Problem 13 - Algebra for JEE ADVANCED

Question

If \(\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}=a^{2}\) \(\left(x_{2}-x_{3}\right)^{2}+\left(y_{2}-y_{3}\right)^{2}=b^{2}\) \(\left(x_{3}-x_{1}\right)^{2}+\left(y_{3}-y_{1}\right)^{2}=c^{2}\) and \(k\left|\begin{array}{lll}x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\\ x_{3} & y_{3} & 1\end{array}\right|=(a+b+c)(b+c-a)(c+a-b)\) \(\times(a+b-c)\), then the value of \(k\) is a. b. 2 C. 4 d. none of these

Step-by-Step Solution

Answer

k = 4

Step 1: Interpret the Given Problem

The problem involves a set of equations related to the distances between three points \((x_1, y_1), (x_2, y_2), (x_3, y_3)\), forming a triangle. We are asked to find the value of \(k\), given an equation involving the determinant of a 3x3 matrix of the coordinates.

Step 2: Recognizing the Triangle Area Formula

The formula \(k \left|\begin{array}{lll}x_{1} & y_{1} & 1 \ x_{2} & y_{2} & 1 \ x_{3} & y_{3} & 1\end{array}\right| = (a+b+c)(b+c-a)(c+a-b)(a+b-c)\) equates the area of a triangle to the semi-perimeter product because the determinant gives twice the area of the triangle.

Step 3: Relate Determinant and Area

The absolute value of the determinant \(\left|\begin{array}{lll}x_{1} & y_{1} & 1 \ x_{2} & y_{2} & 1 \ x_{3} & y_{3} & 1\end{array}\right|\) is twice the area of a triangle formed by the points \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \). The area \(A\) can also be expressed using Heron's Formula: \(A = \sqrt{s(s-a)(s-b)(s-c)}\), where \(s\) is the semi-perimeter \(\frac{a+b+c}{2}\).

Step 4: Equate the Two Expressions for Area

The determinant expression equals \(2A\), or \( k \times 2A = (a+b+c)(b+c-a)(c+a-b)(a+b-c)\). Recognizing that the right side simplifies to \(16A^2\), we equate \(k \times 2A = 16A^2\).

Step 5: Solve for k

Cancel out area \(A\) from both sides, assuming \(A eq 0\), giving \(2kA = 16A^2\), or after simplifying \(k = 8A\). Divide both sides by \(2A\): \(k = 4\). Thus \(k = 4\) satisfies the equation under non-zero area condition.

Key Concepts

Determinant and Area

The determinant of a matrix can be used to calculate certain characteristics of geometric figures, such as the area of a triangle formed by three points. When you have vertices of a triangle with coordinates \((x_1, y_1), (x_2, y_2), (x_3, y_3)\), you can put these into a matrix form:
\[\begin{vmatrix} x_1 & y_1 & 1 \ x_2 & y_2 & 1 \ x_3 & y_3 & 1 \end{vmatrix}\]
The absolute value of this determinant is related to the area of the triangle. Specifically, it equals twice the area of the triangle. Thus, the area \(A\) can be directly found using:

• \( 2A = \left| \begin{array}{ccc} x_1 & y_1 & 1 \ x_2 & y_2 & 1 \ x_3 & y_3 & 1 \end{array} \right| \)
• To find \(A\), divide the determinant's absolute value by 2.

Utilizing this property, the determinant provides a straightforward way of determining the area without additional geometric constructions, making it efficient for calculations.

Heron's Formula

Heron's Formula offers another method for calculating the area of a triangle, particularly when the side lengths are known. If a triangle has side lengths \(a\), \(b\), and \(c\), you can first calculate the semi-perimeter \(s\) with:
\[ s = \frac{a + b + c}{2} \]
Once you have \(s\), Heron's Formula can be applied:
\[ A = \sqrt{s(s-a)(s-b)(s-c)} \]
This beautifully relates the triangle's side lengths to its area, without requiring coordinate geometry. For example, if you know the distances between your triangle's vertices, these can serve as \(a\), \(b\), and \(c\) in Heron's formula.

• This method is particularly useful when you do not have vertex coordinates but do know side lengths.
• Heron's Formula is rooted in geometry and ensures an accurate calculation of area based solely on the triangle’s inherent dimensions.

By comparing the triangle area derived either from determinants or Heron's work, you verify important geometric properties effectively.

Matrix Determinant

In mathematics, the determinant of a matrix is a special number that can give insights into the matrix’s properties. When working with a 3x3 matrix that represents points in a triangle, like so:
\[\begin{vmatrix} x_1 & y_1 & 1 \ x_2 & y_2 & 1 \ x_3 & y_3 & 1 \end{vmatrix}\]
you can identify aspects like area and collinearity. Suppose the determinant equals zero; this implies the points are collinear (they all lie on a single straight line), and thus, the triangle's area is zero.

For a triangle with non-zero area, the determinant provides a relation to the twice-area property. Additionally,

• The determinant reflects how much the shape defined by the matrix is spread out in space.
• Determinants are useful in larger contexts like solving systems of linear equations or changing bases in linear transformations.

Thus, understanding how to compute and interpret determinants plays a vital role in linking algebraic structures with geometric interpretations.