Question

If the sum of squares of all real values of \( \alpha \), for which the lines \( 2x - y + 3 = 0 \), \( 6x + 3y + 1 = 0 \) and \( \alpha x + 2y - 2 = 0 \) do not form a triangle \( p \), then the greatest integer less than or equal to \( p \) is ....

Answer 1

Correct Answer: 32

To find the sum of squares of all real values of \( \alpha \) for which the lines do not form a triangle, we analyze the conditions for parallelism. Given lines are: 
1. \( 2x - y + 3 = 0 \)
2. \( 6x + 3y + 1 = 0 \)
3. \( \alpha x + 2y - 2 = 0 \)
Lines do not form a triangle if any two lines are parallel. For parallel lines, \(\frac{a_1}{a_2} = \frac{b_1}{b_2}\). The line pair combinations are:
1. Lines 1 and 2:
\(\frac{2}{6} = \frac{-1}{3}\) implies not parallel.
2. Lines 1 and 3:
\(\frac{2}{\alpha} = \frac{-1}{2}\) gives \(\alpha = -4\).
3. Lines 2 and 3:
\(\frac{6}{\alpha} = \frac{3}{2}\) gives \(\alpha = 4\).
Real values of \(\alpha\) are \(-4\) and \(4\). Calculate the sum of squares:
Sum \(= (-4)^2 + (4)^2 = 16 + 16 = 32\).
The greatest integer less than or equal to 32 is \(32\). The computed value, 32, is within the given range [32, 32].

Answer 2

Given:
\(2x - y + 3 = 0, \quad 6x + 3y + 1 = 0, \quad ax + 2y - 2 = 0.\)

To not form a triangle, \(ax + 2y - 2 = 0\) must be concurrent or parallel with the other lines.

Solving for concurrent lines: 
\(\frac{2}{6} = \frac{-1}{3} \implies a = \frac{4}{5}.\)

Similarly, for parallel lines:  
\(a = \pm 4.\)

Calculating \(p\):
\(p = \left(\frac{4}{5}\right)^2 + 4^2 + 4^2 = 32.\)

The Correct answer is: 32