Let the points of intersections of the lines x - y + 1 = 0, x - 2y + 3 = 0 and 2x - 5y + 11 = 0 are

Q: Wait, why do we multiply by 4? Can't we just find the vertices of $ABC$?

You absolutely could find the vertices of $ABC$ by using the midpoint property (e.g., side $BC$ is parallel to $EF$ and passes through $D$), but that is a much longer, more tedious path. The medial triangle property is a classic JEE shortcut that saves you time and reduces the chance of calculation errors. Always look for the geometric shortcut!

Q: What if the lines were parallel? Would the triangle still exist?

Great question! If the lines were parallel, they would not intersect at three distinct points, and thus, they would not form a medial triangle. The problem implicitly guarantees that these lines form a valid triangle. In a real exam, if you find the lines are parallel, you know something is wrong with the premise, but here, the geometry is sound.

Q: Is the order of vertices in the area formula important?

The formula uses an absolute value, which is there specifically to handle the sign. Whether you list the vertices in clockwise or counter-clockwise order, the absolute value ensures your final area is always a positive, physical quantity. Don't worry about the order!

Question

Let the points of intersections of the lines $x - y + 1 = 0$, $x - 2y + 3 = 0$ and $2x - 5y + 11 = 0$ are the mid points of the sides of a triangle $ABC$. Then the area of the triangle $ABC$ is .

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, open field, and you are given three lines: 
$L_1: x - y + 1 = 0$
$L_2: x - 2y + 3 = 0$
$L_3: 2x - 5y + 11 = 0$

These lines are not just arbitrary paths; they are the skeletal structure of a hidden triangle. The problem states that the intersection points of these lines are the midpoints of the sides of a larger triangle, $ABC$.

This is a classic JEE Advanced setup—it tests your ability to bridge the gap between algebraic equations and geometric properties. Let us embark on this journey together.

### The Hunt for the Medial Triangle

Our first task is to uncover the vertices of the medial triangle, which we will call $DEF$. These vertices are simply the intersection points of our three lines. We solve the equations pairwise to find them.

For point $D$, the intersection of $L_1$ and $L_2$, we subtract the equations:
$$(x - y + 1) - (x - 2y + 3) = 0$$

This simplifies to $y - 2 = 0$, or $y = 2$. Substituting $y = 2$ back into $L_1$, we find $x = 1$. Thus, $D(1, 2)$.

For point $E$, the intersection of $L_2$ and $L_3$, we multiply $L_2$ by $2$ to get $2x - 4y + 6 = 0$. Subtracting this from $L_3$ gives:
$$(2x - 5y + 11) - (2x - 4y + 6) = 0$$

This yields $-y + 5 = 0$, so $y = 5$. Substituting back, we find $x = 7$. Thus, $E(7, 5)$.

For point $F$, the intersection of $L_1$ and $L_3$, we multiply $L_1$ by $2$ to get $2x - 2y + 2 = 0$. Subtracting this from $L_3$ yields:
$$(2x - 5y + 11) - (2x - 2y + 2) = 0$$

This results in $-3y + 9 = 0$, so $y = 3$. Substituting back, we find $x = 2$. Thus, $F(2, 3)$.

### The Area of the Medial Triangle

Now that we have the coordinates of the medial triangle $DEF$ as $D(1, 2)$, $E(7, 5)$, and $F(2, 3)$, we use the coordinate geometry formula for area:

$$	ext{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$

Substituting our values into the formula:

$$	ext{Area} = \frac{1}{2} |1(5 - 3) + 7(3 - 2) + 2(2 - 5)|$$

$$	ext{Area} = \frac{1}{2} |1(2) + 7(1) + 2(-3)| = \frac{1}{2} |2 + 7 - 6| = \frac{3}{2} 	ext{ square units.}$$

### Final Calculation

A fundamental theorem in geometry states that the triangle formed by joining the midpoints of a triangle divides the original triangle into four congruent triangles. Therefore, the area of the original triangle $ABC$ is exactly $4$ times the area of the medial triangle $DEF$.

$$	ext{Area}(\Delta ABC) = 4 	imes \frac{3}{2} = 6 	ext{ square units.}$$

By visualizing the geometry and applying the right theorems, we have conquered the problem. The final area of triangle $ABC$ is **6 square units**.

Animated Solution for Mathematics - Straight Lines: Let the points of intersections of the lines $x - y + 1 = 0$ , $x - 2 y + 3 = 0$ and $2 x - 5 y + 11 = 0$ are the mid points of the sides of a triangle $A B C$ . Then the area of the triangle $A B C$ is .

Enter Numerical Value:

Visualized Solution

Visualizing the Problem

Strategy to Find Midpoints

Intersection of $L_{1}$ and $L_{2}$

Coordinates of Point $D$

Intersection of $L_{2}$ and $L_{3}$

Coordinates of Point $E$

Intersection of $L_{1}$ and $L_{3}$

Coordinates of Point $F$

Area of $Δ D E F$

Substituting Coordinates

Calculating the Area

Relating $Δ D E F$ to $Δ A B C$

Final Area of $Δ A B C$

The Sigma Insight: Area of Triangle

Analyzing the Setup

The Hunt for the Medial Triangle

The Area of the Medial Triangle

Final Calculation

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Enter Numerical Value:

Visualized Solution

Visualizing the Problem

Strategy to Find Midpoints

Intersection of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​

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

Intersection of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L3​

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

Intersection of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L3​

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

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