The lines $$p\left( {{p^2} + 1} \right)x - y + q = 0$$ and $$\left( {{p^2} + 1} \right){}^2x + \left( {{p^2} + 1} \right)y + 2q$$ $$=0$$ are perpendicular to a common line for :

The perpendicular bisector of the line segment joining P(1, 4) and Q(k, 3) has y-intercept -4. Then a possible value of k is :

Let A $$\left( {h,k} \right)$$, B$$\left( {1,1} \right)$$ and C $$(2, 1)$$ be the vertices of a right angled triangle with AC as its hypotenuse. If the area of the triangle is $$1$$ square unit, then the set of values which $$'k'$$ can take is given by :

If one of the lines of $$m{y^2} + \left( {1 - {m^2}} \right)xy - m{x^2} = 0$$ is a bisector of angle between the lines $$xy = 0,$$ then $$m$$ is :