What is the distance between the point of contact of the line x + 4y + 8 = 0 with the lines x = 4 and y = 8?

Let the line x+4y+8 = 0 and the line at x= 4 intersect at
A(x1,y1).

Then we find the coordinates of A(x1,y1)
which lie both on x+y+8 = and x= 4.So,

x1+4y1+8 = 0 and x1
= 4

Therefore 4+4y1 +8 = 0, 4y1 = 0-8-4 = -12. So y1 =
-12/4 = -3

Therefore x1 = 4 and y1 =
-3.

So A(x1,y1) =
A(4,-3).

Similarly if B(x2,y2) is the point of intersection
of x+4y+8, and y= 8, then the coordinates of B(x2,y2) should satisfy both x+4y+8 = 0 and
y = 8:

x2+4y2+8 = 0 and y2 =
8.

So put y2 = 8 in x2+4y2+8 = 0 and we we
get:

x2 +4*8+8 = 0.

x2 +40. Or
x2 = -40.

So x2 = -40 and y2 =
8.

So B(x2,y2) = B(-40,
8).

Therefore the distance between A(x1,y1) and B(x2,y2) =
sqrt{(x2-x1)^2+(y2-y1)^2} = sqrt{(-40-4)^2+(8-(-3))^2} = sqrt{44^2+11^2} =
sqrt(2057).