Question

The combined equation of the three sides of a triangle is $(x^2-y^2)(2x-3y-6)=0$. If the point $(0,\alpha )$ lies in the interior of this triangle then

• A. $-2<\alpha <0$
• B. $-2<\alpha <2$
• C. $0<\alpha <2$
• D. $\alpha \ge 2$

Answer 1

Answer: (B)

$-2<\alpha <2$

$(x^2-y^2)(2x-3y-6)=0$

$\Rightarrow (x+y)=0$

$(x-y)=0$

$2x-3y-6=0$

Points of intersection of $x=y$ and $2x-3y-6=0$

$\Rightarrow -x-6=0$

$x=-6,y=-6$

Similarly point of intersection of $x+y=0$ and $2x-3y-6=0$

$\Rightarrow 2x-3y-6=0$

$x=\frac{6}{5},y=\frac{-6}{5}$

$\alpha$ should lie between $0$ and $-2$ because $(0,\alpha )$ lies on $y$ axis. Part of the $y$ axis lying inside $△$ is between $0$ and $-2$.