QJEEM19Ma7d58: JEE Main 2019 Mathematics - Class 11 JEEM19Ma7d58 Solution: A point on the straight line, 3x + 5y = 15 which is equidistant …

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.