The sides of a triangle are $3 x+2 y-6=0,2 x-3 y+6=0$ and $x+2 y+2=0$. If $P(0, b)$ lies either on the triangle or inside the triangle, then $b$ lies in the interval

• A

  $[-1,3]$
• B

  $[2,3]$
• Correct answer
  C

  $[-1,2]$
• D

  $[-2,2]$

Solution

Given $3 x+2 y-6=0,2 x-3 y+6=0, x+2 y+2=0$ We can re-write it as follows $\frac{x}{2}+\frac{y}{3}=1, \frac{x}{(-3)}+\frac{y}{2}=1, \frac{x}{-2}+\frac{y}{-2}=1$
$\because \quad P \equiv(0, b)$ The triangle $A B C$ in the above graph is the required triangle. In which, clearly, $(0, b)$ can vary from point $(0,2)$ to $(0,-1)$. So, range of $b$ is $[-1,2]$.

Hence, the correct answer is option: 'C'.