Question

An ant is walking along the cartesian plane. It starts at the point A=(−18,8) walks in a straight line to a point on
the x-axis, walks directly to the right for 7 units and then
walks in a straight line to the point B=(9,13). What is
the shortest distance that the ant could have walked?

Answer 1

Let the first point on the x axis be P(x,0),
he then goes to Q(x-7,0) and finally to B(9,13)

So the path is AP + PQ + QB

Distance (D) = √( (x+18)^2 + 64) + 7 + √( (x-2)^2 + 16)

dD/dx = .......
= (x+18)/√(x^2 + 36x + 388) + (x-2)/√(x^2 - 4x + 173) . leaving it up to you to simplify to this point
= 0 for a max/min of D

(x+18)/√(x^2 + 36x + 388) = (2-x)/√(x^2 - 4x + 173)
squaring both sides and simplifying
x^4 + 32x^3 + 353x^2 + 4932x + 56052 =
x^4 + 32x^3 + 248x^2 - 1408x + 1552
PHeeww!
105x^2 + 6340x +54500 = 0
21x^2 + 1268x + 10900 = 0
believe it or not, it factored, (good sign I did not make an error)
(21x+218)(x-50) = 0
x = -218/21 or appr -10.38 which makes sense
or
x = 50 , clearly not a minimum

so the ant should go to P(-218/21, 0) and proceed as instructed in your problem

to find the shortest distance, sub in x = -218/21 into the distance expression above
(I will let you do this, it is just tedious arithmetic)