Question

The combined equation of three sides of a triangle is $(x^2-y^2)(2x+3y-6)=0$. If $(-2,a)$ is an interior point and $(b,1)$ is an exterior point of the triangle, then

• A. $2<a<\frac{10}{3}$
• B. $-2<a<\frac{10}{3}$
• C. $-1<b<\frac{9}{2}$
• D. $-1<b<1$

Answer 1

Answer: (A), (D)

The correct options are
A $2<a<\frac{10}{3}$
D $-1<b<1$

$(x^2-y^2)(2x+3y-6)=0$
$(x-y)(x+y)(2x+3y-6)=0$
Therefore the lines are
$y=x$
$y=-x$
$2x+3y-6=0$
Since $(-2,a)$ is an interior point
$-4+3a-6<0$
$3a<10$
$a<\frac{10}{3}$ ...(i)
and $x+y>0$
$-2+a>0$
$a>2$ ...(ii)
From i and ii we get $2<a<\frac{10}{3}$
For b consider equations $y=x$ and$y=-x$
$b+1>0$
$b>-1$
and $b-1<0$
$b<1$
Therefore $-1<b<1$