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 1^st qudratant or 2^nd qudratant.