Let a triangle be bounded by the lines L1 : 2x + 5y = 10; L2 : -4x + 3y = 12 and the line L3, which

Q: Why can't we just find the equation of line $L_3$ first? It feels more natural.

That is a classic trap! To find the equation of $L_3$, you would need its slope, which is unknown. You would end up with a variable slope $m$, leading to a much more complex algebraic mess. By using parametric coordinates, we bypass the need for the slope entirely and jump straight to the coordinates of $A$ and $B$.

Q: Why did we use $\alpha$ and $\beta$ instead of just one parameter?

Excellent question. $A$ and $B$ are independent points on two different lines. They are only linked by the fact that they lie on the same line $L_3$ passing through $P$. Using two parameters allows us to define their positions independently before the section formula forces them to satisfy the geometric constraint of being collinear with $P$.

Q: Is there a faster way to find the area without the determinant formula?

While you could use the base-height formula, you would first need to calculate the lengths of the sides and the altitude, which involves square roots and distance formulas—a recipe for calculation errors. The determinant formula (or the shoelace formula) is the most robust and elegant tool for this job. Trust the process!

Question

Let a triangle be bounded by the lines $L_1 : 2x + 5y = 10$; $L_2 : -4x + 3y = 12$ and the line $L_3$, which passes through the point $P(2, 3)$, intersect $L_1$ at $A$ and $L_2$ at $B$. If the point $P$ divides the line-segment $AB$, internally in the ratio $1 : 3$, then the area of the triangle is equal to

Options:

(A) $\frac{110}{13}$

(B) $\frac{132}{13}$

(C) $\frac{142}{13}$

(D) $\frac{151}{13}$

Sigma Percentile · Accepted Answer

### The Geometry of Constraints: A Journey into Triangle $ABC$

Welcome, future engineer. Today, we are not just solving a coordinate geometry problem; we are choreographing a dance between lines.

Imagine you are standing on a coordinate plane. You see two fixed lines, $L_1: 2x + 5y = 10$ and $L_2: -4x + 3y = 12$.

They are rigid, unmoving. But then, there is a third line, $L_3$, which acts like a hinge. It is anchored at point $P(2, 3)$, and as it rotates, it carves out a triangle by intersecting our two fixed lines at points $A$ and $B$.

Our goal is to find the area of this triangle. Let's break this down step by step.

### Phase 1: Finding the Fixed Vertex $C$

Every triangle needs three vertices. We already have $A$ and $B$ moving, but where do $L_1$ and $L_2$ meet? That is our fixed vertex $C$.

To find it, we solve the system of equations:

$$2x + 5y = 10$$

$$-4x + 3y = 12$$

By multiplying the first equation by $2$, we get $4x + 10y = 20$. Adding this to the second equation, the $x$ terms vanish, leaving us with $13y = 32$, or $y = \frac{32}{13}$.

Substituting this back, we find $x = -\frac{15}{13}$. So, our anchor vertex is $C = \left(-\frac{15}{13}, \frac{32}{13}ight)$. This point is the bedrock of our triangle.

### Phase 2: The Power of Parametric Coordinates

Now, how do we handle the moving points $A$ and $B$? If we try to find the equation of $L_3$ with an unknown slope $m$, we will drown in algebra.

Instead, let's use parametric coordinates. Since $A$ lies on $L_1$, its coordinates must satisfy $2x + 5y = 10$. We can write $A$ as $\left(\alpha, \frac{10-2\alpha}{5}ight)$.

Similarly, since $B$ lies on $L_2$, we write $B$ as $\left(\beta, \frac{12+4\beta}{3}ight)$. By introducing these parameters, $\alpha$ and $\beta$, we have turned a geometric problem into a simple algebraic one.

### Phase 3: The Section Formula Bridge

We are told that $P(2, 3)$ divides $AB$ in the ratio $1:3$. This is the bridge connecting our parameters.

The section formula tells us that for a point dividing a segment in ratio $m:n$, the coordinates are $\left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}ight)$. Applying this to our $x$-coordinates:

$$2 = \frac{1 \cdot \beta + 3 \cdot \alpha}{1+3} \Rightarrow 3\alpha + \beta = 8$$

This is our first equation. Now, for the $y$-coordinates, we apply the same logic:

$$3 = \frac{1 \cdot \left(\frac{12+4\beta}{3}ight) + 3 \cdot \left(\frac{10-2\alpha}{5}ight)}{4}$$

After clearing the denominators and simplifying, we arrive at $9\alpha - 10\beta = -15$. Solving this system of two equations is straightforward. Substituting $\beta = 8 - 3\alpha$ into the second equation yields $\alpha = \frac{5}{3}$ and $\beta = 3$.

### Phase 4: The Final Calculation

With $\alpha$ and $\beta$ in hand, we find the coordinates: $A = \left(\frac{5}{3}, \frac{4}{3}ight)$ and $B = (3, 8)$.

Now, we use the determinant formula for the area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$:

$$	ext{Area} = \frac{1}{2} |x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)|$$

Substituting our values, the calculation simplifies beautifully to:

$$	ext{Area} = \frac{1}{2} \left| \frac{264}{13} ight| = \frac{132}{13}$$

There you have it. By refusing to be intimidated by the variable line and instead using the elegance of parametric coordinates, we turned a complex problem into a clear, logical path. Keep practicing this mindset—it is the hallmark of a true problem solver. The final area is **$\frac{132}{13}$**.

Select Answer:

Visualized Solution

Visualizing the Boundary Lines

Finding Vertex $C$

Parametrizing $A$ and $B$

Applying the Section Formula

Solving for $x$ -coordinate

Solving for $y$ -coordinate

Finding $α$ and $β$

Coordinates of $A$ and $B$

Calculating the Area

Final Conclusion

The Sigma Insight: Area of Triangle

The Geometry of Constraints

Phase 1

Phase 2

Phase 3

Phase 4

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Select Answer:

Visualized Solution

Visualizing the Boundary Lines

Finding Vertex .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Parametrizing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Applying the Section Formula

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x-coordinate

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y-coordinate

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β

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