Question

The combined equation of three sides of a triangle is $(x^2-y^2)(2x+3y-6)=0$ such that the points $(-2,a)$ and $(b,1)$ be the interior point and extrerior point of the triangle respectively. If $k=\left[\frac{a}{b}\right],$ then maximum value of $|k|$ is
(where [.] denotes the greatest integer function.)

Answer 1

The lines are,
$y=x,y=-x,2x+3y=6$


When $(-2,a)$ is interior point,
Intersection point of $x=-2$ and $y=-x$ is,
$(-2,2)$
Intersection point of $x=-2$ and $2x+3y=6$ is,
$-4+3y=6\Rightarrow y=\frac{10}{3}(-2,\frac{10}{3})$
So,
$2<a<\frac{10}{3}$

When $(b,1)$ is exterior point,
Intersection point of $y=1$ and $y=-x$ is,
$(-1,1)$
Intersection point of $y=1$ and $y=x$ is,
$(1,1)$
So,
$b>1\text{or}b<-1$

Therefore,
$|k|=\left|\left[\frac{a}{b}\right]\right|=4$
When $a\to \frac{10}{3}\text{and}b\to -1$