Alternative working :

Let coordinates of B be (a,b). Since B is on line 1, sub (a,b) into y - 2x = -1,

b - 2a = -1
b = 2a - 1

So the coordinates can be rewritten as
(a, 2a - 1)


By Pythagoras' theorem,
AB ² + BC ² = AC ²

Since we know length = √((x1 - x2)² + (y1 - y2)²)

Then length² = (x1 - x2)² + (y1 - y2)²


So,

(a - 0)² + (2a -1 - (-1))² + (5 - a)² + (-1 - (2a -1) )² = (5 - 0)² + (-1 -(-1))²


a² + (2a)² + 25 - 10a + a² + (2a)² = 25 + 0²

10a² - 10a = 0
a² - a = 0
a(a - 1) = 0

a = 0 or a = 1

b = 2(0) - 1 or b = 2(1) - 1
b = -1 or b = 1

We have (0,-1) and (1,1)

since A(0,-1) and we know B ≠A,

Then B has to be (1,1).