Question

The combined equation of three sides of a triangle is $(x^2-y^2)(2x+3y-6)=0$. lf $(-2,a)$ and $(b,1)$ are interior points 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)

$2<a<\frac{10}{3}$
$-1<b<1$

The figure shows the triangle enclosed by three lines $x+y=0,x-y=0,2x+3y-6=0$
The line $X=-2$ will cut $x=-y$ at $(-2,2)$, and the line $2x+3y-6=0$ at $(-2,\frac{10}{3})$
From the figure, all the points on $x=-2$, between these two points, will be interior points of the triangle.
Hence, $2<a<\frac{10}{3}$
Similarly the line $y=1$ cuts $x=-y$ at $(-1,1)$, and the line $x=y$ at $(1,1)$.
From the figure, all the points on $y=1$, between these two points, will be interior points of the triangle.
Hence, $-1<b<1$