Key Concepts and Formulas

• Intersection of Lines: The intersection of two lines $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ is found by solving the system of equations.
• Area of a Triangle (Determinant Form): The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by: $\text{Area} = \frac{1}{2} \left| \det \begin{pmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{pmatrix} \right|$
• Area Relationship: Medial Triangle: If $D, E, F$ are the midpoints of the sides of $\triangle ABC$, then $\text{Area}(\triangle DEF) = \frac{1}{4} \text{Area}(\triangle ABC)$, which implies $\text{Area}(\triangle ABC) = 4 \times \text{Area}(\triangle DEF)$.

Step-by-Step Solution

Step 1: Understand the Given Information

We have three lines:

1. $L_1: x - y + 1 = 0$
2. $L_2: x - 2y + 3 = 0$
3. $L_3: 2x - 5y + 11 = 0$

The intersection points of these lines are the midpoints of the sides of a triangle $\triangle ABC$. We need to find the area of $\triangle ABC$.

Step 2: Find the Intersection Points

We find the intersection points by solving the equations of the lines pairwise. These intersection points will be the vertices of the medial triangle.

• Intersection of $L_1$ and $L_2$: We solve the system: $x - y + 1 = 0$ ...(1) $x - 2y + 3 = 0$ ...(2) Subtracting (2) from (1), we get: $(x - y + 1) - (x - 2y + 3) = 0$ $y - 2 = 0$ $y = 2$ Substituting $y = 2$ into (1), we get: $x - 2 + 1 = 0$ $x = 1$ So, the intersection point is $(1, 2)$.

• Intersection of $L_1$ and $L_3$: We solve the system: $x - y + 1 = 0$ ...(3) $2x - 5y + 11 = 0$ ...(4) Multiply (3) by 2: $2x - 2y + 2 = 0$ ...(5) Subtract (5) from (4): $(2x - 5y + 11) - (2x - 2y + 2) = 0$ $-3y + 9 = 0$ $y = 3$ Substituting $y = 3$ into (3), we get: $x - 3 + 1 = 0$ $x = 2$ So, the intersection point is $(2, 3)$.

• Intersection of $L_2$ and $L_3$: We solve the system: $x - 2y + 3 = 0$ ...(6) $2x - 5y + 11 = 0$ ...(7) Multiply (6) by 2: $2x - 4y + 6 = 0$ ...(8) Subtract (8) from (7): $(2x - 5y + 11) - (2x - 4y + 6) = 0$ $-y + 5 = 0$ $y = 5$ Substituting $y = 5$ into (6), we get: $x - 2(5) + 3 = 0$ $x - 10 + 3 = 0$ $x = 7$ So, the intersection point is $(7, 5)$.

Therefore, the vertices of the medial triangle $\triangle DEF$ are $D(1, 2)$, $E(2, 3)$, and $F(7, 5)$.

Step 3: Calculate the Area of the Medial Triangle $\triangle DEF$

Using the determinant formula for the area of a triangle: $\text{Area}(\triangle DEF) = \frac{1}{2} \left| \det \begin{pmatrix} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 7 & 5 & 1 \end{pmatrix} \right|$ $\text{Area}(\triangle DEF) = \frac{1}{2} \left| 1(3 - 5) - 2(2 - 7) + 1(10 - 21) \right|$ $\text{Area}(\triangle DEF) = \frac{1}{2} \left| 1(-2) - 2(-5) + 1(-11) \right|$ $\text{Area}(\triangle DEF) = \frac{1}{2} \left| -2 + 10 - 11 \right|$ $\text{Area}(\triangle DEF) = \frac{1}{2} \left| -3 \right|$ $\text{Area}(\triangle DEF) = \frac{3}{2}$

Step 4: Calculate the Area of the Original Triangle $\triangle ABC$

Since $\text{Area}(\triangle DEF) = \frac{1}{4} \text{Area}(\triangle ABC)$, we have: $\text{Area}(\triangle ABC) = 4 \times \text{Area}(\triangle DEF)$ $\text{Area}(\triangle ABC) = 4 \times \frac{3}{2}$ $\text{Area}(\triangle ABC) = 6$

Common Mistakes & Tips

• Double-check the signs when calculating determinants to avoid errors.
• Remember the relationship between the areas of a triangle and its medial triangle.
• Be careful when solving the systems of linear equations; a small arithmetic error can lead to a wrong answer.

Summary

We found the intersection points of the three given lines, which are the midpoints of the sides of triangle $\triangle ABC$. These intersection points form the vertices of the medial triangle $\triangle DEF$. We calculated the area of $\triangle DEF$ using the determinant formula and then used the relationship $\text{Area}(\triangle ABC) = 4 \times \text{Area}(\triangle DEF)$ to find the area of $\triangle ABC$.

The final answer is \boxed{6}.