Coordinate Geometry Problem

[Coookie]

This is really, really doing my head in.

The distance of a point X (k,2) from the line y = x + 4 is four.
Give the possible values of k.

Can someone take me through it step by step?

I can get up to finding the intersection of y = x + 4 with the perpendicular bisector

x + 4 = -x + k + 2

but I'm completely lost after that.

[PlainSight]

There is probably a much easier way to do this, but i solved using algebra on the conditions that need to be fulfilled.
We know that there must be a distance of 4 between the point k,2 and x,y. We can relate these values by going (k-x)^2 + (2-y)^2 = 4^2
We know that y = x+4
thus this simplifies to (k-x)^2 + (-2-x)^2 = 16
We also know that the closest distance the lines and the point will be will be at a bisector to the line we are given, this has the negative inverse gradient to the line which = -1.
y - y1 = m (x - x1)
y1 is 2 and x1 is k also y = x + 4
x + 4 - 2 = -1(x - k)
x + 2 + =-x + k
thus k = 2x + 2
substitute this into (k-x)^2 + (-2-x)^2 = 16
we get (x+2)^2 + (x+2)^2 = 16 => 2(x^2) + 8x - 8 = 0
quadratic solvable for x. gives x = 0.8284, -4.828
substitute these into k = 2x + 2
giving 3.6568 and - 7.656

[BobO.]

Well, the easiest way to show that line is x-y+4=0 . So if you have a line shown by equation Ax + By + C = 0, and a point X (x0,y0) , then the equation for distance of a point from a line is :
d = (|A*x0 + B*y0 + C|)/sqrt(sqr(A)+sqr(B)) (| are absolutes)
So in your case d=4 ,A is 1 , B is -2, y0 is 2 , C is 4 and x0 is k you are looking for
The solutions are 3.6568 and -7.656 as said before
Hope you understood that , English isn't my native language, so it's kinda difficult to explain math when you learn everything under different names

[GeneralCash]

there is an easier way.

first i'm going to make a new coordinate system s' because i hate dragging a bunch of constants with me. it's the easiest way to make a mistake.

s -> s'
(x,y) -> (x-2,y+2)

this way the center of s' is in the point where your line in the system s intersects the line y=2. i think the reason is obvious enough.

the new lines are y' = x' and y' = 0. now isn't that better?

now notice that dy'/dx' (= dx/dy) = 1, i.e. the line forms a 45° angle with the x'-axis. that means: if you note the projection of the point x'(k',0) as a', the distance between x' and a' is the same as the distance between the system's origin (0,0) and a'. distance is kept intact after the transformation obviously, it remains 4.

you can easily use pitagora's theorem to find k'.

k'^2 = 2*(4^2) = +/-sqrt(32)

all that remains is to reverse the coordinates so you get k.

k = k' - 2 = 2 -/+ sqrt(32)

can't get much simpler.

cba with the calculator. the principle is 100% viable, but i could have made a mistake in calculating somewhere so don't just copypasta.