Question

Can any one solve it

106th one

Answer 1

The correct answer is,  (d) a set of four lines forming a square.

On factorizing the given equation, we can get an idea about the graph.

$\begin{aligned}&x^2y^2-4x^2y-2xy^2+8xy-3y^2+12y&=&\ \ 0\\ \\ \Longrightarrow\ \ &y(x^2y-4x^2-2xy+8x-3y+12)&=&\ \ 0\\ \\ \Longrightarrow\ \ &y(x^2y-2xy-3y-4x^2+8x+12)&=&\ \ 0\\ \\ \Longrightarrow\ \ &y(y(x^2-2x-3)-4(x^2-2x-3))&=&\ \ 0\\ \\ \Longrightarrow\ \ &y(y-4)(x^2-2x-3)&=&\ \ 0\\ \\ \Longrightarrow\ \ &y(y-4)(x^2+x-3x-3)&=&\ \ 0\\ \\ \Longrightarrow\ \ &y(y-4)(x(x+1)-3(x+1))&=&\ \ 0\\ \\ \Longrightarrow\ \ &y(y-4)(x+1)(x-3)&=&\ \ 0\end{aligned}$

So we get that the graph is made by the lines,

$y=0,\ \ y-4=0,\ \ x+1=0,\ \ x-3=0$

And also, we can see that the graph is made by two pairs of parallel lines.

$y=0\parallel y-4=0\ \ \ \&\ \ \ x+1=0\parallel x-3=0$

Find the distance between each parallel lines.

$\text{Distance between \ y=0\ \ \ \ \ \ \ \&\ \ y-4=0 \ \ = \ 4}\\ \\ \text{Distance between \ x+1=0\ \ \&\ \ x-3=0 \ \ = \ 4}$

Since both are same, we can say that a square is made between these lines.

So (d) is the right option.