If x and (x + 30" are complements of each other, find the value of x�

1.	If x and (x + 30" are complements of each other, find the value of x
Answer» $\tt \sqrt{ {(x - 1)}^{2} + {(y - 2)}^{2} } = \sqrt{ {(x - 3)}^{2} + {(y + 4)}^{2}}$ $\tt {(x)}^{2} + 1 - 2x + {(y)}^{2} + 4 - 4y = {(x)}^{2} + 9 - 6y + {(y)}^{2} + 16 + 8y$ $\tt 5 - 2x - 4y = 25 - 6x + 8y$ $\tt 4x = 20 + 12y$ $\tt 4x - 12y = 20$ $\tt x - 3y = 5 - - - (i)$ And $:\implies\tt \sqrt{ {(x - 3)}^{2} + {(y + 4)}^{2} } = \sqrt{ {(x - 5)}^{2} + {(y + 6)}^{2} }$ $:\implies\tt {(x)}^{2} + 9 - 6x + {(y)}^{2} + 16 + 8y = {(x)}^{2} + 25 - 10x + {(y)}^{2} + 36 + 12y$ $:\implies\tt 25 - 6x + 8y = 61 - 10x + 12y$ $:\implies\tt 4x - 4y = 36$ $:\implies\tt x - y = 9 - - - (ii)$