Let R be a rectangle given by the lines x = 0, x = 2, y = 0 and y = 5. Let A(α, 0) and B(0, β), α ∈

Q: Wait, why do we assume the triangle $\Delta OAB$ is the smaller part of the ratio $4:1$?

Great observation! The line $AB$ cuts the rectangle into two pieces. One piece is the triangle $\Delta OAB$ with area $\frac{1}{2} \alpha \beta$, and the other is a quadrilateral. Since $\alpha \in [0, 2]$ and $\beta \in [0, 5]$, the maximum area of the triangle is $\frac{1}{2} \times 2 \times 5 = 5$. Since the total area is $10$, the triangle must be the smaller piece (area $2$) compared to the quadrilateral (area $8$). Thus, the ratio $4:1$ implies the triangle is the $1$ part.

Q: Why can't we just use the coordinates of the vertices to find the locus?

The locus is not about a single set of coordinates, but the path traced by the midpoint as $\alpha$ and $\beta$ vary. By defining the midpoint as $(h, k)$ and relating it to the variables $\alpha$ and $\beta$, we create a general equation that holds true for any valid position of the line $AB$. This is the essence of coordinate geometry!

Q: Is $xy = 1$ always a hyperbola?

Yes, absolutely! The equation $xy = c$ (where $c$ is a non-zero constant) is the standard form of a rectangular hyperbola. In our case, $c=1$. It's a beautiful curve that never touches the axes, reflecting the constraints on $\alpha$ and $\beta$.

Question

Let $R$ be a rectangle given by the lines $x = 0, x = 2, y = 0$ and $y = 5$. Let $A(\alpha, 0)$ and $B(0, \beta)$, $\alpha \in [0, 2]$ and $\beta \in [0, 5]$, be such that the line segment $AB$ divides the area of the rectangle $R$ in the ratio $4 : 1$. Then, the mid-point of $AB$ lies on a

Options:

(A) straight line

(B) parabola

(C) hyperbola

(D) circle

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers! Today, we are embarking on a journey through the elegant world of coordinate geometry.

Imagine you are standing in a room, and on the floor, you have drawn a rectangle $R$ bounded by the lines $x = 0$, $x = 2$, $y = 0$, and $y = 5$. This rectangle is our canvas.

Its vertices are at $(0,0)$, $(2,0)$, $(2,5)$, and $(0,5)$. The base is $2$ units, and the height is $5$ units. A quick calculation tells us the total area is:

$$Area = 2 	imes 5 = 10 	ext{ square units}$$

This is our baseline, our constant in a world of moving parts.

### The Slicing Line

Now, imagine a line segment $AB$ that slices through this rectangle. Point $A$ is at $(\alpha, 0)$ on the x-axis, and point $B$ is at $(0, \beta)$ on the y-axis.

As we move $A$ and $B$, the line segment $AB$ dances across the rectangle, constantly changing the shape of the pieces it creates. The problem states that this line divides the rectangle into two regions with an area ratio of $4:1$.

The region near the origin is a right-angled triangle $\Delta OAB$. Its area is given by the classic formula:

$$	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} \alpha \beta$$

Since the total area is $10$ and the ratio is $4:1$, the smaller area must be $\frac{1}{5} 	imes 10 = 2$. Therefore, we have the beautiful, simple relation:

$$\frac{1}{2} \alpha \beta = 2 \Rightarrow \alpha \beta = 4$$

### The Locus of the Midpoint

Now, let's focus on the midpoint $M(h, k)$ of the segment $AB$. Using the midpoint formula, we know that:

$$h = \frac{\alpha + 0}{2} \quad 	ext{and} \quad k = \frac{0 + \beta}{2}$$

This gives us the coordinates of our midpoint in terms of $\alpha$ and $\beta$. To find the locus, we express $\alpha$ and $\beta$ in terms of $h$ and $k$:

$$\alpha = 2h \quad 	ext{and} \quad \beta = 2k$$

Now, we substitute these into our earlier relation, $\alpha \beta = 4$. This yields:

$$(2h)(2k) = 4$$

Dividing both sides by $4$, we arrive at the final, elegant equation:

$$hk = 1$$

### The Final Revelation

Replacing $h$ and $k$ with the general coordinates $x$ and $y$, we get the final locus:

$$xy = 1$$

Look at that equation: $xy = 1$. It is the hallmark of a **rectangular hyperbola**.

As the line segment $AB$ moves while maintaining the area ratio, its midpoint doesn't just wander aimlessly; it traces a precise, beautiful hyperbolic path. This is the power of coordinate geometry—taking a seemingly complex dynamic problem and reducing it to a fundamental curve.

Select Answer:

Visualized Solution

Visualizing the Rectangle $R$

Total Area of Rectangle $R$

Introducing Points $A$ and $B$

Area of Triangle $O A B$

Applying the Area Ratio

Solving for $α β$

Defining the Midpoint $M$

Expressing $α$ and $β$

Substituting into the Product Equation

The Final Locus

Identifying the Curve

The Sigma Insight: Rectangular Hyperbola

Analyzing the Setup

The Slicing Line

The Locus of the Midpoint

The Final Revelation

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