A straight line through a fixed point (2, 3) intersects the coordinate axes at distinct points P and

Q: Why do we use the intercept form $\frac{x}{h} + \frac{y}{k} = 1$ instead of the slope-intercept form $y = mx + c$?

Great question! While $y = mx + c$ is standard, it struggles with vertical lines (where slope is undefined). The intercept form is elegant here because the problem explicitly defines the line by its intercepts $P$ and $Q$. It directly links the geometry of the rectangle to the equation, making the substitution of the fixed point $(2, 3)$ much cleaner.

Q: Is it mathematically rigorous to just replace $h$ and $k$ with $x$ and $y$ at the end?

Yes, it is! In locus problems, $h$ and $k$ represent the coordinates of a *variable* point $R$ that satisfies a specific condition. By treating $h$ and $k$ as the coordinates of $R$ and then replacing them with the general variables $x$ and $y$, we are essentially defining the set of all points $(x, y)$ that satisfy the condition imposed by the fixed point. It is the standard procedure for finding a locus.

Q: Wait, why do we assume the intercepts are $h$ and $k$? What if the line passes through the origin?

The problem states the line intersects the coordinate axes at *distinct* points $P$ and $Q$. If the line passed through the origin, it would intersect the axes at the same point (the origin), violating the 'distinct' condition. Thus, we are safe to assume non-zero intercepts $h$ and $k$.

Question

A straight line through a fixed point $(2, 3)$ intersects the coordinate axes at distinct points $P$ and $Q$. If $O$ is the origin and the rectangle $OPRQ$ is completed, then the locus of $R$ is :

Options:

(A) 3x + 2y = 6xy

(B) 3x + 2y = 6

(C) 2x + 3y = xy

(D) 3x + 2y = xy

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a Cartesian plane with a fixed anchor point at $(2, 3)$. A line is pinned to this point, but it is free to rotate like a propeller.

As it rotates, it carves out different intercepts on the $x$ and $y$ axes, which we call $P$ and $Q$. The rectangle $OPRQ$ forms as the line pivots, and our goal is to find the equation of the path traced by the corner $R$.

### The Power of Intercepts

To solve this, we use the intercept form of a line. Since the line cuts the axes at $P(h, 0)$ and $Q(0, k)$, the equation of the line is:

$$ \frac{x}{h} + \frac{y}{k} = 1 $$

This equation is the heartbeat of our solution. Because the line is forced to pass through our fixed anchor $(2, 3)$, these coordinates must satisfy the equation.

Substituting $x = 2$ and $y = 3$, we obtain the constraint:

$$ \frac{2}{h} + \frac{3}{k} = 1 $$

This equation represents the mathematical "DNA" of our line.

### The Birth of the Locus

Now, consider the rectangle $OPRQ$. With $O$ at $(0, 0)$, $P$ at $(h, 0)$, and $Q$ at $(0, k)$, the vertex $R$ must be at $(h, k)$.

This is the point whose path we are tracking. To find the locus, we transition from the specific variables $h$ and $k$ to the general coordinates $(x, y)$ of any point on the path.

Substituting $h = x$ and $k = y$ into our constraint, we get:

$$ \frac{2}{x} + \frac{3}{y} = 1 $$

### The Final Elegance

To express this in a standard form, we multiply the entire expression by $xy$ to clear the denominators:

$$ xy \left( \frac{2}{x} \right) + xy \left( \frac{3}{y} \right) = xy(1) $$

This simplifies beautifully to:

$$ 2y + 3x = xy $$

Rearranging the terms, we arrive at the final equation of the locus:

$$ **xy - 3x - 2y = 0** $$

If you were to graph this, you would see a hyperbola. It is fascinating to observe that a simple, rotating straight line, constrained by a single point, creates such an elegant curve. You have successfully mapped the motion of $R$.

Animated Solution for Mathematics - Straight Lines: A straight line through a fixed point $(2, 3)$ intersects the coordinate axes at distinct points $P$ and $Q$ . If $O$ is the origin and the rectangle $O P R Q$ is completed, then the locus of $R$ is :

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Visualized Solution

Setting up the Coordinate System

Drawing the Variable Line

Defining the Intercepts $P$ and $Q$

Completing the Rectangle $O P R Q$

Equation of Line $P Q$ in Intercept Form

Applying the Fixed Point Condition

Simplifying the Equation

Finding the Locus of $R$

Final Answer

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Setup

The Power of Intercepts

The Birth of the Locus

The Final Elegance

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Visualized Solution

Setting up the Coordinate System

Drawing the Variable Line

Defining the Intercepts .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

Completing the Rectangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OPRQ

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