Question

find the shortest distance between y= X+10 and Y=6√X

Answer 1

recall that if you have
Ax + By + C = 0 , and some external point (a,b)
then the distance from (a,b) to the line is
| aA + bB + C| / √(A^2 + B^2)

so let P( a, 6Va) be the point closest to the line
x - y + 10 = 0

Let the distance be D
D = | a - 6√a + 10|/√(1+1)
= (1/√2) |a - 6√a + 10|

dD/da = (1/√2)|1 - 3/√a + 0| = 0 for a min of D

1 - 3/√a = 0
√a = 3
a = 9

so point P is (9, 6√9) or (9,18)

distance form (9,18) to x-y+10=0
is
|9 - 18 + 10|/√2 =<b> 1/√2 or √2/2 or appr .7071</b>

(you might have noticed that by coincidence that is also sin45°)

check:
take a point just to the right and just to the left of (9,18)
e.g. (8.9 , 17.8997)
distance to line = |8.9 - 17.8997+10|√2 = appr .7073 , ahh , just a bit longer
e.g. , (9.1, 18.0997..)
distance to line = |9.1 - 18.0997 + 10}|/√2 = appr .7073.. , again, just a bit longer
My answer sounds very plausable.