Question

Ki are possible values of K for which lines \(Kx + 2y + 2 = 0\), \(2x + Ky + 3 = 0\), \(3x + 3y + K = 0\) are concurrent, then \(∑k_i\) has value.

• A. 0
• B. -2
• C. 2
• D. 5

Answer 1

The Correct Option is A

To determine the possible values of K for which the lines \(Kx + 2y + 2 = 0\), \(2x + Ky + 3 = 0\), and \(3x + 3y + K = 0\) are concurrent, we need to check if the determinant of the coefficient matrix is zero.

The coefficient matrix is: \(\begin{vmatrix} K & 2 & 0\\ 2 & K & 3\\ 3 & 3 & K \end{vmatrix}\)

Taking the determinant of this matrix and setting it equal to zero, we get:

\(K(K^2 - 6) - 6(K - 6) + 18(3 - 3K) = 0\)

Simplifying the equation, we have: \(K^3 - 6K^2 - 6K + 216 = 0\)

Factoring the left-hand side, we find: \((K - 6)(K^2 + 12K - 36) = 0\)

Setting each factor equal to zero, we obtain two possible values for K: \(K = 6\) and \(K = -6 ± 2\sqrt3\)

The sum of these possible values of K is: \(6 + (-6 + 2\sqrt3) + (-6 - 2\sqrt3) = 6 - 6 = 0\)

Therefore, the value of \(∑k_i\) is 0.

Answer 2

Given:
\(Kx + 2y + 2 = 0\)
\(2x + Ky + 3 = 0\)
\(3x + 3y + K = 0\)

Formulate the coefficient matrix:
\(\begin{vmatrix}   K & 2 & 2 \\   2 & K & 3 \\   3 & 3 & K   \end{vmatrix}\)

Calculate the determinant and set it to zero:
\(\begin{aligned}   \text{Det} &= K \left( K^2 - 6 \right) - 2 \left( 2K - 6 \right) + 2 \left( 6 - 3K \right) \\   &= K^3 - 6K^2 - 6K + 36   \end{aligned}\)
\(K^3 - 6K^2 - 6K + 36 = 0\)
\((K - 6)(K^2 + 12K - 6) = 0\)
\(K - 6 = 0 \Rightarrow K = 6\)
Solve \(K^2 + 12K - 6 = 0\) using the quadratic formula:
\(K = \frac{-12 \pm \sqrt{144 + 24}}{2} = \frac{-12 \pm \sqrt{168}}{2} = \frac{-12 \pm 4\sqrt{42}}{2} = -6 \pm 2\sqrt{42}\)

Therefore, the roots are \(K = 6\), \(K = -6 + 2\sqrt{42}\), and \(K = -6 - 2\sqrt{42}\).

Calculate \(\sum K_i\):
\(\sum K_i = 6 + (-6 + 2\sqrt{42}) + (-6 - 2\sqrt{42}) = 6 - 6 = 0\)

So, the correct option is (A): 0.