Q.Find the value of $k$ if the area of a triangle whose vertices are $(k,0)$, $(4,0)$ and $(0,2)$ is 4 square units. OR If $A = \begin{bmatrix} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{bmatrix}$, then find $|A|$.

Concept and Intuition

The area of a triangle whose vertices are given by coordinates is a fundamental concept in coordinate geometry. While you might be familiar with the base-height formula or Heron's formula, when coordinates are involved, a powerful tool is the determinant.

The determinant method for calculating the area of a triangle arises from vector geometry. If we consider two vectors forming two sides of a triangle, say $\vec{AB}$ and $\vec{AC}$, then the area of the triangle is half the magnitude of their cross product, i.e., $\frac{1}{2} |\vec{AB} \times \vec{AC}|$. When these vectors are expressed in coordinates, this cross product magnitude simplifies to a determinant.

Alternatively, you can think of it as a generalization of the "shoelace formula" for polygon areas. The determinant essentially calculates a signed area, where the sign depends on the order of vertices (clockwise or counter-clockwise). Since area is always positive, we take the absolute value of the determinant.

Step-by-Step Solution

1. Identify the vertices and the standard area formula: The vertices of $\triangle ABC$ are given as $(a, b)$, $(c, d)$, and $(e, f)$. Using the determinant formula for the area of a triangle, we can write:

2. Isolate the determinant from the area formula: From the formula above, we can multiply both sides by 2:

Let's denote the determinant inside the absolute value as $D$:

So, we have $2\Delta = |D|$.

3. Consider the given expression:

We need to evaluate $\begin{vmatrix} a & c & e \\ b & d & f \\ 1 & 1 & 1 \end{vmatrix}^2$.

Let's call the determinant in this expression $D'$.

4. Relate $D'$ to $D$ using determinant properties: A fundamental property of determinants states that the determinant of a matrix is equal to the determinant of its transpose. That is, $\det(A) = \det(A^T)$. If we compare $D$ and $D'$, we can see that $D'$ is the transpose of $D$. …

The area of a triangle in coordinate geometry is a fundamental concept. While you might be familiar with the base-height formula, when the vertices are given as coordinates, a more direct formula exists. This formula can be elegantly expressed using a determinant, which is what this question explores.

The core idea is that a determinant involving the coordinates of the vertices provides a value that is directly proportional to the area of the triangle. The sign of this determinant tells us about the orientation of the vertices (whether they are listed in a clockwise or counter-clockwise order), while its absolute value gives twice the area. Since area is always a positive quantity, we take the absolute value of the determinant expression.

1. Recall the Area Formula for a Triangle with Given Vertices The area $A$ of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by the formula:

The absolute value is crucial here because area must be non-negative. The expression inside the absolute value can be positive or negative depending on the order in which the vertices are taken.

> [!FORMULA]
> The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$ is:
> $$A = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$

2. Define the Determinant in Question

Let's consider the determinant given in the options:

3. Expand the Determinant We expand this $3 \times 3$ determinant along the first row:

Now, evaluate the $2 \times 2$ determinants:

Rearranging the terms to match the area formula's structure:

This can be rewritten as:

Notice that this is exactly the expression inside the absolute value in the area formula from Step 1.

4. Relate the Determinant to the Area

From Step 1, we have $A = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.

From Step 3, we found that $D = x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)$.

Therefore, we can write:

Multiplying both sides by 2, we get: …

The problem gives you three points: $(-2,0)$, $(2,0)$, and $(0,k)$. The area is 4 square units. You need to find $k$.

The key idea is the area of a triangle by coordinates. If you know the coordinates of the three vertices, you don't need to draw anything — you can compute the area directly using a simple determinant formula. This works because the area is half the absolute value of the cross product of two side vectors, which in coordinate form becomes a neat expression.

Why does this work? Imagine the triangle in the plane. The expression inside the absolute value is actually twice the signed area — it gives a positive or negative number depending on the order of the points. Taking absolute value and halving gives the actual area. This is much faster than using base-height, especially when the triangle isn't aligned with the axes.

Let's apply it step by step.

1. Label the points. Let:

   • $(x_1, y_1) = (-2, 0)$
   • $(x_2, y_2) = (2, 0)$
   • $(x_3, y_3) = (0, k)$

2. Plug into the formula. The area $A$ is:

3. Simplify inside the absolute value. Compute each term:

   • First term: $(-2)(0 - k) = (-2)(-k) = 2k$
   • Second term: $2(k - 0) = 2k$
   • Third term: $0(0 - 0) = 0$

   So the sum is $2k + 2k + 0 = 4k$.

4. Set the area equal to 4. You have:

Multiply both sides by 2:

5. Solve for $k$. The absolute value equation $|4k| = 8$ means $4k = 8$ or $4k = -8$. So:
   • $k = 2$
   • $k = -2$ …