + 1^2) = 2 / √5....(I)
The distance from the in-between line (2x + y - 3 = 0) to 2x + y + 8 - p = 0 is |8 - p - (-3)| / √(2^2 + 1^2) = |11 - p| / √5....(II)
Equalizing (I) and (II) to one another yields 2 / √5 = |11 - p| / √5 ⇒ 2 = |11 - p|...(III)
For the equation (III), If p > 11 then 2 = -(11 -p) ⇒ p = 13 but B(-4, 13) is already on the line 2x + y - 5 = 0 given in the problem. Hence, p < 11 and 2 = 11 - p, so p = 9.
In conclusion, A(-4, 9) is on the line 2x + y + 8 - 9 = 0, of which its final standard form is 2x + y - 1 = 0.

Solution - 2:
Setting the distance from the line passing through (-4, p) to the in-between line (2x + y - 3 = 0) and the distance from the in-between line to 2x + y - 5 = 0 directly gives;
|(-4)·2 + p·1 - 3| / √(2^2 + 1^2) = |-5 - (-3)| / √(2^2 + 1^2)
|p - 11| = 2 and is solved as shown in the first solution.

Graph:

https://i72.servimg.com/u/f72/19/97/10/39/parale10.png

Original Turkish version of the question:

https://i.servimg.com/u/f72/19/97/10/39/symmet10.png
https://scontent-otp1-1.xx.fbcdn.ne...=874e0b497c11962820057748d95f4ad6&oe=5E16BC21
https://www.facebook.com/photo.php?...&set=gm.1685826884888181&type=3&theater&ifg=1