Q.Find the value of $k$ for which the equation $(k+1)x^2+4xy+(k-1)y^2=0$ represents a pair of perpendicular straight lines.

The perpendicularity condition for a homogeneous pair depends only on the coefficients of $x^2$ and $y^2$ summing to zero — setting up and solving that single equation gives $k$ directly.

For a homogeneous pair $ax^2+2hxy+by^2=0$ to represent perpendicular lines, the condition is $a+b=0$. Here, comparing with the given equation, $a=(k+1)$ and $b=(k-1)$ (the $xy$ coefficient, $4=2h$, does not affect this particular condition).

Setting $a+b=0$:

$(k+1)+(k-1) = 0 \quad \Rightarrow \quad 2k = 0 \quad \Rightarrow \quad k=0$ …

Trying to make the perpendicularity condition depend on the $xy$-coefficient (here, $4$) is a common misunderstanding — perpendicularity of a homogeneous pair depends only on $a+b=0$, never on $h$; a very different mista …