The line 3x-4y=5 intersects the $x$-axis and $y$-axis at points $A$ and $B,$ respectively. Find the distance $AB.$
3x - 4y = 5
x axis intersection , let y = 0
3x = 5
x = 5/3
y axis intersection, let x =0
-4y = 5
y = -5/4
AB = sqrt [ (5/3)^2 + (-5/4)^2 ] = sqrt [ 25/9 + 25/ 16 ] = [ sqrt [ (9 + 16)(25) / (16 * 9) ] =
sqrt [ 25^2 / 144] = 25 / 12
+368 
