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

We have been asked to locate those points which simultaneously satisfy the straight-line equation $$3x + 5y = 15$$ and are equidistant from the two coordinate axes.

First, we recall the mathematical fact that the perpendicular distance of a point $$P(x , y)$$ from the $$y$$-axis is $$|x|$$, while the perpendicular distance from the $$x$$-axis is $$|y|$$. A point is equidistant from the two axes exactly when these two absolute values are equal. Hence we must have

$$|x| = |y|.$$

This equality of absolute values can occur in two algebraic ways:

(i) $$y = x$$    or    (ii) $$y = -x.$$

Thus every point that is equidistant from the axes lies on one of the two lines $$y = x$$ or $$y = -x$$. We now examine each possibility in turn, always keeping the additional requirement that the point must also lie on the given straight line $$3x + 5y = 15$$.

Case 1 : $$y = x$$

We substitute $$y = x$$ in the equation $$3x + 5y = 15$$.

Using substitution we get

$$3x + 5x = 15.$$

Combining like terms on the left side,

$$8x = 15.$$

Solving for $$x$$, we divide both sides by $$8$$:

$$x = \dfrac{15}{8}.$$

Because $$y = x$$ in this case, we have

$$y = \dfrac{15}{8}.$$

Both coordinates are positive, i.e. $$x > 0$$ and $$y > 0$$, so the point

$$\left(\dfrac{15}{8},\,\dfrac{15}{8}\right)$$

lies in the first quadrant.

Case 2 : $$y = -x$$

Again we substitute, now using $$y = -x$$ in the same line equation:

$$3x + 5(-x) = 15.$$

That gives

$$3x - 5x = 15.$$

Combining like terms,

$$-2x = 15.$$

Dividing by $$-2$$, we obtain

$$x = -\dfrac{15}{2}.$$

Since $$y = -x$$ here,

$$y = -\left(-\dfrac{15}{2}\right) = \dfrac{15}{2}.$$

Now we observe that $$x < 0$$ while $$y > 0$$; hence the point

$$\left(-\dfrac{15}{2},\,\dfrac{15}{2}\right)$$

falls in the second quadrant.

There are no valid solutions in the third or fourth quadrants, because our two exhaustive cases have already accounted for all possibilities of $$|x| = |y|$$.

Therefore the points satisfying both conditions—lying on $$3x + 5y = 15$$ and being equidistant from the coordinate axes—occur only in the first and second quadrants.

Hence, the correct answer is Option A.