A point which is equidistant from both the axes lies on either y=x and y=−x. Since, point lies on the line 3x+5y=15 Then the required point 3x+5y=15 x+y=0 x=−
15
2
y=
15
2
⇒(x,y)=(−
15
2
,
15
2
){2nd quadrant } 3x+5y=15 or
x−y=0
15
x=
15
8
y=
15
8
⇒(x,y)=(
15
8
,
15
8
){1st quadrant } Hence, the required point lies in 1st and 2nd quadrant.