In a triangle ABC, coordinates of A are (1,2) and the equations of the medians through B and C are r

Q: Wait, why do we assume the medians intersect at the centroid? Is that always true?

Yes, absolutely! By definition, the centroid of a triangle is the point where all three medians intersect. Since the problem gives you the equations of two medians, their intersection point is by definition the centroid. It is a fundamental property of triangles that you can always rely on.

Q: Why can't we just use the slope-intercept form to find the vertices?

You could, but it would be much more tedious. Parameterization is a powerful tool in coordinate geometry. By expressing the coordinates of $B$ and $C$ in terms of a single variable based on the line equations, you reduce the number of unknowns, making the algebra much cleaner and less prone to errors.

Q: Is there a faster way to find the area once we have the vertices?

Once you have the coordinates of all three vertices, the coordinate geometry area formula (often called the shoelace formula) is the most direct and reliable method. It is robust and works for any triangle, regardless of its orientation or shape.

Question

In a triangle ABC, coordinates of A are (1, 2) and the equations of the medians through B and C are respectively, $x + y = 5$ and $x = 4$. Then area of $\Delta ABC$ (in sq. units) is:

Options:

(A) 12

(B) 9

(C) 4

(D) 5

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a coordinate plane, looking at a triangle $ABC$. You know one corner, $A(1, 2)$, but the other two corners, $B$ and $C$, are shrouded in mystery.

However, you have two clues: the equations of the medians passing through $B$ and $C$. This is a detective story where we use the properties of geometry to uncover the hidden vertices.

### The Centroid Discovery

The first step in our investigation is to find the meeting point of our two clues. The medians of a triangle intersect at a very special point called the Centroid, denoted by $G$.

By solving the system of equations $x + y = 5$ and $x = 4$, we find the intersection. Substituting $x = 4$ into the first equation, we get $4 + y = 5$, which simplifies to $y = 1$.

Thus, our Centroid $G$ is located at $(4, 1)$. We have found our first major landmark.

### The Art of Parameterization

Now, we need to find the vertices $B$ and $C$. We know $B$ lies on the line $x + y = 5$.

Instead of treating $B$ as two unknown variables $(x_1, y_1)$, we can use the line equation to express $y_1$ as $5 - x_1$. So, $B$ becomes $(x_1, 5 - x_1)$.

Similarly, $C$ lies on the vertical line $x = 4$. This means its x-coordinate is fixed at $4$, so $C$ is $(4, y_2)$. We have successfully reduced our unknowns to just two variables, $x_1$ and $y_2$.

### The Centroid Bridge

The Centroid formula is our bridge between the vertices and the center. It states that:

$$G = \left( \frac{x_A + x_B + x_C}{3}, \frac{y_A + y_B + y_C}{3} ight)$$

Let's apply this to the x-coordinates:

$$4 = \frac{1 + x_1 + 4}{3}$$

Multiplying by $3$, we get $12 = 5 + x_1$, which gives us $x_1 = 7$. Now we know $B = (7, 5 - 7) = (7, -2)$.

Next, we apply it to the y-coordinates:

$$1 = \frac{2 + (5 - x_1) + y_2}{3}$$

Substituting $x_1 = 7$, we get $1 = \frac{2 + (-2) + y_2}{3}$, which simplifies to $1 = \frac{y_2}{3}$, so $y_2 = 3$. Our vertices are $A(1, 2)$, $B(7, -2)$, and $C(4, 3)$.

### The Final Calculation

With the vertices in hand, the area is just a calculation away. Using the area formula:

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

We substitute our values:

$$	ext{Area} = \frac{1}{2} |1(-2 - 3) + 7(3 - 2) + 4(2 - (-2))|$$

This becomes:

$$	ext{Area} = \frac{1}{2} |1(-5) + 7(1) + 4(4)| = \frac{1}{2} |-5 + 7 + 16| = \frac{1}{2} |18| = 9$$

The area of our triangle is **$9$ square units**. We have solved the mystery!

Animated Solution for Mathematics - Straight Lines: In a triangle ABC, coordinates of A are (1, 2) and the equations of the medians through B and C are respectively, $x + y = 5$ and $x = 4$ . Then area of $Δ A B C$ (in sq. units) is:

Select Answer:

Visualized Solution

Visualize the Given Data

Identify the Centroid $G$

Parameterize Vertex $B$

Parameterize Vertex $C$

Apply Centroid Formula for $x$

Solve for $x_{1}$

Apply Centroid Formula for $y$

Solve for $y_{2}$

Calculate Area of $Δ A B C$

Final Computation

Summary & Key Takeaway

The Sigma Insight: Centroid, Incenter, Orthocenter, and Circumcenter

Analyzing the Setup

The Centroid Discovery

The Art of Parameterization

The Centroid Bridge

The Final Calculation

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Select Answer:

Visualized Solution

Visualize the Given Data

Identify the Centroid .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } G

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

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

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

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x1​

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

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y2​

Calculate Area of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔABC