Official solution
Let the point is P(x, y).
According to the question, the point P(x, y) is equidistance from both x and y axis.
\[ \therefore \] |x| = |y|
\[ \Rightarrow \] x = \[ \pm \] y
So the point P lies on the either x = y or x = - y line. And point P(x, y) also lies on the straight line 3x + 5y = 15.
Form the graph, you can see the point P can either be on 1st qudratant or 2nd qudratant.