Question

Let the lines $y-k_{1}x-\beta =0$ and $y-k_{2}x-\beta =0,(k_1\ne k_2),k_1,k_2\in R$ intersect at $P$ and the lines $x-p_{1}y-\alpha =0$ and $x-p_{2}y-\alpha =0,(p_1\ne p_2),p_1,p_2\in R$ intersect at $Q$. If the points $P$ and $Q$ always lies on or inside the triangle formed by the lines $2x-3y-6=0,3x-y+3=0$ and $3x+4y-12=0$, then

• A. $\alpha \in [-1,3]$
• B. $\alpha \in [-2,4]$
• C. $\beta \in [-3,-2)$
• D. $\beta \in [-2,3]$

Answer 1

Answer: (A), (D)

The correct options are
A $\alpha \in [-1,3]$
D $\beta \in [-2,3]$

$y-k_{1}x-\beta =0$ and $y-k_{2}x-\beta =0$
$\Rightarrow (k_2-k_1)x=0\Rightarrow x=0(k_1\ne k_2)\Rightarrow P=(0,\beta )$
$x-p_{1}y-\alpha =0$ and $x-p_{2}y-\alpha =0$
$\Rightarrow Q=(\alpha ,0)$

Plotting the lines $2x-3y-6=0,3x-y+3=0$ and $3x+4y-12=0$,


As the points $P$ and $Q$ always lies on or inside the triangle formed by the lines, therefore
$\alpha \in [-1,3]\beta \in [-2,3]$