The pair of straight lines $x^2 - 4xy + y^2 = 0$ together with the line $x + y + 4 = 0$ form a triangle which is :

The area of triangle formed by the lines $x + y - 3 = 0 , x - 3y + 9 = 0$ and $3x - 2y + 1= 0$

If the vertices $P$ and $Q$ of a triangle $PQR$ are given by $(2, 5)$ and $(4, -11)$ respectively, and the point $R$ moves along the line $N: 9x + 7y + 4 = 0$, then the locus of the centroid of the triangle $PQR$ is a straight line parallel to

If in triangle $ABC$ ,$ A \equiv (1, 10) $, circumcentre $\equiv$ $\left( { - \,\,{\textstyle{1 \over 3}}\,\,,\,\,{\textstyle{2 \over 3}}} \right)$ and orthocentre $\equiv$ $\left( {{\textstyle{{11} \over 3}}\,\,,\,\,{\textstyle{4 \over 3}}} \right)$ then the co-ordinates of mid-point of side opposite to $A$ is :

A square of side a lies above the $x$ -axis and has one vertex at the origin. The side passing through the origin makes an angle $\alpha ,(0 < \alpha < \frac{\pi }{4})$ with the positive direction of $x$-axis. The equation of its diagonal not passing through the origin is

The circumcentre of a triangle lies at the origin and its centroid is the mid point of the line segment joining the points $(a^2 + 1 , a^2 + 1 )$ and $(2a, - 2a)$, $a \ne 0$. Then for any $a$ , the orthocentre of this triangle lies on the line