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…
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
$[-1,3]$
$[2,3]$
$[-1,2]$
$[-2,2]$
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 $[-1,2]$.
