Question

find the coordinates of the points which are simultaneously 6 units from the points (6,6) and equidistant from the x and y axis. There are two ways to do this. I would like them both.

Answer 1

method 1:

to be equidistant from both the x and the y axes, point would have to lie on y = x
So the point would have to have the form (x,x)
distance from (6,6) has to be 6
√( (x-6)^2 + (x-6)^2 = 6
√( 2x^2 - 24x + 72) = 6
square both sides
2x^2 - 24x + 72 = 36
x^2 - 12x + 18 = 8
complete the square:
x^2 - 12x + ... = -18 + ...
x^2 - 12x + 36 = -18+36
(x-6)^2 = 18
x-6 = ± 18
x = 6 ± √18

there are two such points,
(6+√18 , 6+√18) and (6-√18 , 6-√18)

method 2

To be equidistant from (6,6) and have that length be 6, would require our points to lie on a circle with radius 6 and centre (6,6)
equation:
(x-6)^2 + (y-6)^2 = 36
but it must also lie on y = x
so solving y = x with the circle --->
(x-6)^2 + (x-6)^2 = 36
2x^2 - 24x + 72 = 36

we reach the same equation as in method 1