A point on the straight line,

A point which is equidistant from both the axes lies on

either $y=x$ and $y=-x$.

Since, point lies on the line $3 x+5 y=15$

Then the required point

$3 x+5 y=15$

$\frac{x+y=0}{x=-\frac{15}{2}}$

$y=\frac{15}{2} \Rightarrow(x, y)=\left(-\frac{15}{2}, \frac{15}{2}\right)\left\{2^{\text {nd }}\right.$ quadrant $\}$

or $3 x+5 y=15$

$\frac{x-y=0}{x=\frac{15}{8}}$

$y=\frac{15}{8} \Rightarrow(x, y)=\left(\frac{15}{8}, \frac{15}{8}\right)\left\{1^{\text {st }}\right.$ quadrant $\}$

Hence, the required point lies in $1^{\text {st }}$ and $2^{\text {nd }}$ quadrant.