For a point P in the plane, let d1(P) and d2(P) be the distance of the point P from the lines x - y

Q: Wait, why can we just drop the absolute value for $d_2(P)$?

Great question! The distance formula for $L_2: x + y = 0$ is $d_2(P) = \frac{|x + y|}{\sqrt{2}}$. Because the problem explicitly states that point $P$ lies in the first quadrant, we know that both $x$ and $y$ are strictly positive. Therefore, their sum $x + y$ is always positive, allowing us to remove the absolute value bars without changing the value.

Q: Why does this result in an L-shaped region?

When we solve the inequality for the two cases ($x \ge y$ and $y > x$), we find that in the first case, $x$ is bounded between $\sqrt{2}$ and $2\sqrt{2}$, and in the second case, $y$ is bounded between $\sqrt{2}$ and $2\sqrt{2}$. When you plot these constraints in the first quadrant, you are essentially looking at a large square with side $2\sqrt{2}$ and removing a smaller square with side $\sqrt{2}$ from the corner near the origin. This leaves you with an L-shaped region.

Q: Is there a faster way to solve this than splitting cases?

You could potentially use a coordinate rotation where $u = \frac{x-y}{\sqrt{2}}$ and $v = \frac{x+y}{\sqrt{2}}$, which transforms the lines into the axes of a new coordinate system. However, for JEE-level problems, splitting the cases based on the absolute value is the most reliable and standard method to avoid errors. It keeps the geometry clear and manageable.

Question

For a point $P$ in the plane, let $d_1(P)$ and $d_2(P)$ be the distance of the point $P$ from the lines $x - y = 0$ and $x + y = 0$ respectively. The area of the region $R$ consisting of all points $P$ lying in the first quadrant of the plane and satisfying $2 \le d_1(P) + d_2(P) \le 4$, is .........

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the origin of a Cartesian plane. You see two lines stretching out: $L_1: x - y = 0$ and $L_2: x + y = 0$.

These lines are the angle bisectors of your axes, meeting at a perfect $90^{\circ}$ angle. We are exploring a region $R$ in the first quadrant defined by the sum of distances from these two lines.

### The Distance Toolkit

To begin, we must translate our geometric intuition into the language of algebra. The perpendicular distance from any point $P(x, y)$ to a line $ax + by + c = 0$ is given by the formula:

$$d = \frac{|ax + by + c|}{\sqrt{a^2 + b^2}}$$

For our first line, $L_1: x - y = 0$, the distance $d_1(P)$ is:

$$d_1(P) = \frac{|x - y|}{\sqrt{1^2 + (-1)^2}} = \frac{|x - y|}{\sqrt{2}}$$

For our second line, $L_2: x + y = 0$, the distance $d_2(P)$ is:

$$d_2(P) = \frac{|x + y|}{\sqrt{1^2 + 1^2}} = \frac{|x + y|}{\sqrt{2}}$$

Because we are restricted to the first quadrant, $x > 0$ and $y > 0$. This means $x + y$ is always positive, so we can write $d_2(P) = \frac{x + y}{\sqrt{2}}$.

### The Inequality of Space

We are given the condition $2 \le d_1(P) + d_2(P) \le 4$. Substituting our expressions, we get:

$$2 \le \frac{|x - y|}{\sqrt{2}} + \frac{x + y}{\sqrt{2}} \le 4$$

Multiplying by $\sqrt{2}$ clears the denominators, leaving us with:

$$2\sqrt{2} \le |x - y| + x + y \le 4\sqrt{2}$$

The absolute value $|x - y|$ acts as a gatekeeper, forcing us to consider two distinct scenarios based on the relationship between $x$ and $y$.

### The Case Analysis

**Case 1: Below or on the line $y = x$.** 
In this region, $x \ge y$, so $x - y \ge 0$. The absolute value opens positively: $|x - y| = x - y$.

Our inequality becomes:

$$2\sqrt{2} \le (x - y) + x + y \le 4\sqrt{2}$$

The $y$ terms cancel out, leaving $2\sqrt{2} \le 2x \le 4\sqrt{2}$, which simplifies to $\sqrt{2} \le x \le 2\sqrt{2}$.

**Case 2: Above the line $y = x$.** 
Here, $y > x$, so $x - y < 0$. The absolute value opens with a negative sign: $|x - y| = y - x$.

Our inequality becomes:

$$2\sqrt{2} \le (y - x) + x + y \le 4\sqrt{2}$$

The $x$ terms cancel out, leaving $2\sqrt{2} \le 2y \le 4\sqrt{2}$, which simplifies to $\sqrt{2} \le y \le 2\sqrt{2}$.

### The Final Synthesis

When we combine these two cases, we see a symmetric L-shaped region. For $x \ge y$, $x$ is trapped between $\sqrt{2}$ and $2\sqrt{2}$. For $y > x$, $y$ is trapped between $\sqrt{2}$ and $2\sqrt{2}$.

This region is the area inside a large square of side $2\sqrt{2}$ minus the area of a smaller square of side $\sqrt{2}$. The area of the outer square is:

$$(2\sqrt{2})^2 = 8$$

The area of the inner square is:

$$(\sqrt{2})^2 = 2$$

Subtracting these, we find the area of our region $R$ is **$8 - 2 = 6$**.

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

Distance to the Line $x - y = 0$

Distance to the Line $x + y = 0$

Combining the Distances

Case 1: Below or on the Line $y = x$

Case 2: Above the Line $y = x$

Understanding the L-Shaped Region

Area of the Region $R$

The Sigma Insight: Distance of a Point from a Line

Analyzing the Setup

The Distance Toolkit

The Inequality of Space

The Case Analysis

The Final Synthesis

Similar Questions

A rectangle is formed by the lines $x = 0, y = 0, x = 3$ and $y = 4$ . Let the line L be perpendicular to $3 x + y + 6 = 0$ and divide the area of the rectangle into two equal parts. Then the distance of the point $(\frac{1}{2}, - 5)$ from the line $L$ is equal to :

Let the point $P (α, β)$ be at a unit distance from each of the two lines $L_{1} : 3 x - 4 y + 12 = 0$ , and $L_{2} : 8 x + 6 y + 11 = 0$ . If $P$ lies below $L_{1}$ and above $L_{2}$ , then $100 (α + β)$ is equal to

Let $R$ be the point $(3, 7)$ and let $P$ and $Q$ be two points on the line $x + y = 5$ such that $P QR$ is an equilateral triangle. Then the area of $Δ P QR$ is :

If the line, $2 x - y + 3 = 0$ is at a distance $\frac{1}{5}$ and $\frac{2}{5}$ from the lines $4 x - 2 y + α = 0$ and $6 x - 3 y + β = 0$ , respectively, then the sum of all possible values of $α$ and $β$ is

Let $A$ be the set of all points $(α, β)$ such that the area of triangle formed by the points $(5, 6), (3, 2)$ and $(α, β)$ is 12 square units. Then the least possible length of a line segment joining the origin to a point in $A$ , is :

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

A straight line through the origin $O$ meets the parallel lines $4 x + 2 y = 9$ and $2 x + y + 6 = 0$ at points $P$ and $Q$ respectively. Then the point $O$ divides the segment $P Q$ in the ratio

Let the range of the function $f (x) = 6 + 16 cos x \cdot cos (\frac{π}{3} - x) \cdot cos (\frac{π}{3} + x) \cdot sin 3 x \cdot cos 6 x,$ $x \in R$ be $[α, β]$ . Then the distance of the point $(α, β)$ from the line $3 x + 4 y + 12 = 0$ is:

Lines are drawn parallel to the line $4 x - 3 y + 2 = 0$ , at a distance $3/5$ from the origin. Then which one of the following points lies on any of these lines ?

Let $R$ be the interior region between the lines $3 x - y + 1 = 0$ and $x + 2 y - 5 = 0$ containing the origin. The set of all values of $a$ , for which the points $(a^{2}, a + 1)$ lie in $R$ , is :

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

Distance to the Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x−y=0

Distance to the Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x+y=0