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

Q: Wait, why do we use $|x| = |y|$ instead of just $x = y$?

That is a fantastic question! If you only use $x = y$, you are only looking at the first and third quadrants. But the distance to the axes is always positive. A point like $(-5, 5)$ is 5 units from the x-axis and 5 units from the y-axis. Here, $x = -5$ and $y = 5$, so $x \neq y$, but $|x| = |y|$ holds true. The modulus captures all possibilities.

Q: Why are there two cases?

The equation $|x| = |y|$ is a geometric definition of two lines. One line, $y = x$, bisects the first and third quadrants. The other line, $y = -x$, bisects the second and fourth quadrants. Since our given line $3x + 5y = 15$ passes through multiple quadrants, it is bound to intersect both of these lines.

Q: How do I know which quadrant the points lie in?

Once you solve for the coordinates $(x, y)$, look at the signs. If $x > 0$ and $y > 0$, it is the first quadrant. If $x 0$, it is the second quadrant. If $x 0$ and $y < 0$, it is the fourth quadrant. It is as simple as checking the signs!

Question

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

Options:

(A) $1^{st}$ and $2^{nd}$ quadrants

(B) $4^{th}$ quadrant

(C) $1^{st}$, $2^{nd}$ and $4^{th}$ quadrant

(D) $1^{st}$ quadrant

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a vast, infinite coordinate plane. You have a straight line, defined by the equation $3x + 5y = 15$, stretching across this plane like a bridge.

Your mission is to find the specific locations on this bridge that are perfectly balanced—equidistant from the $x$-axis and the $y$-axis. This is a quest to understand the symmetry of the Cartesian system.

### Decoding the Condition

The phrase "equidistant from the coordinate axes" is the key to the entire problem. If you are at a point $(x, y)$, your perpendicular distance to the $x$-axis is $|y|$, and your perpendicular distance to the $y$-axis is $|x|$.

For these distances to be equal, we must satisfy the condition:
$$|x| = |y|$$

This is the heartbeat of our problem. It is a simple, elegant constraint that defines the geometry of our search.

### The Power of Modulus

When we see the equation $|x| = |y|$, we must recognize that the modulus function splits the solution into two distinct paths. These paths represent the lines where this balance occurs.

Case 1 is $y = x$, a line that cuts through the first and third quadrants at a $45^{\circ}$ angle. Case 2 is $y = -x$, a line that cuts through the second and fourth quadrants.

Any point that is equidistant from the axes must lie on one of these two lines.

### The Intersection: Where Paths Cross

Now, we bring our original line, $3x + 5y = 15$, into the mix to find where it intersects our lines of balance.

For Case 1, we substitute $y = x$ into the main equation:
$$3x + 5(x) = 15$$
$$8x = 15$$
$$x = \frac{15}{8}$$

Since $y = x$, our $y$-coordinate is also $\frac{15}{8}$. This point, $(\frac{15}{8}, \frac{15}{8})$, sits in the first quadrant.

For Case 2, we substitute $y = -x$ into the main equation:
$$3x + 5(-x) = 15$$
$$3x - 5x = 15$$
$$-2x = 15$$
$$x = -\frac{15}{2}$$

Since $y = -x$, our $y$-coordinate becomes the negative of our $x$-coordinate, which is $\frac{15}{2}$. This point, $(-\frac{15}{2}, \frac{15}{2})$, resides in the second quadrant.

### Final Reflection

We have found our two points of balance:
$$\left( \frac{15}{8}, \frac{15}{8} ight) \quad 	ext{and} \quad \left( -\frac{15}{2}, \frac{15}{2} ight)$$

The first point is in the first quadrant, and the second is in the second quadrant. We have successfully navigated the geometry and solved the algebra to find the **points of balance**.

Animated Solution for Mathematics - Straight Lines: A point on the straight line, $3 x + 5 y = 15$ which is equidistant from the coordinate axes will lie only in :

Select Answer:

Visualized Solution

Visualizing the Given Line $3 x + 5 y = 15$

The Equidistant Condition $∣ x ∣ = ∣ y ∣$

Two Possible Cases: $y = x$ and $y = - x$

Case 1: Intersection with $y = x$

Solving for Case 1

Locating the First Point

Case 2: Intersection with $y = - x$

Solving for Case 2

Finding the Y-coordinate

Locating the Second Point

Final Conclusion

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Setup

Decoding the Condition

The Power of Modulus

The Intersection

Final Reflection

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Animated Solution for Mathematics - Straight Lines: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A point on the straight line, 3x+5y=15 which is equidistant from the coordinate axes will lie only in :

Select Answer:

Visualized Solution

Visualizing the Given Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 3x+5y=15

The Equidistant Condition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣x∣=∣y∣

Two Possible Cases: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=x and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=−x

Case 1: Intersection with .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=x