If the sum of the distances of a point from two perpendicular lines in a plane is 1, then its locus

Q: Wait, why can't we just write $x + y = 1$? Why do we need the modulus signs?

That is a fantastic question! Remember, distance is a physical quantity—it is always non-negative. If we just wrote $x + y = 1$, we would be assuming that $x$ and $y$ are always positive. But the point $P$ can exist anywhere in the plane, even where coordinates are negative. The modulus function $|x|$ ensures that even if $x$ is negative, the distance remains positive. Without it, we would miss three-quarters of the possible locus!

Q: Why do we have to break this into four cases? Can't we just solve it in one go?

I understand the desire for a shortcut! However, the modulus function is piecewise defined. It behaves differently depending on the sign of the input. To 'unlock' the equation and see the geometry, we must analyze the behavior of $x$ and $y$ in each quadrant. It is not just about solving an equation; it is about mapping the entire territory of the plane.

Q: How do we know for sure it's a square and not just a rhombus or a parallelogram?

Great eye! A square is a special type of rhombus. To confirm it is a square, look at the vertices: $(1, 0), (0, 1), (-1, 0),$ and $(0, -1)$. The distance between $(1, 0)$ and $(0, 1)$ is $\sqrt{(1-0)^2 + (0-1)^2} = \sqrt{1+1} = \sqrt{2}$. Because of the symmetry of the equation $|x| + |y| = 1$, all sides will have this same length, and the diagonals will be perpendicular and equal. That confirms it is a perfect square!

Question

If the sum of the distances of a point from two perpendicular lines in a plane is 1, then its locus is

Options:

(A) square

(B) circle

(C) straight line

(D) two intersecting lines

Sigma Percentile · Accepted Answer

### Analyzing the Setup The first step in any JEE Advanced problem is to simplify the environment. We are given two perpendicular lines, so we can define our own coordinate system to make the math intuitive. Let us align these two lines with the standard Cartesian axes. We define the first line as the $Y$-axis ($x=0$) and the second line as the $X$-axis ($y=0$). By making this choice, we have transformed an abstract geometric problem into a concrete algebraic one. We are now looking for the locus of a point $P(x, y)$ in the $XY$-plane. ### Translating English to Math The problem states that the sum of the distances of point $P$ from these lines is $1$. The distance of a point $P(x, y)$ from the $Y$-axis is $|x|$, and the distance from the $X$-axis is $|y|$. We use the modulus function because distance is a scalar quantity and must be positive. Thus, our condition becomes the elegant equation: $$|x| + |y| = 1$$ ### The Modulus Challenge To visualize $|x| + |y| = 1$, we must peel back the layers of the modulus function by considering the four quadrants of the Cartesian plane. In the **First Quadrant** ($x > 0, y > 0$), the equation becomes: $$x + y = 1$$ This is a straight line with intercepts at $(1, 0)$ and $(0, 1)$. In the **Second Quadrant** ($x < 0, y > 0$), the equation becomes: $$-x + y = 1$$ This is a line segment connecting $(0, 1)$ and $(-1, 0)$. In the **Third Quadrant** ($x < 0, y < 0$), the equation becomes: $$-x - y = 1 \quad \Rightarrow \quad x + y = -1$$ This connects $(-1, 0)$ and $(0, -1)$. In the **Fourth Quadrant** ($x > 0, y < 0$), the equation becomes: $$x - y = 1$$ This connects $(0, -1)$ and $(1, 0)$. ### The Geometric Revelation We have created four line segments forming a closed, symmetric shape. The vertices are $(1, 0)$, $(0, 1)$, $(-1, 0)$, and $(0, -1)$. If you calculate the length of any side, such as the segment from $(1, 0)$ to $(0, 1)$, you get: $$\sqrt{(1-0)^2 + (0-1)^2} = \sqrt{2}$$ Because the equation is perfectly symmetric, all four sides are equal to $\sqrt{2}$. Furthermore, the diagonals are perpendicular and equal in length. **Conclusion:** The locus is a **square**. Whenever you encounter an equation of the form $|x| + |y| = c$, you are looking at a square rotated by $45^\circ$.

Animated Solution for Mathematics - Straight Lines: If the sum of the distances of a point from two perpendicular lines in a plane is 1, then its locus is

Select Answer:

Visualized Solution

Choosing the Reference Frame

The Moving Point $P (x, y)$

The Locus Condition

Handling the Modulus

Quadrant I: $x > 0, y > 0$

Quadrant II: $x < 0, y > 0$

Quadrant III: $x < 0, y < 0$

Quadrant IV: $x > 0, y < 0$

Conclusion: The Locus is a Square

The Sigma Insight: Distance of a Point from a Line

Analyzing the Setup

Translating English to Math

The Modulus Challenge

The Geometric Revelation

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Animated Solution for Mathematics - Straight Lines: If the sum of the distances of a point from two perpendicular lines in a plane is 1, then its locus is

Select Answer:

Visualized Solution

Choosing the Reference Frame

The Moving Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(x,y)

The Locus Condition

Handling the Modulus

Quadrant I: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x>0,y>0

Quadrant II: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x<0,y>0

Quadrant III: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x<0,y<0

Quadrant IV: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x>0,y<0