Let $b, d>0$. The locus of all points $P(r, \theta)$ for which the line $P$ (where, $O$ is the origin) cuts the line $r \sin \theta=b$ in $Q$ such that $P Q=d$ is
$(r-d) \sin \theta=b$
$(r \pm d) \sin \theta=b$
$(r-d) \cos \theta=b$
$(r \pm d) \cos \theta=b$
The equation of the base of an equilateral triangle is $x + y = 2$ and the vertex is $(2, -1)$. The length of the side of the triangle is
If the middle points of the sides $BC,\, CA$ and $AB$ of the triangle $ABC$ be $(1, 3), \,(5, 7)$ and $(-5, 7)$, then the equation of the side $AB$ 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
Co-ordinates of the orthocentre of the triangle whose vertices are $A(0, 0) , B(3, 4)$ and $C(4, 0)$ is
The orthocentre of the triangle formed by the lines $xy = 0$ and $x + y = 1$ is