Question

The three lines of a triangle are given by \((x^2 - y^2)(2x + 3y - 6) = 0\). If the point \((-2,\lambda)\) lies inside and \((\mu,1)\) lies outside the triangle, then

• A. \(\lambda \in (1,\frac{10}{3}),\ \mu \in (-3,5)\)
• B. \(\lambda \in (2,\frac{10}{3}),\ \mu \in (-1,1)\)
• C. \(\lambda \in (-1,\frac{9}{2}),\ \mu \in (-2,\frac{10}{3})\)
• D. None of the above

Hint:

For triangle region problems, use sign method for all three lines.

Answer 1

The Correct Option is D

Concept: Triangle formed by lines: \[ x-y=0,\quad x+y=0,\quad 2x+3y-6=0 \]

Step 1: Region check.
For a point to lie inside triangle, it must satisfy consistent sign conditions for all three lines.

Step 2: Check point \((-2,\lambda)\).
Substitute into: \[ x-y = -2 - \lambda \] \[ x+y = -2 + \lambda \] \[ 2x+3y-6 = -4 + 3\lambda -6 = 3\lambda -10 \] Sign conditions restrict \(\lambda\), but none of given intervals satisfy all simultaneously.

Step 3: Check \((\mu,1)\).
\[ x-y = \mu -1,\quad x+y = \mu +1,\quad 2x+3y-6 = 2\mu -3 \] Again, none of the given options satisfy required outside conditions. Conclusion: \[ {\text{None of the above}} \]