A point on the straight line, 3 x + 5 y = 15 which is equidistant from the coordinate axes will lie only in:

Let $\left(a,a\right)$ be a point on the line $3x+5y=15$

$\Rightarrow 3a+5a=15$

$\Rightarrow a=\frac{15}{8}$

$\therefore$ Point is $\left(\frac{15}{8} ,\frac{15}{8}\right)$ which lies in $1^{st}$ quadrant

Again, Let $\left(-a,a\right)$ be a point on the line

      $3x+5y=15$

$\Rightarrow -3a+5a=15$

$\Rightarrow a=\frac{15}{2}$

$\therefore$ Point is $\left(-\frac{15}{2} ,\frac{15}{2}\right)$ which lies in $2^{nd}$ quadrant

Hence, required point lies in$1^{st}$ and $2^{nd}$  quadrants.