If the line y = mx bisects the area enclosed by the lines x = 0, y = 0, x = 3/2 and the curve y = 1

Q: Wait, why can we just use the triangle area formula for the line $y=mx$? Shouldn't we integrate that too?

You absolutely *could* integrate $\int_{0}^{3/2} mx dx$, and you would get the exact same result! But in JEE, time is your currency. Recognizing that the area under a linear function $y=mx$ from $x=0$ to $x=b$ is simply a right-angled triangle with base $b$ and height $mb$ saves you precious seconds and reduces the chance of calculation errors.

Q: How do I know the line $y=mx$ doesn't intersect the parabola before $x = \frac{3}{2}$?

That is a brilliant intuition! In this specific problem, we are given that the line bisects the area. If the line intersected the curve earlier, the area calculation would be much more complex. We assume the standard configuration where the line cuts through the region cleanly. Always check your boundary conditions, but trust the problem statement's implication of a clean bisection.

Q: Why did we multiply the triangle area by 2 instead of dividing the total area by 2?

It is mathematically identical! Whether you write $\text{Area}_{triangle} = \frac{1}{2} \times \text{Total Area}$ or $2 \times \text{Area}_{triangle} = \text{Total Area}$, you are expressing the same physical reality. I prefer the second form because it clears the fractions early, making the algebra cleaner and less prone to those pesky arithmetic slips.

Question

If the line $y = mx$ bisects the area enclosed by the lines $x = 0, y = 0, x = \frac{3}{2}$ and the curve $y = 1 + 4x - x^2$, then $12m$ is equal to .

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer! Today, we aren't just solving a calculus problem; we are learning to balance the scales of geometry and analysis. When you look at a problem involving a curve and a line, don't just see equations. See a landscape—a region of space that we are about to divide with surgical precision.

Imagine you are standing on the Cartesian plane. You have a downward-opening parabola defined by $y = 1 + 4x - x^2$. We are trapping this curve between the y-axis ($x=0$) and the vertical wall at $x = \frac{3}{2}$.

To understand the total magnitude of this space, we must call upon the power of definite integration. We are summing up an infinite number of infinitesimally thin vertical strips from $x=0$ to $x=\frac{3}{2}$.

### The Total Area

Our integral setup is:

$$A = \int_{0}^{3/2} (1 + 4x - x^2) dx$$

Now, let's perform the integration term by term. The integral of $1$ is $x$, the integral of $4x$ is $2x^2$, and the integral of $-x^2$ is $-\frac{x^3}{3}$. We evaluate this from $0$ to $\frac{3}{2}$.

Substituting the upper limit, we get:

$$A = \left[ \frac{3}{2} + 2\left(\frac{3}{2}ight)^2 - \frac{(\frac{3}{2})^3}{3} ight] - [0]$$

Take a breath here. Do not rush. Squaring $\frac{3}{2}$ gives $\frac{9}{4}$, and multiplying by $2$ gives $\frac{9}{2}$. Cubing $\frac{3}{2}$ gives $\frac{27}{8}$, and dividing by $3$ gives $\frac{9}{8}$.

So, our total area is:

$$A = \frac{3}{2} + \frac{9}{2} - \frac{9}{8} = 6 - \frac{9}{8} = \frac{39}{8}$$

### The Geometric Slice

Now, the problem introduces a line, $y=mx$, passing through the origin. This line acts like a blade, slicing our region into two equal halves. The area *under* this line, bounded by the x-axis and the vertical line $x = \frac{3}{2}$, forms a perfect right-angled triangle.

Why is this beautiful? Because we don't need calculus for a triangle! The base is $\frac{3}{2}$. The height, at the point where $x = \frac{3}{2}$, is simply the y-value of the line: $y = m(\frac{3}{2}) = \frac{3m}{2}$.

Using the classic formula, $	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$, we get:

$$	ext{Area}_{triangle} = \frac{1}{2} 	imes \left(\frac{3}{2}ight) 	imes \left(\frac{3m}{2}ight) = \frac{9m}{8}$$

### The Synthesis

We are at the finish line. The problem states that the line $y=mx$ bisects the total area. This means the area of our triangle must be exactly half of the total area $A$ we calculated earlier.

$$\frac{9m}{8} = \frac{1}{2} 	imes \frac{39}{8}$$

Look at the elegance of the cancellation! We can multiply both sides by $8$ immediately:

$$9m = \frac{39}{2}$$

Cross-multiplying the $2$ gives us $18m = 39$. Dividing both sides by $3$, we find $6m = 13$. The question asks for $12m$. Simply multiply by $2$:

$$12m = 26$$

And there it is. We didn't just calculate a number; we balanced the area of a complex curve with the simplicity of a triangle. The final answer is **26**.

Animated Solution for Mathematics - Definite Integration: If the line $y = m x$ bisects the area enclosed by the lines $x = 0, y = 0, x = \frac{3}{2}$ and the curve $y = 1 + 4 x - x^{2}$ , then $12 m$ is equal to .

Enter Numerical Value:

Visualized Solution

Visualizing the Region

Setting up the Integral

Integrating the Expression

Substituting the Upper Limit

Simplifying the Terms

Calculating Total Area

The Bisecting Line $y = m x$

Area of the Triangle

Calculating Triangle Area

Equating the Areas

Solving for $m$

Solving for $12 m$

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The Total Area

The Geometric Slice

The Synthesis

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