Problem 7 - Algebra 2 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

Using determinants, find the value of \(k\) so that the points \((k, 2-2 k),(-k+1,2 k)\) and \((-4-k,\), \(6-2 k\) ) may be collinear.

Step-by-Step Solution

Answer

The values for \(k\) are \(k = \frac{5}{4}\) or \(k = -\frac{1}{2}\).

Step 1: Understanding Collinearity with Determinants

Three points \((x_1, y_1), (x_2, y_2), (x_3, y_3)\) are collinear if the determinant of the matrix formed by these points and a column of ones is zero. The matrix is: \[ \begin{vmatrix} x_1 & y_1 & 1 \ x_2 & y_2 & 1 \ x_3 & y_3 & 1 \end{vmatrix} = 0 \] For our points \( (k, 2-2k), (-k+1, 2k), (-4-k, 6-2k) \), construct this determinant.

Step 2: Setting Up the Determinant Equation

Substitute the given points into the determinant equation: \[\begin{vmatrix} k & 2-2k & 1 \ -k+1 & 2k & 1 \ -4-k & 6-2k & 1 \end{vmatrix} = 0 \] Now, expand this determinant.

Step 3: Expanding the Determinant

Expand the determinant using the first row: \[ k \begin{vmatrix} 2k & 1 \ 6-2k & 1 \end{vmatrix} - (2-2k) \begin{vmatrix} -k+1 & 1 \ -4-k & 1 \end{vmatrix} + \begin{vmatrix} -k+1 & 2k \ -4-k & 6-2k \end{vmatrix} = 0 \] Calculate each of these smaller 2x2 determinants.

Step 4: Calculating 2x2 Determinants

Calculate each of the 2x2 determinants: - For the first: \[ \begin{vmatrix} 2k & 1 \ 6-2k & 1 \end{vmatrix} = 2k(1) - (6-2k)(1) = 2k - 6 + 2k = 4k - 6 \] - For the second: \[ \begin{vmatrix} -k+1 & 1 \ -4-k & 1 \end{vmatrix} = (-k+1)(1) - (-4-k)(1) = -k + 1 + 4 + k = 5 \] - For the third: \[ \begin{vmatrix} -k+1 & 2k \ -4-k & 6-2k \end{vmatrix} = (-k+1)(6-2k) - (-4-k)(2k) \] Simplify: \[ (-k+1)(6-2k) = -6k + 2k^2 + 6 - 12k = 2k^2 - 18k + 6 \] \[ (-4-k)(2k) = -8k - 2k^2 \] Therefore, \[ 2k^2 - 18k + 6 + 8k + 2k^2 = 4k^2 - 10k + 6 \]

Step 5: Forming the Final Equation

Substitute the calculated values back into the equation from the determinant expansion: \[ k(4k - 6) - (2-2k)(5) + (4k^2 - 10k + 6) = 0 \] Simplify this equation to solve for \(k\): \[ 4k^2 - 6k - 10 + 10k + 4k^2 - 10k + 6 = 0 \] Combine like terms.

Step 6: Solving the Simplified Equation

Combine the terms to form: \[ 8k^2 - 6k + 6 - 10 = 0 \] This simplifies to \[ 8k^2 - 6k - 4 = 0 \] This is a quadratic equation in \(k\). Simplify it further or solve using the quadratic formula.

Step 7: Solving the Quadratic Equation

Simplify further: divide through by 2: \[ 4k^2 - 3k - 2 = 0 \] Use the quadratic formula: \[ k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \(a = 4, b = -3, c = -2\). Substitute these values into the formula.

Step 8: Finding the Roots Using the Quadratic Formula

Calculate: \[ k = \frac{3 \pm \sqrt{(-3)^2 - 4 \times 4 \times (-2)}}{2 \times 4} \] Which evaluates to: \[ k = \frac{3 \pm \sqrt{9 + 32}}{8} \] Thus: \[ k = \frac{3 \pm 7}{8} \] Therefore, the possible values for \(k\) are: \[ k = \frac{10}{8} = \frac{5}{4} \quad \text{or} \quad k = \frac{-4}{8} = -\frac{1}{2} \]

Key Concepts

Collinearity

Collinearity occurs when three or more points lie on the same straight line. For three points

• \((x_1, y_1)\)
• \((x_2, y_2)\)
• \((x_3, y_3)\)

to be collinear, their slope between any two pairs of points must be equal. However, there's a more straightforward method involving matrices and determinants.
A determinant can be used to check for collinearity by forming a special matrix that includes the coordinates of the points and a column of ones. Consider the following matrix:\[\begin{vmatrix} x_1 & y_1 & 1 \ x_2 & y_2 & 1 \ x_3 & y_3 & 1 \end{vmatrix} = 0\]
If this determinant equals zero, the points are collinear. This is because a zero determinant signifies the rows of the matrix are linearly dependent, which in geometric terms means the points align on a straight line.

Quadratic Equation

A quadratic equation is a polynomial equation of the second degree, which typically takes the form:\[ax^2 + bx + c = 0\]
where \(a\), \(b\), and \(c\) are constant coefficients with \(a eq 0\). The solutions to the quadratic equation are known as the roots, and they can be found using various methods such as factoring, completing the square, or the quadratic formula.
The quadratic formula is given by:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
In our exercise, after setting up the determinant to find collinearity, we derived the quadratic equation \(4k^2 - 3k - 2 = 0\). By using the quadratic formula with \(a = 4\), \(b = -3\), and \(c = -2\), we calculated the possible values of \(k\) where the points become collinear. Quadratic equations are crucial in many mathematical and engineering applications because they frequently describe parabolic paths or can be used to model various phenomena.

2x2 Determinant Expansion

The expansion of a determinant is a method used to compute its value, especially when it's a matrix larger than 2x2. However, understanding the simpler 2x2 determinant is fundamental.
For any 2x2 matrix:\[\begin{bmatrix} a & b \ c & d \end{bmatrix}\]The determinant can be calculated as:\[ad - bc\]
In the context of our original exercise, we needed to expand the 3x3 determinant, which included calculating smaller 2x2 determinants as intermediate steps. This is often how 3x3 or larger determinants are computed—by breaking them down into more manageable 2x2 determinants. In step 4, we calculated:

• \(4k - 6\)
• 5
• \(4k^2 - 10k + 6\)

These values are achieved by applying the formula \(ad - bc\) on each 2x2 submatrix. The ability to compute these quickly and correctly is critical in solving for variables such as \(k\) in determinant problems.