Finding the position in a triangle for the given challenge
Your current algorithm is very inefficient. I would use the following algorithmic approach:

def answer(x, y):
    y_diff = y - 1
    corner = x + y_diff
    id = corner * (corner + 1) // 2
    id -= y_diff
    return str(id)


This method takes advantage of basic algebra. It uses the sum of the arithmetic progression of x to calculate the id of the bottom right corner, and then subtracts the difference of the input coordinate from the bottom right corner. The reason it take the y difference is because moving 45 degrees diagonally down right is equivalent to moving down (y-1) units and moving right (y-1) units. Is is much more efficient than your current method: instead of an increasing numbering of iterations through range, it does one multiplication.

Benchmark

Testing your code compared to mine, mine runs about 10,000 to 100,000 times faster for random values of x and y between 1 and 100000. This is a very significant performance difference.

Additionally, in benchmarking, I discovered a slight bug in your current implementation. I didn't analyze your implementation deeply to understand it, but it seems that your current implementation returns an output corresponding to the reversed inputs (so for example, if I input (3, 2), the output is correct for (2, 3).) I suspect this may have been missed in your personal testing because the input is reversed of traditional coordinate order, i.e. you ask for y coordinate input before you ask for x coordinate input. I fixed this in my testing by switching all occurrences of x (and related variables) with y and vice versa.

Some notes


  Write a function answer(x, y) which returns the prisoner ID of the bunny at location (x, y).


So it seems your answer should be in the form:

def answer(x, y):
    # ... algorithm
    return str(output)


Also, use consistent spacing:


in_sum =i + 2 to in_sum = i + 2


And take advantage of +=:


sum_in_x = sum_in_x + range(x)[i] to sum_in_x += range(x)[i]
sum_in_y = sum_in_y + in_sum to sum_in_y += in_sum