Key Concepts and Formulas

• Distance of a point $(x_0, y_0)$ from the x-axis is $|y_0|$, and from the y-axis is $|x_0|$.
• If a point is equidistant from the coordinate axes, then $|x| = |y|$. This implies $x = y$ or $x = -y$.
• The quadrant of a point $(x, y)$ is determined by the signs of $x$ and $y$.

Step-by-Step Solution

Step 1: Interpreting the condition "equidistant from the coordinate axes"

Let the point be $P(x, y)$. The problem states that $P(x, y)$ is equidistant from the x-axis and the y-axis. This means the distance from $P$ to the x-axis equals the distance from $P$ to the y-axis.

• Distance from $P(x, y)$ to the x-axis = $|y|$.
• Distance from $P(x, y)$ to the y-axis = $|x|$.

Therefore, we have the equation: $|x| = |y|$ This equation implies two possibilities:

1. $x = y$
2. $x = -y$

Explanation: The absolute value ensures we consider distances, which are always non-negative. The equation $|x| = |y|$ represents two lines: $y = x$ and $y = -x$. These lines are angle bisectors of the quadrants.

Step 2: Incorporating the given line equation

The problem also states that the point $P(x, y)$ lies on the line $3x + 5y = 15$. This means the point must satisfy both the condition of being equidistant from the axes AND lie on the given line. Therefore, we need to find the intersection of the line $3x + 5y = 15$ with both $y = x$ and $y = -x$.

Explanation: We are looking for the simultaneous solutions to the equations. This is a standard approach for finding points satisfying multiple conditions.

Step 3: Solving for the intersection points in each case

Case 1: Intersection of $3x + 5y = 15$ and $y = x$

Substitute $y = x$ into the equation $3x + 5y = 15$: $3x + 5(x) = 15$ $3x + 5x = 15$ $8x = 15$ $x = \frac{15}{8}$ Since $y = x$, we also have $y = \frac{15}{8}$. So, the first point is $P_1 \left( \frac{15}{8}, \frac{15}{8} \right)$.

Determining the Quadrant for $P_1$: Since $x = \frac{15}{8} > 0$ and $y = \frac{15}{8} > 0$, the point $P_1$ lies in the 1st Quadrant.

Case 2: Intersection of $3x + 5y = 15$ and $y = -x$

Substitute $y = -x$ into the equation $3x + 5y = 15$: $3x + 5(-x) = 15$ $3x - 5x = 15$ $-2x = 15$ $x = -\frac{15}{2}$ Since $y = -x$, we have $y = - \left( -\frac{15}{2} \right) = \frac{15}{2}$. So, the second point is $P_2 \left( -\frac{15}{2}, \frac{15}{2} \right)$.

Determining the Quadrant for $P_2$: Since $x = -\frac{15}{2} < 0$ and $y = \frac{15}{2} > 0$, the point $P_2$ lies in the 2nd Quadrant.

Step 4: Concluding the quadrants

We found two points that satisfy both conditions:

1. $P_1 \left( \frac{15}{8}, \frac{15}{8} \right)$, which is in the 1st Quadrant.
2. $P_2 \left( -\frac{15}{2}, \frac{15}{2} \right)$, which is in the 2nd Quadrant.

Therefore, the point satisfying the given conditions will lie only in the 1st and 2nd quadrants.

Common Mistakes & Tips:

• Missing the x = -y case: Forgetting the absolute value in $|x| = |y|$ and only considering $x = y$ is a common error.
• Sign Errors: Be careful with signs when substituting and solving for x and y, especially when dealing with $y = -x$.
• Quadrant Awareness: Always double-check the signs of the coordinates to correctly identify the quadrant.

Summary:

The problem requires finding points on the line $3x + 5y = 15$ that are equidistant from the x and y axes. This means $|x| = |y|$, which gives us two cases: $y = x$ and $y = -x$. Solving the system of equations formed by $3x + 5y = 15$ and each of these cases yields two points, one in the 1st quadrant and one in the 2nd quadrant.

Final Answer: The final answer is \boxed{1 st and 2 nd quadrants}, which corresponds to option (A).