A point on the straight line, 3x + 5y = 15 which is equidistant...

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

$⇒ 3a + 5a = 15$

$⇒ a = \dfrac{15}{8}$

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

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

$3x + 5y = 15$

$⇒ - 3a + 5a = 15$

$⇒ a = \dfrac{15}{2}$

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

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