A straight line L through the origin meets the lines x + y = 1 and x + y = 3 at P and Q respectively

Q: Wait, why do we use $y=mx$ for the line $L$? What if the line is vertical?

That is a fantastic observation! In coordinate geometry, we often assume $y=mx$ for simplicity. If the line were vertical, its equation would be $x=0$. However, the problem states $L$ passes through the origin and meets $x+y=1$ and $x+y=3$. If $x=0$, then $y=1$ and $y=3$. This is a specific case, but the general locus is defined by the variable slope $m$. For the purpose of finding the locus, $y=mx$ covers all non-vertical lines, which is sufficient to define the path.

Q: Why does the parameter $m$ disappear in the final equation?

The parameter $m$ represents the 'freedom' of the line $L$ to rotate. When we find the locus, we are looking for the path that the point $R$ traces regardless of the specific value of $m$. By eliminating $m$, we are essentially saying: 'For any valid $m$, the point $R$ must satisfy this specific linear equation.' The disappearance of $m$ confirms that the path is a fixed geometric object—a straight line.

Q: Is there a geometric way to solve this without all this algebra?

Yes, there is! You could use the concept of homothety (scaling). Since $P$ and $Q$ are intersections of a line through the origin with two parallel lines, $Q$ is just a scaled version of $P$. However, in JEE Advanced, algebraic elimination is often the most robust and least error-prone method. Stick to the algebra, but keep the geometric intuition in your back pocket to verify your result!

Question

A straight line $L$ through the origin meets the lines $x + y = 1$ and $x + y = 3$ at $P$ and $Q$ respectively. Through $P$ and $Q$ two straight lines $L_1$ and $L_2$ are drawn, parallel to $2x - y = 5$ and $3x + y = 5$ respectively. Lines $L_1$ and $L_2$ intersect at $R$. Show that the locus of $R$, as $L$ varies, is a straight line.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the origin of a coordinate plane, watching a line $L$ rotate like the hand of a clock. As it sweeps through the plane, it slices through two parallel lines, $x + y = 1$ and $x + y = 3$.

These two lines are our fixed stage, and the rotating line $L$ is our dynamic actor. Our goal is to track the intersection point $R$ of two other lines, $L_1$ and $L_2$, which are born from the points of intersection $P$ and $Q$.

### The Stage and the Actor

First, let us define our stage. We have two parallel lines, $x + y = 1$ and $x + y = 3$. Because they share the same slope, they remain perfectly equidistant.

Now, our actor, the line $L$, passes through the origin. We define its equation as $y = mx$, where $m$ is the variable slope. As $m$ changes, the line $L$ rotates, and the points $P$ and $Q$ slide along our parallel lines.

### Finding the Coordinates

To find the intersection point $P$, we solve $y = mx$ and $x + y = 1$ simultaneously. Substituting $y = mx$ into the line equation gives $x + mx = 1$, which simplifies to $x(1 + m) = 1$.

Thus, the coordinates of $P$ are:
$$P = \left( \frac{1}{1+m}, \frac{m}{1+m} ight)$$

Applying the same logic to $Q$ with the line $x + y = 3$, we find $x(1 + m) = 3$. This gives us the coordinates of $Q$ as:
$$Q = \left( \frac{3}{1+m}, \frac{3m}{1+m} ight)$$

Notice the elegance here: $Q$ is simply $3P$. The origin acts as a center of scaling, a beautiful geometric symmetry.

### The New Paths

Now, we draw $L_1$ through $P$ with a slope of $2$ (parallel to $2x - y = 5$) and $L_2$ through $Q$ with a slope of $-3$ (parallel to $3x + y = 5$). Let the intersection of these two lines be $R(X, Y)$.

Using the point-slope form, the equation for $L_1$ is:
$$Y - \frac{m}{1+m} = 2\left(X - \frac{1}{1+m}ight)$$

Similarly, for $L_2$, we have:
$$Y - \frac{3m}{1+m} = -3\left(X - \frac{3}{1+m}ight)$$

### The Algebraic Symphony

Now, we face the challenge: eliminating the parameter $m$. Let us rearrange the equation for $L_1$:
$$Y - 2X = \frac{m}{1+m} - \frac{2}{1+m} = \frac{m-2}{1+m}$$

By writing $m-2$ as $(m+1) - 3$, we get $Y - 2X = 1 - \frac{3}{1+m}$. This gives us the isolated term:
$$\frac{3}{1+m} = 1 - Y + 2X$$

Now, let us look at $L_2$:
$$Y + 3X = \frac{3m}{1+m} + \frac{9}{1+m} = \frac{3m+9}{1+m}$$

Writing $3m+9$ as $3(m+1) + 6$, we get $Y + 3X = 3 + \frac{6}{1+m}$. This gives us:
$$\frac{6}{1+m} = Y + 3X - 3$$

### The Grand Finale

Here is the moment of truth. We observe that $\frac{6}{1+m} = 2 	imes \left(\frac{3}{1+m}ight)$. Substituting our expression from $L_1$ into $L_2$, we get:
$$Y + 3X - 3 = 2(1 - Y + 2X)$$

Expanding this, we find:
$$Y + 3X - 3 = 2 - 2Y + 4X$$

Bringing all terms to one side, we arrive at the final locus:
$$X - 3Y + 5 = 0$$

The parameter $m$ has vanished, leaving us with a clean, linear equation. The locus of $R$ is a straight line. You have successfully navigated the complexity and found the underlying simplicity. **The final answer is $X - 3Y + 5 = 0$.**

Visualized Solution

The Fixed Geometry

The Variable Line and Intersections

Coordinates of Point $P$

Coordinates of Point $Q$

Introducing Lines $L_{1}$ and $L_{2}$

The Intersection Point $R$

Equation of Line $L_{1}$

Equation of Line $L_{2}$

Isolating the Parameter in $L_{1}$

Isolating the Parameter in $L_{2}$

Eliminating the Parameter ' $m$ '

The Final Locus Equation

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Setup

The Stage and the Actor

Finding the Coordinates

The New Paths

The Algebraic Symphony

The Grand Finale

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Visualized Solution

The Fixed Geometry

The Variable Line and Intersections

Coordinates of Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Coordinates of Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

Introducing Lines .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​

The Intersection Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R

Equation of Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​

Equation of Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​

Isolating the Parameter in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​

Isolating the Parameter in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​

Eliminating the Parameter ' .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m'